Category: Inter 2nd Year

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 6 Binomial Theorem to solve questions creatively.

Intermediate 2nd Year Maths 2A Binomial Theorem Formulas

→ Let n be a positive integer and x, a be real numbers then
(x + a)ⁿ = ⁿC₀. xⁿ. a⁰ + ⁿC₁. x^{n – 1}. a¹ + ⁿC₂. x^{n – 2}. a² + ……… + ⁿC_r. x^{n – r}. a^r + ……… + ⁿC_n. x⁰. aⁿ = \(\sum_{r=0}^{n}\) ⁿC_r.x^{n – r}. a^r. and (x – a)ⁿ – ⁿC₀. xⁿ – (x)ⁿ – ⁿC₁. x^{n – 1}a + ⁿC₂ x^{n – 2}. a² …….. + ( – 1)^{r n}C_r. x^{n – r}. a^r + ……… + (- 1)^{n n}C_n.aⁿ

→ The expansion of (x + a)ⁿ contains (n + 1) terms.

→ In the expansion, the coefficients ⁿC₀, ⁿC₁, ⁿC₂, ….. ⁿC_n are called binomial coefficients and these are simply denoted by C₀, C₁, C₂, …….. C_n,.

→ In the expansion, (r + 1)^th term is called the general term. It is denoted by T_{r + 1}.
∴ T_{r + 1} = ^cC_r. x^{n – r}. a^r, (0 ≤ r ≤ n)

→ The number of terms in the expansion of (a + b + c)ⁿ = \(\frac{(n+1)(n+2)}{2}\)

→ If n is even in the expansion of (x + a)ⁿ, the middle term = T_{\(\left(\frac{n}{2}+1\right)\)}

→ If n is odd in the expansion of (x + a)ⁿ, it has two middle terms which are T_{\(\left(\frac{n+1}{2}\right)\)}, T_{\(\left(\frac{n+3}{2}\right)\)}.

→ If \(\frac{(n+1)|x|}{|x|+1}\) = p, a positive integer then p^th and (p + 1)^th terms are the numerically greatest terms in the expansion of (1 + x)ⁿ.

→ If \(\frac{(n+1)|x|}{|x|+1}\) = P + F where p is a positive integer and 0 < F < 1 then (p + 1)^th the numerically greatest term in the expansion of (1 + x)ⁿ

→ C₀ + C₁ + C₂ + ………. + C_n = 2ⁿ

→ C₀ – C₁ + C₂ – C₃ + ……… + (- 1)ⁿC_n = 0

→ C₀ + C₂ + C ₄ + …………… = C₁ + C₃ + C₅ + …………….. = 2^{n – 1}

→ \(\sum_{r=0}^{n}\) ⁿC_r = 2ⁿ

→ \(\sum_{r=0}^{n}\) r. ⁿC_r = n. 2^{n – 1}

→ \(\sum_{r=2}^{n}\) r(r – 1). ⁿC_r = n(n – 1). 2^{n – 2}

→ \(\sum_{r=1}^{n}\) r² . ⁿC_r = n(n + 1). 2^{n – 2}

→ a. C₀ + (a + d). C₁ + (a + 2d). C₂ + ……… + (a + nd). C_n = (2a + nd) 2^{n – 1}

→ C₀C_r + C₁C_{r + 1} + C₂C_{r + 2} + ………… + C_{n – r}. C_n = ²ⁿC_{n + r}

→ If f(x) = (a₀ + a₁x + a₂x² + ……… a_mx^m)ⁿ then

• Sum of the coefficients = f(1)
• Sum of the coefficients of even powers of x is \(\frac{f(1)+f(-1)}{2}\)
• Sum of the coefficients of odd powers of x is \(\frac{f(1)-f(-1)}{2}\)

→ Let n be a positive integer and x is a ,real number such that |x| < 1 then

→ (1 – x)^-n = 1 + nx + \(\frac{n(n+1)}{2 !}\) x² + \(\frac{n(n+1)(n+2)}{3 !}\) x² + ……… + ……… + \(\frac{n(n+1)(n+2) \ldots \ldots(n+r-1)}{r !}\) x^r + ……. to ∞

→ (1 + x)^-n = 1,- nx + \(\frac{n(n+1)}{2 !}\) x² + …….. + \(\frac{(-1)^{r} n(n+1)(n+2) \ldots(n+r-1)}{r !}\) x^r + …….. ∞

→ If |x| < 1, then for p, q ∈ N

→ (1 – x)^-p/q = 1 + \(\frac{p}{1 !}\left(\frac{x}{q}\right)\) + \(\frac{p(p+q)}{2 !}\left(\frac{x}{q}\right)^{2}\) + ………. + \(\frac{p(p+q) \ldots \ldots(p+(r-1) q)}{r !}\) \(\left(\frac{x}{q}\right)^{r}\) + …….. ∞

→ (1 + x)^-p/q = 1 – \(\frac{p}{1 !}\left(\frac{x}{q}\right)\) + \(\frac{p(p+q)}{2 !}\left(\frac{x}{q}\right)^{2}\) + ………. + \(\frac{(-1)^{r} p(p+q) \ldots(p+(r-1) q)}{r !}\) \(\left(\frac{x}{q}\right)^{r}\) + …….. ∞

→ (1 + x)^-p/q = 1 + \(\frac{p}{1 !}\left(\frac{x}{q}\right)\) + \(\frac{(p)(p-q)}{1 .2}\left(\frac{x}{q}\right)^{2}\) + …………. + \(\frac{(p)(p-q)(p-2 q) \ldots \ldots .[p-(r-1) q]}{(r) !}\) \(\left(\frac{x}{\cdot q}\right)^{r}\) + ……… ∞

→ (1 – x)^-p/q = 1 – \(\frac{p}{1 !}\left(\frac{x}{q}\right)\) + \(\frac{(p)(p-q)}{1.2}\left(\frac{x}{q}\right)^{2}\) – …………. + (- 1)^r \(\) \(\left(\frac{x}{q}\right)^{r}\) + ……… ∞

Binomial Theorem for integral index:
If n is a positive integer then (x + a)ⁿ = ⁿC_o xⁿ + ⁿC₁ x^n-1 a + ⁿC₂ x^n-2 a² + . … + ⁿC_r x^n-ra^r + …… + ⁿC_naⁿ

→ The expansion of (x + a)ⁿ contains (n + 1) terms.

→ In the expansion, the sum of the powers of x and a in each term is equal to n.

→ In the expansion, the coefficients ⁿC₀, ⁿC₁. ⁿC₂………….. ⁿC_n are called binomial coefficients and these are simply denoted by C₀, C₁, C₂ …. C_N.
ⁿC₀ = 1, ⁿC_N = 1, ⁿC₁ = n, ⁿC_r = ⁿC_n-r

→ In the expansion, (r + 1)^th term is called the general term. It is denoted by
T_r+1. Thus T_r+1 = ⁿC_rx_n-ra_r

→ (x + a)ⁿ = \(\sum_{r=0}^{n}\)ⁿC_rx_n-ra_r

→ (x + a)ⁿ = \(\sum_{r=0}^{n}\)ⁿC_rx_n-r(-a)_r = \(\sum_{r=0}^{n}\)(-1)^{n n}C_rx_n-r(-a)_r = ⁿC₀xⁿ – ⁿC₁x^n-1a + ⁿC₁x^n-2a² – ……….. + (-1)^{n n}C_n aⁿ

→ (1 + x)ⁿ = \(\sum_{r=0}^{n}\)ⁿC_rx_r = ⁿC₀ + ⁿC₀x + ……….. + ⁿC_n xⁿ = C₀ + C₁x + C₂x² + ………. + C_nxⁿ

→ Middle term(s) in the expansion of (x + a)ⁿ.

• If n is even, then (\(\frac{n}{2}\) + 1)th term is the middle term
• If n is odd, then \(\frac{n+1}{2}\) th and \(\frac{n+3}{2}\) th terms are the middle terms.

→ Numerically greatest term in the expansion of (1 + x)ⁿ :

• If \(\frac{(n+1)|x|}{|x|+1}\) = p, a integer then plu1 and (p + 1) th terms are the numerically greatest terms in the expansion of (1 + x).
• If \(\frac{(n+1)|x|}{|x|+1}\) = p + F where pis a positive integer and 0< F < 1 then (p+1) th term is the numerically greatest term in the expansion of (1 + x).

→ Binomial Theorem for rational index: If n is a rational number and
|x| < 1, then 1 + nx + \(\frac{n(n-1)}{2 !}\) x² + \(\frac{n(n-1)(n-2)}{3 !}\)x³ + ………… = (1 + x)ⁿ

→ If |x| < 1 then

• (1 + x)^-1= 1 – x + x² – x³ + … + (-1)^rx^r + …….
• (1 – x)^-1 = 1 + x + x² + x³ + … + x^r + …….
• (1 + x)^-2 = 1 – 2x + 3x² – 4x³ + … + (-1)^r (r + 1)x^r + ………..
• (1 – x)^-2 = 1 + 2x + 3x² + 4x³ + … +(r + 1)x^r + ….
• (1 – x)^-n = 1 – nx + \(\frac{n(n-1)}{2 !}\) x² – \(\frac{n(n-1)(n-2)}{3 !}\)x³ + …………..
• (1 – x)^-n = 1 + nx + \(\frac{n(n-1)}{2 !}\) x² + \(\frac{n(n-1)(n-2)}{3 !}\)x³ + ………

→ If |x| < 1 an dn is a positive integer, then

• (1 – x)^-n = 1 + ⁿC₁x + ⁽ⁿ⁺¹⁾C₂x² + ⁽ⁿ⁺²⁾C₃x³ + ………….
• (1 + x)^-n = 1 – ⁿC₁x + ⁽ⁿ⁺¹⁾C₂x² – ⁽ⁿ⁺²⁾C₃x³ + ………….

→ When |x| < 1
(1 – x)^-p/q = 1 + \(\frac{p}{1 !}\left(\frac{x}{q}\right)+\frac{p(p+q)}{2 !}\left(\frac{x}{q}\right)^{2}+\frac{p(p+q)(p+2 q)}{3 !}\left(\frac{x}{q}\right)^{3}\) + …………………∞

→ When |x| < 1
(1 + x)^-p/q = 1 – \(\frac{p}{1 !}\left(\frac{x}{q}\right)_{+} \frac{p(p+q)}{2 !}\left(\frac{x}{q}\right)^{2} \quad \frac{p(p+q)(p+2 q)}{3 !}\left(\frac{x}{q}\right)^{3}\) + …………………∞

Binomial Theorem:
Let n be a positive integer and x, a be real numbers, then (x + a)ⁿ = ⁿC₀.xⁿa° + ⁿC₁.xⁿa¹ + ⁿC₂.x^n-1a² + …………… + ⁿC_r.x^n-ra^r + ……….. + ⁿC_n.x⁰aⁿ
Proof :
We prove this theorem by u sing the principle of mathematical induction (on n).
When n = 1, (x + a)¹ =(x + a)¹ = x + a = ¹C₀x¹a°+ ¹C₁x°a¹
Thus the theorem is true for n = 1
Assume that the theorem is true for n = k ≥ 1 (where k is a positive integer). That is
(x+a)^k = ^kC₀x^k.a⁰ + ^kC₁x^k-2a² + ^kC₂.x^k-2a² + …+ ^kC_r.x^k-r.a^r + ………….. + ^kCx⁰a^k

Now we prove that the theorem is true when n = k + 1 also
(x + a)^k+1 = (x + a)(x + a)^k

Therefore the theorem is true for n = k + 1
Hence, by mathematical induction, it follows that the theorem is true of all positive integer n

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 5 Permutations and Combinations to solve questions creatively.

Intermediate 2nd Year Maths 2A Permutations and Combinations Formulas

→ Fundamental principle: If a work can be performed in m different ways and another work can be performed in n different ways, then the two works simultaneously can be performed in mn different ways.

→ If n is a non-negative integer then

• 0 ! = 1
• n ! = n (n – 1) ! if n > 0

→ The number of permutations of n dissimilar things taken ‘r at a time is denoted by ⁿP_r and ⁿP_r = \(\frac{n !}{(n-r) !}\) for 0 ≤ r ≤ n.

→ If n, r positive integers and r ≤ n, then

• ⁿP_r = n. ^{n – 1}P_{r – 1} if r ≥ 1
• ⁿP_r = n.(n – 1) ^{n – 2}P_{r – 2} if r ≥ 2
• ⁿP_r = ^{n – 1}P_r + r. ^{(n – 1)}P_{(r – 1)}

→ The number of injections that can be defined from set A into set B is ^n(B)p_n(A) n(A) ≤ n(B)

→ The number of bijections that can be defined from set A into set B having same number of elements with A is n(A)!

→ The number of permutations of1n’ dissimilar things taken ‘r’ at a time.

• Containing a particular thing is (r) ^{n – 1}P_{r – 1}
• Not containing a particular thing is ^{n – 1}P_r
• Containing a particular thing in a particular place is ^{n – 1}P_{r – i}.

→ The number of functions that can be defined from set A into set B in [n(B)]^n(A)

→ The sum of the r – digited numbers that can be formed using the given ‘n’ distinct non-zero digits (r ≤ n ≤ 9) is ^{(n – 1)}P_{(r – 1)} × sum of all n digits × 111 …… 1 (r times)

→ In the above, if ‘0’ is one among the given n digits, then the sum is ^{(n – 1)}P_{(r – 1)} × sum of the digits × 111 … 1 (r times) ^{(n – 2)}P_{(r – 2)} × sum of the digits × 111 … 1 [(r – 1) times]

→ The number of permutations of n dissimilar things taken r at a time when repetitions are allowed any number of times is n^r.

→ The number of circular permutations of n dissimilar things is (n – 1) !

→ In case of hanging type circular permutations like garlands of flowers, chains of beads etc., the number of circular permutations of n things is \(\frac{1}{2}\) [(n – 1) !]

→ If in the given n things, p alike things are of one kind, q alike things are of the second kind, r alike things are of the third kind and the rest are dissimilar, then the number of permutations (of these n things) is \(\frac{n !}{p ! q ! r !}\)

→ The number of combinations of n things taken r at a time is denoted by ⁿC_r and ⁿC_r = \(\frac{n !}{(n-r) ! r !}\) for 0 ≤ r ≤ n and \(\frac{{ }^{n} C_{r}}{{ }^{n} C_{r-1}}=\frac{n-r+1}{r}\) and ⁿP_r = r! ⁿC_r

→ If n, r are integers and 0 ≤ r ≤ n then ⁿc_r = ⁿC_{r – 1}

→ Let n, r, s are integers and 0 ≤ r ≤ n, 0 ≤ s ≤ n. If ⁿC_r = ⁿC_s then r = s or r + s = n.

→ The number of ways of dividing ‘m + n’ things (m ≠ n) into two groups containing m, n things is ^{(m + n)}C_m = ^{(m + n)}C_m = \(\frac{(m+n) !}{m ! n !}\)

→ The number of ways of dividing (m + n + p) things (m, n, p are distinct) into 3 groups of m, n, p things is \(\frac{(m+n+p) !}{m ! n ! p !}\).

→ The number of ways of dividing mn things into m equal groups is \(\frac{(m n) !}{(n !)^{m} m !}\)

→ The number of ways if distributing mn things equally to m persons is \(\frac{(m n) !}{(n !)^{m}}\)

→ If p alike things are of one kind, q alike things are of the second kind and r alike things are of the third kind, then the number of ways of selecting one or more things out of them is (p + 1) (q + 1) (r +1) – 1.

→ If m is a positive integer and m = p₁^α₁, p₁^α₂ …… p_k^α_k where p₁, p₂ …… p_k are distinct primes and α₁, α₂, …. α_k are non-negative integers, then the number of divisors of m is (α₁ + 1) (α₂ + 1) ……… (α_k + 1). [This includes 1 and m].

→ The total number of combinations of n different things taken any number of times is 2ⁿ.

→ The total number of combinations of n different things taken one or more at a timers 2ⁿ – 1.

→ The number of diagonals in a regular polygon of n sides is \(\frac{n(n-3)}{2}\)

→ Permutations are arrangements of things taken some or all at a time.

→ In a permutation, order of the things is taken into consideration.

→ ⁿp_r represents the number of permutations (without repetitions) of n dissimilar things taken r at a time.

→ Fundamental Principle: If an event is done in ‘m’ ways and another event is done in ‘n’ ways, then the two events can be together done in mn ways provided the events are independent.

→ ⁿp_r = \(\frac{n !}{(n-r) !}\) = n(n – 1)(n – 2) …………. (n – r + 1)

→ ⁿp₁ = n, ⁿp₂ = n(n-1), ⁿp₃ = n(n-1)(n-2).

→ ⁿp_r = n!

→ \(\frac{{ }^{\mathrm{n}} \mathrm{p}_{\mathrm{r}}}{{ }^{\mathrm{n}} \mathrm{p}_{\mathrm{r}-1}}\)ⁿp_n = n!

→ ⁿ⁺¹p_r = r. ⁿp_r-1 + ⁿp_r

→ ⁿp_r = r. ^n-1p_r-1 + ⁿp_r

→ ⁿp_r = r.^n-1p_r-1 + ^n-1p_r

→ The number of permutations of n things taken r at a time containing a particular thing is r × ^n-1p_r-1

→ The number of permutations of n things taken r at a time not containing a particular thing is ^n-1p_r

• The number of permutations of n things taken r at a time allowing repetitions is n^r.
• The number of permutations of n things taken not more than r at a time allowing repetitions is \(\frac{n\left(n^{\mathrm{r}}-1\right)}{(n-1)}\)

→ The number of permutations of n things of which p things are of one kind and q things are of another kind etc., is \(\frac{n !}{p ! q ! \ldots \ldots \cdots}\)

→ The sum of all possible numbers formed out of all the ‘n’ digits without zero is (n-1)! (sum of all the digits) (111 ……….. n times).

→ The sum of all possible numbers formed out of all the ‘n’ digits which includes zero is [(n -1)! (sum of all the digits) (111 …………… n times)] – [( n – 2)! (sum of all the digits)(111 ………….. (n -1) times]

→ The sum of all possible numbers formed by taking r digits from the given n digits which do not include zero is ^n-1p_r-1 (sum of all the digits)(111 …………… r times).

→ The sum of all possible numbers formed by taking r digits from the given n digits which include zero is ^n-1p_r-1(sum of all the digits)(111 ……………….. r times) – ^n-2p_r-2(sum of all the digits)(111 ……………….. (r-1) times).

• The number of permutations of n things when arranged round a circle is (n -1)!
• In case of necklace or garland number of circular permutations is \(\frac{(n-1) !}{2}\)

→ Number of permutations of n things taken r at a time in which there is at least one repetition is n^r – ⁿp_r.

→ The number of circular permutations of ‘n’ different things taken ‘r’ at a time is \(\frac{{ }^{n} \mathrm{P}_{\mathrm{r}}}{\mathrm{r}}\)

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 4 Theory of Equations to solve questions creatively.

Intermediate 2nd Year Maths 2A Theory of Equations Formulas

→ If n is a non-negative integer and a₀, a₁, a₂, ……….. a_n are real or complex numbers and a₀ ≠ 0, then the expression f(x) = a₀xⁿ + a₁x^{n – 1} + a₂x^{n – 2} + ……. + a_n is called a polynomial in x of degree n.

→ f(x) = a₀xⁿ + a₁x^{n – 1} + a₂x^{n – 2} + ……. + a_n = 0 is called a polynomial equation in x of degree n (a₀ ≠ 0). Every non-constant polynomial equation has atleast one root.

→ If f(α) = 0 then α is called a root of the equation f(x) = 0.

→ If f(α) = 0 then (x – α) is a factor of f(x).

Relation between roots and coefficients of an equation:
→ If α β γ are the roots of x³ + p₁x² + p₂x + p₃ = 0 then sum of the roots s₁ = α + β + γ = – p₁.
Sum of the products of two roots taken at a time s₂ = αβ + βγ + γα = p₂.
Product of all the roots, s₃ = αβγ = – p₃.

→ If α, β, γ, δ are the roots of x⁴ + p₁x³ + p₂x² + p₃x + p₄ = 0 then sum of the roots s₁ = α + β + γ + δ = – p₁.
Sum of the products of roots taken two at a time
s₂ = αβ + αγ + αδ + βγ + βδ + γδ = p₂.
Sum of the products of roots taken three at a time .
s₃ = αβγ + βγδ + γδα + δαβ = – p₃.
Product of the roots, s₄ = αβγδ = p₄.

→ For a cubic equation, when the roots are

• In A.P., then they are taken as a – d, a, a + d.
• In G.P., then they are taken as \(\frac{a}{r}\), a, ar.
• In H.P., then they are taken as \(\frac{1}{a-d}, \frac{1}{a^{\prime}}, \frac{1}{a+d}\).

→ For a bi quadratic equation, if the roots are

• In A.P., then they are taken as a – 3d, a – d, a + d, a + 3d.
• In C.P., then they are taken as \(\frac{1}{a-3 d}, \frac{1}{a-d}, \frac{1}{a+d}, \frac{1}{a+3 d}\).

→ In an equation with real coefficients, imaginary roots occur in conjugate pairs.

→ In an equation with rational coefficients, irrational roots occur in pairs of conjugate surds.

→ The equation whose roots are those of the equation f(x) = 0 with contrary signs is f(- x) = 0.

→ The equation whose roots are multiplied by kfa 0) of those of the proposed equation f(x) = 0 is f \(\left(\frac{x}{k}\right)\) = 0.

→ The equation whose roots are reciprocals of the roots of f(x) = 0 is f \(\left(\frac{1}{x}\right)\) = 0.

→ The equation whose roots are exceed by h than those of f(x) = 0 is f(x – h) = 0.

→ The equation whose roots are diminished by h than those of f(x) 0 is f(x + h) = 0.

→ The equation whose roots are the square of the roots of f(x) = 0 is obtained by eliminating square root from f(√x) = 0.

→ If f(x) = p₀xⁿ + p₁x^{n – 1} + p₂x^{n – 2} + ……. + p_n = 0 then to eliminate the second term,
f(x) = 0 can be transformed to f(x + h) = 0 where h = \(\frac{-p_{1}}{n \cdot p_{0}}\).

→ If an equation is unaltered by changing x into \(\frac{1}{x}\) then it is a reciprocal equation.

→ A reciprocal equation f(x) = p₀xⁿ + p₁x^{n – 1} + …… + pⁿ = 0 is said to be a reciprocal equation of first class p_i = p_{n – i} for all i.

→ A reciprocal equation f(x) = p₀xⁿ + p₁x^{n – 1} + …… + pⁿ = 0 is said to be a reciprocal equation of second class if p_i = – p_{n – i} for all i.

→ For an odd degree reciprocal equation of class one, – 1 is a root and for an odd degree reciprocal equation of class two, 1 is a root.

→ For an even degree reciprocal equation of class two, 1 and – 1 are roots.

→ If f(x) = 0 is an equation of degree ‘n’ then to eliminate r^th term, f(x) = 0 can be transformed to f(x + h) = 0 where h is a constant such that f^{(n – r + 1)}(h) = 0 i.e.,(n – r + 1)^th derivative of f(h) is zero.

→ Every n^th degree equation has exactly n roots real or imaginary.

→ Relation between, roots and coefficients of an equation.

(i) If α, β, γ are the roots of x³ + p₁x² + p₂x + p₃ = 0 the sum of the roots s₁ = α + β + γ = -p₁.
Sum of the products of two roots taken at a time s₂ = αβ + βγ + γα = -p₂
Product of all the roots, s₃ = αβγ= – p₃.

(ii) If α, β, γ, δ are the roots of x⁴ + p₁x³ + p₂x² + p₃x + p₄ = 0 then

• Sum of the roots s₁ = a+P+y+S = -p₁.
  s₂ = αβ + αγ + αδ + βα + βδ + γδ = p₂.
• Sum of the products of roots taken three at a time
  s₃ = αβγ + βγδ + γδα + δαβ = – p₃.
• Product of the roots, s₄ = αβγδ = p₄

→ For the equation xⁿ + p₁x^n-1 + p₂x^n-2 + ……… + p_n = 0

• Σ α₂ = p₁² – 2p₂
• Σ α₃ = -p₁³ + 3p₁p₂ – 3p₃
• Σ α₄ =p₁⁴ – 4p₁²p² + 2p₂² + 4p₁p₃ – 4p₄
• Σ α₂β = 3p₃ – p₁p₂
• Σ α₂βγ = p₁p₃ – 4p₄

Note: For the equation x³ + p₁x² + p₂x + p₃ = 0 Σα₂β₂ — p₂ -2p₁p₃

→ To remove the second term from a nth degree equation, the roots must be diminished by \(\frac{-\mathrm{a}_{1}}{\mathrm{na}_{0}}\) and the resultant equation will not contain the term with x^n-1.

→ If α₁ , α₂ ………………. , α_n are the roots of f(x) = 0, the equation

• Whose roots are \(\) is f\(\left(\frac{1}{x}\right)\) = 0
• Whose roots are kα₁, kα₂ …,kα_n is f\(\left(\frac{x}{h}\right)\) = 0
• Whose roots are α₁ – h, α₂ – h, …. α_n – h is f(x + h) = 0.
• Whose roots are α₁ + h, α₂ + h, ………….. α_n + h is f(x – h) = 0
• Whose roots are α₁², α₂²…. α₁² is f (f√y) = 0

→ In any equation with rational coefficients, irrational roots occur in conjugate pairs.

→ In any equation with real coefficients, complex roots occur in conjugate pairs.

→ If α is r – multiple root of f(x) = 0, then a is a (r – 1) – multiple root of f¹(x) = 0 and (r-2) – Multiple root of f ¹¹(x) = 0 and non multiple root of f^r-1(x) =0.

→ If f(x) = xⁿ + p₁x^n-1 + …………. + p_n-1x + p_n and f(a) and f(b) are of opposite sign, then at least
one real root of f(x) =0 lies between a and b.

(a) For a cubic equation, when the roots are

• In A.P., then they are taken as a – d, a, a + d
• In G.P., then are taken as \(\frac{a}{r}\), a, ar
• In H.P., then they are taken as \(\frac{1}{a-d}, \frac{1}{a}, \frac{1}{a+d}\)

(b) For a bi quadratic equation, if the roots are

• In A.P., then they are taken as a – 3d, a + d, a + 3d
• In G.P., then they are \(\frac{a}{d^{3}}, \frac{a}{d}\), ad, ad³
• In H.P., then they are taken as \(\frac{1}{a-3 d}, \frac{1}{a-d}, \frac{1}{a+d}, \frac{1}{a+3 d}\)

→ It an equation is unaltered by changing x into \(\frac{1}{x}\), then it is a reciprocal equation.

• A reciprocal equation f (x) = p₀xⁿ + p₁x^n-1 + ……………….. + p_n = 0 is said to be a reciprocal equation of first class p_i = p_n-i for all i.
• A reciprocal equation f (x) = p₀xⁿ + p₁x^n-1 + ……………….. + p_n = 0 = 0 is said to be a reciprocal equation of second class p_i = p_n-i for all i.
• For an odd degree reciprocal equation of class one, -1 is a root and for an odd degree reciprocal equation of class two, 1 is a root.
• For an even degree reciprocal equation of class two, 1 and -1 are roots.

→ If f(x) = 0 is an equation of degree ‘n’ then to eliminate rth term, .f(x) = 0 can be transformed to f(x+h) = 0 where h is a constant such that f^(n-r+1)(h) =0 i.e., (n – r + 1)^th derivative of f(h) is zero.

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 3 Quadratic Expressions to solve questions creatively.

Intermediate 2nd Year Maths 2A Quadratic Expressions Formulas

→ If a, b, c are real or complex numbers and a ≠ 0, then the expression ax² + bx + c is called a quadratic expression in the variable x.
Eg: 4x² – 2x + 3

→ If a, b, c are real or complex numbers and a ≠ 0, then ax² + bx + c = 0 is called a quadratic equation in x.
Eg: 2x² – 5x + 6 = 0

→ A complex number α is said to be a root or solution of the quadratic equation ax² + bx + c = 0 if aα² + bα + c = 0

→ The roots of ax² + bx + c = 0 are \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

→ If α, β are roots of ax² + bx + c = 0, then α + β = \(\frac{-b}{a}\) and αβ = \(\frac{C}{a}\)

→ The equation of whose roots are α, β is x² – (α + β)x + αβ = 0

→ Nature of the roots: ∆ = b² – 4ac is called the discriminant of the quadratic equation ax² + bx + c = 0. Let α, β be the roots of the quadratic equation ax² + bx + c = 0
Case 1: If a, b, c are real numbers, then

• ∆ = 0 ⇔ α = β = \(\frac{-b}{2 a}\) (a repeated root or double root)
• ∆ > 0 ⇔ α and β are real and distinct.
• ∆ < 0, ⇔ α and β are non- real complex numbers conjugate to each other.

Case 2: If a, b, c are rational numbers, then

• ∆ = 0 ⇔ α and β are rational and equal (ei) α = \(\frac{-b}{2 a}\), a double root or a repeated root.
• ∆ > 0 and is a square of a rational number ⇔ α and β are rational and distinct.
• ∆ > 0 but not a square of a rational number ⇔ α and β are conjugate surds.
• ∆ < 0, ⇔ α and β are non- real ⇔ α and β are non-real con conjugate complex numbers.

→ Let a, b and c are rational numbers, α and β be the roots of the equations ax² + bx + c = 0. Then

• α, β are equal rational numbers if ∆ = 0.
• α, β are distinct rational numbers if ∆ is the square of a non zero rational numbers.
• α, β are conjugate surds if ∆ > 0 and ∆ is not the square of a nonzero square of a rational number.

→ If a₁x² + b₁x + c₁ = 0, a₂x² + b₂x + c₂ = 0 have two same roots, then \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)

→ If α, β are roots of ax² + bx + c = 0,

• the equation whose roots are \(\frac{1}{\alpha}, \frac{1}{\beta}\) is f \(\left(\frac{1}{x}\right)\) = 0. If c ≠ 0 (ie) αβ ≠ 0
• the equation whose roots are α + k, β + k is f(x – k) = 0
• the equation whose roots are kα and kβ is f\(\left(\frac{x}{k}\right)\) = 0
• the equation whose roots are equal but opposite in sign is f(-x) = 0
  (ie) the equation whose roots are – α, – β is f(-x) = 0.

→ If the roots of ax² + bx + c = 0 are complex roots then for x ∈ R, ax² + bx + c and ‘a’ have the same sign.

→ If α and β (α < β) are the roots of ax² + bx + c = 0 then

• ax² + bx + c and ‘a’ are of opposite sign when α < x < β
• ax² + bx + c and ‘a’ are of the same sign if x < α or x > β.

→ Let f(x) = ax² + bx + c be a quadratic function

• If a > 0 then f(x) has minimum value at x = \(\frac{-b}{2 a}\) and the minimum value is given by \(\frac{4 a c-b^{2}}{4 a}\)
• If a < 0 then f(x) has maximum value at x = \(\frac{-b}{2 a}\) and the maximum value is given by \(\frac{4 a c-b^{2}}{4 a}\)

→ A necessary and sufficient condition for the quadratic equation a₁x² + b₁x + c₁ = 0 and a₂x² + b₂x + c₂ = 0 to have a common root is (c₁a₂ – c₂a₁)² = (a₁b₂ – a₂b₁) (b₁c₂ – b₂c₁).

→ If a₁b₂ – a₂b₁ = 0 then common root of a₁x² + b₁x + c₁ = 0, a₂x² + b₂x + c₂ = 0 is \(\frac{c_{1} a_{2}-c_{2} a_{1}}{a_{1} b_{2}-a_{2} b_{1}}\).

→ The standard form of a quadratic ax² + bx + c = 0 where a, b, c ∈ R and a ≠ 0

→ The roots of ax² + bx + c = 0 are \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

→ For the equation ax² + bx + c = 0, sum of the roots = \(-\frac{b}{a}\), product of the roots = \(\frac{c}{a}\).

→ If the roots of a quadratic are known, the equation is x² – (sum of the roots)x +(product of the roots)= 0

→ “Irrational roots” of a quadratic equation with “rational coefficients” occur in conjugate pairs. If p + √q is a root of ax² + bx + c = 0, then p – √q is also a root of the equation.

→ “Imaginary” or “Complex Roots” of a quadratic equation with “real coefficients” occur in conjugate pairs. If p + iq is a root of ax² + bx + c = 0. Then p – iq is also a root of the equation.

→ Nature of the roots of ax² + bx + c = 0

→ Two equations a₁x² + b₁x + c₁ = 0, a₂x² + b₂x + c₂ = 0 have exactly the same roots if \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)

→ The equations a₁x² + b₁x + c₁ = 0, a2x² + b₂x + c₂ = 0 have a common root, if (c₁a₂ – c₂a₁)² = (a₁b₂ – a₂b₁)(b₁c₂ – b₂c₁) and the common root is \(\frac{c_{1} a_{2}-c_{2} a_{1}}{a_{1} b_{2}-a_{2} b_{1}}\) if a₁b₂ ≠ a₂b₁

→ If f(x) = 0 is a quadratic equation, then the equation whose roots are

• The reciprocals of the roots of f(x) = 0 is f\(\left(\frac{1}{x}\right)\) = 0
• The roots of f(x) = 0, each ‘increased’ by k is f(x – k) = 0
• The roots of f(x) = 0, each ‘diminished’ by k is f(x + k) = 0
• The roots of f(x) = 0 with sign changed is f(-x) = 0
• The roots of f(x) = 0 each multiplied by k(≠0) is f\(\left(\frac{x}{k}\right)\) = 0

→ Sign of the expression ax² + bx + c = 0

• The sign of the expression ax² + bx + c is same as that of ‘a’ for all values of x if b² – 4ac ≤ 0 i.e. if the roots of ax² + bx + c = 0 are imaginary or equal.
• If the roots of the equation ax² + bx + c = 0 are real and different i.e b² – 4ac > 0, the sign of the expression is same as that of ‘a’ if x does not lie between the two roots of the equation and opposite to that of ‘a’ if x lies between the roots of the equation.

→ The expression ax² + bx + c is positive for all real values of x if b² – 4ac < 0 and a > 0.

→ The expression ax² + bx + c has a maximum value when ‘a’ is negative and x = –\(\frac{\mathrm{b}}{2 \mathrm{a}}\). Maximum value of the expression = \(\frac{4 a c-b^{2}}{4 a}\)

→ The expression ax² + bx + c has a maximum value when ‘a’ is positive and x = –\(\frac{\mathrm{b}}{2 \mathrm{a}}\). Minimum value of the expression = \(\frac{4 a c-b^{2}}{4 a}\)

Theorem 1:
If the roots of ax² + bx + c = 0 are imaginary, then for x ∈ R , ax² + bx + c and a have the same sign.
Proof:
The root are imaginary
b² – 4ac < 0 4ac – b² > 0
\(\frac{a x^{2}+b x+c}{a}=x^{2}+\frac{b}{a} x+\frac{c}{a}=\left(x+\frac{b}{2 a}\right)^{2}+\frac{c}{a}-\frac{b^{2}}{4 a^{2}}=\left(x+\frac{b}{2 a}\right)^{2}+\frac{4 a c-b^{2}}{4 a^{2}}\)
∴ For x ∈ R, ax² + bx + c = 0 and a have the same sign.

Theorem 2.
If the roots of ax² + bx + c = 0 are real and equal to α = \(\frac{-b}{2 a}\), then α ≠ x ∈ R ax² + bx + c and a will have same sign.
Proof:
The roots of ax² + bx + c = 0 are real and equal
⇒ b² = 4ac ⇒ 4ac – b² = 0
\(\frac{a x^{2}+b x+c}{a}\) = x + \(\frac{b}{a}\)x + \(\frac{c}{a}\)
= \(\left(x+\frac{b}{2 a}\right)^{2}+\frac{c}{a}-\frac{b^{2}}{4 a^{2}}\)
= \(\left(x+\frac{b}{2 a}\right)^{2}+\frac{4 a c-b^{2}}{4 a^{2}}\)
= \(\left(x+\frac{b}{2 a}\right)^{2}\) > 0 for x ≠ \(\frac{-b}{2 a}\) = α
For α ≠ x ∈ R, ax² + bx + c and a have the same sign.

Theorem 3.
Let be the real roots of ax² + bx + c = 0 and α < β. Then
(i) x ∈ R, α < x< β ax² + bx + c and a have the opposite signs
(ii) x ∈ R, x < α or x> β ax² + bx + c and a have the same sign.
Proof:
α, β are the roots of ax² + bx + c = 0
Therefore, ax² + bx + c = a(x – α)(x – β)
\(\frac{a x^{2}+b x+c}{a}\) = (x – α)(x – β)

(i) Suppose x ∈ R, α < x < β
⇒ x < α < β then x – α < 0, x – β < 0 ⇒ (x – α)(x – β) > 0 ⇒ \(\frac{a x^{2}+b x+c}{a}\) > 0
⇒ ax² + bx + c, a have a same sign

(ii) Suppose x ∈ R, x > β, x > β > α then x – α > 0, x – β > 0
⇒ (x – α)(x – β) > 0 ⇒ \(\frac{a x^{2}+b x+c}{a}\) > 0
⇒ ax² + bx + c, a have same sign
∴ x ∈ R, x < α or x > β ⇒ ax² + bx + c and a have the same sign.

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 2 De Moivre’s Theorem to solve questions creatively.

Intermediate 2nd Year Maths 2A De Moivre’s Theorem Formulas

Statement:
→ If ‘n’ is an integer, then (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ
If n’ is a rational number, then one of the values of
(cos θ + i sin θ)ⁿ is cos nθ + i sin nθ

n^th roots of unity:
→ n^th roots of unity are {1, ω, ω² …….. ω^{n – 1}}.
Where ω = \(\left[\cos \frac{2 k \pi}{n}+i \sin \frac{2 k \pi}{n}\right]\) k = 0, 1, 2 ……. (n – 1).
If ω is a n^th root of unity, then

• ωⁿ = 1
• 1 + ω + ω² + ………… + ω^{n – 1} = 0

Cube roots of unity:
→ 1, ω, ω² are cube roots of unity when

• ω³ = 1
• 1 + ω + ω² = 0
• ω = \(\frac{-1+i \sqrt{3}}{2}\), ω² = \(\frac{-1-i \sqrt{3}}{2}\)
• Fourth roots of unity roots are 1, – 1, i, – i

→ If Z₀ = r₀ cis θ₀ ≠ 0, then the n^th roots of Z₀ are α_k = (r₀)^1/n cis\(\left(\frac{2 k \pi+\theta_{0}}{n}\right)\) where k = 0, 1, 2, ……… (n – 1)

→ If n is any integer, (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ

→ If n is any fraction, one of the values of (cosθ + i sinθ)ⁿ is cos nθ + i sin nθ.

→ (sinθ + i cosθ)ⁿ = cos(\(\frac{n \pi}{2}\) – nθ) + i sin(\(\frac{n \pi}{2}\) – nθ)

→ If x = cosθ + i sinθ, then x + \(\frac{1}{x}\) = 2 cosθ, x – \(\frac{1}{x}\) = 2i sinθ

→ xⁿ + \(\frac{1}{x^{n}}\) = 2cos nθ, xⁿ – \(\frac{1}{x^{n}}\) = 2i sin nθ

→ The nth roots of a complex number form a G.P. with common ratio cis\(\frac{2 \pi}{n}\) which is denoted by ω.

→ The points representing n^th roots of a complex number in the Argand diagram are concyclic.

→ The points representing n^th roots of a complex number in the Argand diagram form a regular polygon of n sides.

→ The points representing the cube roots of a complex number in the Argand diagram form an equilateral triangle.

→ The points representing the fourth roots of complex number in the Argand diagram form a square.

→ The nth roots of unity are 1, w, w²,………. , w^n-1 where w = cis\(\frac{2 \pi}{n}\)

→ The sum of the nth roots of unity is zero (or) the sum of the nth roots of any complex number is zero.

→ The cube roots of unity are 1, ω, ω² where ω = cis\(\frac{2 \pi}{3}\), ω² = cis\(\frac{4 \pi}{3}\) or
ω = \(\frac{-1+i \sqrt{3}}{2}\)
ω² = \(\frac{-1-i \sqrt{3}}{2}\)
1 + ω + ω² = 0
ω³ = 1

→ The product of the n^th roots of unity is (-1)^n-1 .

→ The product of the n^th roots of a complex number Z is Z(-1)^n-1 .

→ ω, ω² are the roots of the equation x² + x + 1 = 0

Use these Inter 2nd Year Maths 2A Formulas PDF Chapter 1 Complex Numbers to solve questions creatively.

Intermediate 2nd Year Maths 2A Complex Numbers Formulas

Definition of a complex number:
→ A number of the type z = x + yi, where x, y ∈ R and i = √- 1 i.e., i² = – 1 is called a complex number ‘x’ is called real part of z, and ‘y’ is called imaginary part of z. We write x = Re(z) and y = Im(z). A number z = x + yi is said to be purely real iff y = 0 (x ≠ 0) and purely imaginary iff x = 0 (y ≠ 0)
A complex number a + ib is an ordered pair of real numbers. It is denoted by (a, b), a ∈ R, b ∈ R.

→ Two complex numbers z₁ = (a, b), z₂ = (c, d) are said to be equal iff a = c and b = d.

→ If z₁ = (a, b), z₂ = (c, d) then

• z₁ + z₁ = (a + c, b + d)
• z₁ – z₂ = (a – c, b – d)
• z₁. Z₂ = (ac – bd, ad + bc) and
• \(\frac{z_{1}}{z_{2}}=\left(\frac{a c+b d}{c^{2}+d^{2}}, \frac{b c-a d}{c^{2}+d^{2}}\right)\)

Modulus and Amplitude:
→ The modulus of a complex number z = x + iy is defined as a non-negative real number r = \(\sqrt{x^{2}+y^{2}}\). It is denoted by |z|.

→ Any real number θ satisfying the equation cos θ = \(\frac{x}{r}\), sin θ = \(\frac{y}{r}\) is called an amplitude or argument of z. The unique argument θ of z satisfying – π < θ ≤ π is called the principal argument of z and is denoted by Arg z.

→ The polar form or modulus amplitude form of the complex number
z = x + iy is r(cos θ + i sin θ)

Conjugate of a complex number:
→ If z = x + iy i.e., x, y ∈ R, then the complex number x – iy is called conjugate of z and is written as z̅. The sum and product of a complex number and its Conjugate is always purely real.

Some properties of modulus, amplitude and conjugate:

• (z̅) = z
• z + z̅ = 2 Re (z) and z – z̅ = 2 Im (z)
• \(\overline{Z_{1} Z_{2}}\) = \(\overline{Z_{1}}\) \(\overline{Z_{2}}\)
• \(\) and \(\) (z₂ ≠ 0) and |z₁z₂| = |z₁| |z₂|
• zz̅ = |z|²
• |z| = |z̅|;
• |z₁ + z₂| ≤ |z₁| + |z₂|, |z₁ – z₂| ≤ |z₁| + |z₂|
• |z₁ – z₂| > | |z₁| – |z₂| |
  In (vii) and (viii) equality holds iff amp (z₁) – amp (z₂) is an integral multiple of 2π.
• amp (z₁ z₂) = amp (z₁) + amp (z₂) + nπ, for some n ∈ {- 1, 0, 1}
• amp \(\left(\frac{Z_{1}}{Z_{2}}\right)\) = amp (z₁) – amp (z₂) + nπ, for some n ∈ {- 1, 0, 1}
• \(\frac{1}{{cis} \alpha}\) = cis(- α)
• cis α cis β = cis (α + β)
• \(\frac{{cis} \alpha}{{cis} \beta}\) = cis (α – β)

De-Moivre’s theorem:

• If n is any integer, then (cos θ + i sin θ)ⁿ = cos nθ + i sin nθ
• If n = \(\frac{p}{q}\), where p and q are integers having no common factor and q > 1, then cos nθ + i sin nθ is one of the q values of (cos θ + i sin θ)ⁿ
• If z₀ = r₀ cis θ₀ ≠ 0, then the n^th roots of z₀ are α_k = r₀^1/n cis \(\left(\frac{2 k \pi+\theta_{0}}{n}\right)\),
  k = 0, 1, 2, 3 … (n-1)

Cube roots of unity:

• The cube roots of unity are 1, ω = \(\frac{-1+\sqrt{3 i}}{2}\) and ω² = \(\frac{-1-\sqrt{3 i}}{2}\)
• 1 + ω + ω² = 0 and w³ = 1; 1 + ω = – ω ², 1 + ω² = – ω, ω + ω² = – 1
• Either of the two non-real cube roots of unity is square of the other.
• For either of the two non – real cube roots a.and of unity α + β = – 1, αβ = – 1, α² = β, β² = α and α³ = β³ = 1
• (-1)^1/3 = – 1, – ω – ω²
• The n^th roots of unity are cis\(\left(\frac{2 k \pi}{n}\right)\), k = 0, 1, 2, ….. (n – 1)

Formulae:

→ Modulus of Z = \(\sqrt{x^{2}+y^{2}}\)

→ If \(\sqrt{a+i b}\) = (x + iy), then x = \(\sqrt{\frac{\sqrt{a^{2}+b^{2}}+a}{2}}\) and y = \(\sqrt{\frac{\sqrt{a^{2}+b^{2}}-a}{2}}\)

→ Conjugate of a + ib = a – ib

→ Conjugate of a – ib = a + ib

→ Any number of the form x + iy where x, y ∈ R and i² = -1 is called a Complex Number.

→ In the complex number x + iy, x is called the real part and y is called the imaginary part of the complex number.

→ A complex number is said to be purely imaginary if its real part is zero and is said to be purely real if its imaginary part is zero.
(a) Two complex numbers are said to be equal if their real parts are equal and their imaginary parts are equal.
(b) In the set of complex numbers, there is no meaning to the phrases one complex is greater than or less than another i.e. If two complex numbers are not equal, we say they are unequal.
(c) a+ ib > c + id is meaningful only when b = 0, d = 0.

→ Two complex numbers are conjugate if their sum and product are both real. They are of the form a + ib, a – ib.

→ cisθ₁ cisθ₂ = cis(θ₁ + θ₂), \(\frac{{cis} \theta_{1}}{{cis} \theta_{2}}\) = cis(θ₁ – θ₂), \(\frac{1}{\cos \theta+i \sin \theta}\) = cosθ – isinθ

→ \(\frac{a_{1}+i b_{1}}{a_{2}+i b_{2}}=\frac{\left(a_{1} a_{2}+b_{1} b_{2}\right)+i\left(a_{2} b_{1}-a_{1} b_{2}\right)}{a_{2}^{2}+b_{2}^{2}}\)

→ \(\frac{1+i}{1-i}\) = i, \(\frac{1-i}{1+i}\) = i

→ \(\sqrt{x^{2}+y^{2}}\) is called the modulus of the complex number x + iy and is denoted by r or |x + iy|

→ Any value of 0 obtained from the equations cos θ = \(\frac{x}{r}\), sin θ = \(\frac{y}{r}\) is called an amplitude of the complex number.

→ The amplitude lying between -π and π is called the principal amplitude of the complex number. Rule for choosing the principal amplitude.

→ If θ is the principal amplitude, then -π < 0 < π

→If α is the principle amplitude of a complex number, general amplitude = 2nπ + α where n ∈ Z.

• Amp (Z₁ Z₂) = Amp Z₁ + AmpZ₂
• Amp z + Amp z̅ = 2π (when z is a negative real number) = 0 (otherwise)

→ r(cos θ + i sin θ) is the modulus amplitude form of x + iy.

→ If the amplitude of a complex number is \(\frac{\pi}{2}\), its real part is zero.

→ If the amplitude of a complex number is \(\frac{\pi}{4}\), its real part is equal to its imaginary part.

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A De Moivre’s Theorem Solutions Exercise 2(b) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A De Moivre’s Theorem Solutions Exercise 2(b)

I. Find all the values of the following.

Question 1.
(i) (1 – i√3)^1/3
Solution:

(ii) (-i)^1/6
Solution:

(iii) (1 + i)^2/3
Solution:

(iv) (-16)^1/4
Solution:

(v) (-32)^1/5
Solution:

Question 2.
If A, B, C are angles of a triangle such that x = cis A, y = cis B, z = cis C, then find the value of xyz.
Solution:
∴ A, B, C are angles of a triangle
⇒ A + B + C = 180° ………..(1)
x = cis A, y = cis B, Z = cis C
xyz = cis (A + B + C)
= cos (A + B + C) + i sin (A + B + C)
= cos (180°) + i sin (180°)
= -1 + i(0)
= -1
∴ xyz = -1

Question 3.
(i) If x = cis θ, then find the value of \(\left[x^{6}+\frac{1}{x^{6}}\right]\)
Solution:
∵ x = cos θ + i sin θ
⇒ x⁶ = (cos θ + i sin θ)⁶ = cos 6θ + i sin 6θ
⇒ \(\frac{1}{x^{6}}\) = cos 6θ – i sin 6θ
∴ \(x^{6}+\frac{1}{x^{6}}\) = 2 cos 6θ

(ii) Find the cube roots of 8.
Solution:
Let x³ = 8
⇒ x = 8^1/3
⇒ x = (2³)^1/3 (1)^1/3
⇒ x = 2 (1)^1/3
Since cube roots of unity are 1, ω, ω²
∴ The cube roots or 8 are 2, 2ω, 2ω²

Question 4.
If 1, ω, ω² are the cube roots of unity, then prove that
(i) \(\frac{1}{2+\omega}-\frac{1}{1+2 \omega}=\frac{1}{1+\omega}\)
Solution:
ω is a cube root of unity.
1 + ω + ω² = 0 and ω³ = 1

(ii) (2 – ω) (2 – ω²) (2 – ω¹⁰) (2 – ω¹¹) = 49
Solution:
∵ 1, ω, ω² are the cube roots of unity.
ω³ = 1 and 1 + ω + ω² = 0
2 – ω¹⁰ = 2 – ω⁹ . ω
= 2 – (ω³)³ ω
= 2 – (1)³ ω
= 2 – ω
and 2 – ω¹¹ = 2 – (ω³)³ . ω²
= 2 – (1)³ ω²
= 2 – ω²
(2 – ω) (2 – ω²) = 4 – 2ω – 2ω² + ω³
= 4 – 2(ω + ω²) + 1
= 4 – 2(-1) + 1
= 4 + 2 + 1
= 7
∴ (2 – ω) (2 – ω²) (2 – ω¹⁰) (2 – ω¹¹) = (2 – ω) (2 – ω²) (2 – ω) (2 – ω²)
= ((2 – ω) (2 – ω²))²
= 7²
= 49

(iii) (x + y + z) (x + yω + zω²) (x + yω² + zω) = x³ + y³ + z³ – 3xyz
Solution:
∵ 1, ω, ω² are the cube roots of unity.
⇒ 1 + ω + ω² = 0 and ω³ = 1
Now consider,
(x + yω + zω²) (x + yω² + zω)
= x² + xyω² + zxω + xyω + y²ω³ + yzω² + zxω² + yzω⁴ + z²ω³
= x² + y² (1) + z² (1) + xy (ω + ω²) + yz (ω⁴ + ω²) + zx (ω + ω²)
= x² + y² + z² + xy (-1) + yz (ω + ω²) + zx (-1)
= x² + y² + z² – xy – yz – zx ……….(1)
L.H.S = (x + y + z) (x + yω + zω²) (x + yω² + zω)
= (x + y + z) (x² + y² + z² – xy – yz – zx) [by (1)]
= x³ + y³ + z³ – 3xyz
= R.H.S

Question 5.
Prove that -ω, and -ω² are the roots of z² – z + 1 =0, where ω and ω² are the complex cube roots of unity.
Solution:
Since ω and ω² are the complex cube roots of unity
∴ 1 + ω + ω² = 0 and ω² = 1
z² – z + 1 = (-ω)² – (-ω) + 1
= ω² + ω + 1
= 0
∴ -ω is a root of the equation z² – z + 1 = 0
z² – z + 1 = (-ω²)² – (-ω²) + 1
= ω⁴ + ω² + 1
= ω³ . ω + ω² + 1
= ω + ω²+ 1
= 0
∴ -ω² is a root of the equation z² – z + 1 = 0

Question 6.
If 1, ω, ω² are the cube roots of unity, then find the values of the following.
(i) (a + b)³ + (aω + bω²)³ + (aω² + bω)³
Solution:
Since 1, ω, ω² are the cube roots of unity
∴ 1 + ω + ω² = 0 and ω³ = 1
Now (a + b)³ = a³ + 3a²b + 3ab² + b³ ……..(1)
(aω + bω²)³ = [ω(a + bω)]³
= ω³ (a + bω)³
= (1) (a + bω)³
= a³ + 3a²bω + 3ab²ω² + b³ω³
= a³ + 3a²bω + 3ab²ω² + b³ ……….(2)
∵ ω³ = 1
and (aω² + bω)³ = [ω(aω + b)]³
= ω³ (aω + b)³
= (1) (aω + b)³
= a³ω³ + 3a²bω² + 3ab²ω + b³
= a³(1) + 3a²bω² + 3ab²ω + b³
∴ (aω² + bω)³ = a³ + 3a²bω² + 3ab²ω + b³ ……….(3)
Adding (1), (2) and (3)
(a + b)³ + (aω + bω²)³ + (aω² + bω)³ = 3a³ + 3a²b (1 + ω + ω²) + 3ab² (1 + ω + ω²) + 3b³
= 3(a³ + b³) + 3a²b (0) + 3ab² (0)
= 3(a³ + b³)
∴ (a + b)³ + (aω + bω²)³ + (aω² + bω)³ = 3 (a³ + b³)

(ii) (a + 2b)² + (aω² + 2bω)² + (aω + 2bω²)²
Solution:
(a + 2b)² = a² + 4ab + 4b² ……….(1)
(aω² + 2bω)² = a²ω⁴ + 4abω³ + 4b²ω²
= a²ω³ω + 4ab (1) + 4b²ω²
= a²ω + 4ab + 4b²ω² ………..(2)
and (aω + 2bω²)² = a²ω² + 4abω³ + 4b²ω⁴
= a²ω² + 4ab (1) + 4b²ω³ω
= a²ω² + 4ab + 4b² (1) ω
∴ (aω + 2bω²)² = a²ω² + 4ab + 4b²ω ……….(3)
By Adding (1), (2) and (3)
(a + 2b)² + (aω² + 2bω)² + (aω + 2bω²)²
= a² (1 + ω + ω²) + 12ab + 4b² (1 + ω + ω²)
= a² (0) + 12ab + 4b² (0)
= 12ab
∴ (a + 2b)² + (aω² + 2bω)² + (aω + 2bω²)² = 12ab

(iii) (1 – ω + ω²)³
Solution:
(1 – ω + ω²)³ = (-ω – ω)³
= (-2ω)³
= -8ω³
= -8(1)
= -8 (∵ 1 + ω + ω² = 0)

(iv) (1 – ω) (1 – ω²) (1 – ω⁴) (1 – ω⁸)
Solution:
1 – ω⁴ = 1 – (ω³) ω
= 1 – (1) ω
= 1 – ω
1 – ω⁸ = 1 – (ω³)² ω²
= 1 – (1) ω²
= 1 – ω²
∴ (1 – ω) (1 – ω²) (1 – ω⁴)(1 – ω⁸) = (1 – ω) (1 – ω²) (1 – ω) (1 – ω²)
= [(1 – ω) (1 – ω²)]²
= (1 – ω – ω² + ω³)²
= [1 – (ω + ω²) + 1] (∵ 1 + ω + ω² = 0)
= [1 – (-1) + 1]²
= (3)²
= 9
∴ (1 – ω) (1 – ω²) (1 – ω⁴) (1 – ω⁸) = 9

(v) \(\left[\frac{a+b \omega+c \omega^{2}}{c+a \omega+b \omega^{2}}\right]+\left[\frac{a+b \omega+c \omega^{2}}{b+c \omega+a \omega^{2}}\right]\)
Solution:
∴ 1, ω, ω² are the cube roots of unity
⇒ ω³ = 1 and 1 + ω + ω² = 0 ………..(1)

(vi) (i + ω)³ + (1 + ω²)³
Solution:
(i + ω)³ + (1 + ω²)³ = (-ω²)³ +(-ω)³
= -ω⁶ – ω³
= -1 – 1
= -2
∴ (1 + ω)³ + (1 + ω²)³ = -2

(vii) (1 – ω + ω²)⁵ + (1 + ω – ω²)⁵
Solution:
(1 – ω + ω²)⁵ + (1 + ω – ω²)5 = (-ω – ω)⁵ + (-ω² – ω²)⁵
= (-2ω)⁵ + (-2ω²)⁵
= -32ω⁵ – 32ω¹⁰
= -32(ω⁵ + ω¹⁰)
= -32(ω² + ω)
= -32(-1)
= 32
∴ (1 – ω + ω²)⁵ + (1 + ω – ω²)⁵ = 32

II.

Question 1.
Solve the following equations.
(i) x⁴ – 1 = 0
Solution:
x⁴ – 1 = 0
⇒ x⁴ = 1
⇒ x⁴ = cos 0° + i sin 0°
⇒ x⁴ = cos 2kπ + i sin 2kπ
⇒ x = (cos 2kπ + i sin 2kπ)^1/4
= cos \(\frac{k \pi}{2}\), k = 0, 1, 2, 3
i.e., cos 0° + i sin 0°, cos \(\frac{\pi}{2}\) + i sin \(\frac{\pi}{2}\), cos π + i sin π, cos \(\frac{3 \pi}{2}\) + i sin \(\frac{3 \pi}{2}\),
i.e., 1, i, -1, -i = ±1, ±i

(ii) x⁵ + 1 = 0
Solution:
x⁵ + 1 = 0
⇒ x⁵ = -1
⇒ x⁵ = cos π + i sin π
⇒ x⁵ = cos(2k + 1) π + i sin(2k + 1) π, k ∈ z
⇒ x = [cos(2k + 1) π + i sin(2k + 1) π]^1/5
⇒ x = cis \(\frac{(2 k+1) \pi}{5}\), k = 0, 1, 2, 3, 4

(iii) x⁹ – x⁵ + x⁴ – 1 = 0
Solution:
x⁹ – x⁵ + x⁴ – 1 = 0
⇒ x⁵ (x⁴ – 1) + 1 (x⁴ – 1) = 0
⇒ (x⁴ – 1) (x⁵ + 1) = 0
⇒ x⁴ – 1 = 0
Solving the roots are ±1, ±i
(see the above problem)
x⁵ + 1 = 0
Solving the roots are cis \(\frac{(2 k+1) \pi}{5}\)
k = 0, 1, 2, 3, 4 (see the above problem)
∴ The roots of the given equation are ±1, ±i, cis (2k + 1) \(\frac{\pi}{5}\), k = 0, 1, 2, 3, 4
i.e., ±1, ±i, cis(\(\pm \frac{\pi}{5}\)), cis(\(\pm \frac{3 \pi}{5}\))

(iv) x⁴ + 1 = 0
Solution:
x⁴ + 1 = 0
⇒ x⁴ = -1
⇒ x⁴ = cos π + i sin π
∴ x⁴ = cos(2kπ + π) + i sin(2kπ + π),
∴ x = [cis(2k + 1)π]^1/4
= cis(2k + 1) \(\frac{\pi}{4}\), where k = 0, 1, 2, 3
∴ x = \({cis} \frac{\pi}{4}, {cis}\left(\frac{3 \pi}{4}\right), {cis}\left(\frac{5 \pi}{4}\right)\) and \({cis}\left(\frac{7 \pi}{4}\right)\)
These four values of x are the solutions to the given equation.

Question 2.
Find the common roots of x¹² – 1 = 0 and x⁴ + x² + 1 = 0
Solution:
Consider x¹² – 1 = 0
⇒ x¹² = 1
⇒ x¹² = (cos 0 + i sin 0)
⇒ x¹² = (cos 2kπ + i sin 2kπ), k is a positive integer
⇒ x = (cos 2kπ + i sin 2kπ)^1/2

Question 3.
Find the number of 15th roots of unity, which are also the 25th roots of unity.
Solution:
The number of common roots = H.C.F of {15, 25} = 5

Question 4.
If the cube roots of unity are 1, ω, ω2, then find the roots of the equation (x – 1)³ + 8 = 0.
Solution:
Given (x – 1)³ + 8 = 0
⇒ (x – 1)³ = -8
⇒ (x – 1)³ = (-2)³ (1)³
⇒ (x – 1) = (-2) (1)^1/3
⇒ x – 1 = -2, -2ω, -2ω²
⇒ x = 1 – 2, 1 – 2ω, 1 – 2ω²
⇒ x = -1, 1 – 2ω, 1 – 2ω²

Question 5.
Find the product of all the values of (1 + i)4/5.
Solution:

Question 6.
If z² + z + 1 =0, where z is a complex number, prove that
\(\left(z+\frac{1}{z}\right)^{2}+\left(z^{2}+\frac{1}{z^{2}}\right)^{2}+\left(z^{3}+\frac{1}{z^{3}}\right)^{2}\) + \(\left(z^{4}+\frac{1}{z^{4}}\right)^{2}+\left(z^{5}+\frac{1}{z^{5}}\right)^{2}+\left(z^{6}+\frac{1}{z^{6}}\right)\) = 12
Solution:
Given z² + z + 1 = 0
⇒ z = \(\frac{-1 \pm \sqrt{1-4.1 .1}}{2}\)
= \(\frac{-1 \pm i \sqrt{3}}{2}\)
= \(\frac{-1+i \sqrt{3}}{2}, \frac{-1-i \sqrt{3}}{2}\)
= ω, ω²
∴ 1 + ω + ω² = 0 and ω³ = 1
If z = ω then

= (ω + ω²)² + (ω² + ω)² + (1 + 1)² + (ω + ω²)² + (ω² + ω)² + (1 + 1 )²
= (-1)² + (-1)² + 4 + (-1)² + (-1)² + 4
= 1 + 1 + 4 + 1 + 1 + 4
= 12
Similarly If z = ω² then
\(\left(z+\frac{1}{z}\right)^{2}+\left(z^{2}+\frac{1}{z^{2}}\right)^{2}+\left(z^{3}+\frac{1}{z^{3}}\right)^{2}\) + \(\left(z^{4}+\frac{1}{z^{4}}\right)^{2}+\left(z^{5}+\frac{1}{z^{5}}\right)^{2}+\left(z^{6}+\frac{1}{z^{6}}\right)\) = 12

III.

Question 1.
If 1, α, α², α³, ……., α^n-1 be the nth roots of unity, then prove that 1^p + α^p + (α²)^p + (α³)^p + …… + (α^n-p)^2p = 0
= 0; if p ≠ kn
= n; if p ≠ kn, where p, k ∈ N
Solution:
nth roots of unity are 1, α, α², ………., α^n-1


∴ Each term of the series in (1) is 1
Hence the sum of the series 1 + αp + (α²)p + (α³)p + …….. + (α^n-1)^p
= 1 + 1 + 1 + ……… + 1 (n times)
= n(1)
= n

Question 2.
Prove that the sum of 99th powers of the roots of the equation x⁷ – 1 = 0 is zero and hence deduce the roots of x⁶ + x⁵ + x⁴ + x³ + x² + x + 1 = 0.
Solution:

Question 3.
If n is a positive integer, show that \((p+i q)^{1 / n}+(p-i q)^{1 / n}=2\left(p^{2}+q^{2}\right)^{1 / 2 n}\) . \(\cos \left(\frac{1}{n}, \tan \frac{q}{p}\right)\)
Solution:

Question 4.
Show that one value of \(\left(\frac{1+\sin \frac{\pi}{8}+i \cos \frac{\pi}{8}}{1+\sin \frac{\pi}{8}-i \cos \frac{\pi}{8}}\right)^{8 / 3}\) is -1.
Solution:

Question 5.
Solve (x – 1)ⁿ = xⁿ, n is a positive integer.
Solution:
Since x = 0 is not a solution of the given equation, it is equivalent to the equation \(\left(\frac{x-1}{x}\right)^{n}=1\)
Clearly \(\left(\frac{x-1}{x}\right)^{n}=1\)
⇒ \(\frac{x-1}{x}\) is an nth root of unity other than one.
Suppose that ω is an nth root of unity and ω ≠ 1.
Then, \(\frac{x-1}{x}\) = ω
⇒ x – 1 = xω
⇒ (1 – ω) x = 1
⇒ x = \(\frac{1}{1-\omega}\), (∵ ω ≠ 1) ……….(1)

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A De Moivre’s Theorem Solutions Exercise 2(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A De Moivre’s Theorem Solutions Exercise 2(a)

I.

Question 1.
If n is an integer then show that (1 + i)²ⁿ + (1 – i)²ⁿ = 2ⁿ⁺¹ cos \(\frac{n \pi}{2}\)
Solution:

Question 2.
Find the values of the following:
(i) (1 + i√3)³
Solution:
Let 1 + i√3 = r (cos θ + i sin θ)
Equating real and imaginary parts
r cos θ = 1 and r sin θ = √3
⇒ r² (1) = 1 + 3 = 4
⇒ r = 2
cos θ = \(\frac{1}{2}\) and sin θ = \(\frac{\sqrt{3}}{2}\)
⇒ θ = \(\frac{\pi}{3}\)

(ii) (1 – i)⁸
Solution:

(iii) (1 + i)¹⁶
Solution:

(iv) \(\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^{5}-\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^{5}\)
Solution:

II.

Question 1.
If α, β are the roots of the equation x² – 2x + 4 = 0 then for any n ∈ N show that αⁿ + βⁿ = 2ⁿ⁺¹ cos(\(\frac{n \pi}{3}\))
Solution:

Question 2.
If cos α + cos β + cos γ = 0 = sin α + sin β + sin γ then show that
(i) cos 3α + cos 3β + cos 3γ = 3 cos(α + β + γ)
(ii) sin 3α + sin 3β + sin 3γ = 3 sin(α + β + γ)
(iii) cos(α + β) + cos(β + γ) + cos(γ + α) = 0
Solution:
Given that cos α + cos β + cos γ = 0 and sin α + sin β + sin γ = 0
From the hypothesis, we have
(cos α + cos β + cos γ) + i (sin α + sin β + sin γ) = 0
i.e., (cos α + i sin α) + (cos β + i sin β) + (cos γ + i sin γ) = 0
We know that if a + b + c = 0, then a³ + b³ + c³ = 3abc
Hence (cos α + i sin α)³ + (cos β + i sin β)³ + (cos γ + i sin γ)³ = 3 (cos α + i sin α) (cos β + i sin β) + (cos γ + i sin γ)
i.e., cos 3α + i sin 3α + cos 3β + i sin 3β + cos 3γ + i sin 3γ = 3[cos(α + β + γ) + i sin(α + β + γ)] …….(1)
On equating the real parts on both sides of equation (1) we get
(i) cos 3α + cos 3β + cos 3γ = 3 cos(α + β + γ)
On equation thenmaginary parts on both sides of equation (1) we get,
(ii) sin 3α + sin 3β + sin 3γ = 3 sin(α + β + γ)
(iii) Let a = cos α + i sin α then \(\frac{1}{a}\) = cos α – i sin α
b = cos β + i sin β and \(\frac{1}{b}\) = cos β – i sin β
c = cos γ + i sin γ and \(\frac{1}{c}\) = cos γ – i sin γ
Now \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) = (cos α + cos β + cos γ) – i (sin α + sin β + sin γ)
⇒ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) = 0 – i(0)
⇒ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) = 0
⇒ bc + ca + ab = 0
⇒ [cos(β + γ) + i sin(β + γ)] + [cos(γ + α) + i sin(γ + α)] + [cos(α + β) + i sin(α + β)] = 0
⇒ [cos(α + β) + cos(β + γ) + cos(γ + α)] + i [sin(α + β) + sin(β + γ) + sin(γ + α)] = 0
By equation real parts on both sides
cos (α + β) + cos (β + γ) + cos (γ + α) = 0

Question 3.
If n is an integer and z = cis θ, (θ ≠ (2n + 1) π/2), then show that \(\frac{z^{2 n}-1}{z^{2 n}+1}\) = i tan nθ.
Solution:
z = cis θ = cos θ + i sin θ, θ ≠ (2n + 1) π/2

(∵ i² = -1)
= i tan(nθ)
= R.H.S

Question 4.
If (1 + x)ⁿ = a₀ + a₁x + a₂x² + ……… + anxn, then show that
(i) a₀ – a₂ + a₄ – a₆ + …….. = \(2^{n / 2} \cos \frac{n \pi}{4}\)
(ii) a₁ – a₃ + a₅ – a₇ + ……… = \(2^{n / 2} \sin \frac{n \pi}{4}\)
Solution:

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Complex Numbers Solutions Exercise 1(d) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Complex Numbers Solutions Exercise 1(d)

I.

Question 1.
(i) Find the equation of the perpendicular bisector of the line segment joining the points 7 + 7i, 7 – 7i in the Argand diagram.
Solution:
A(7, 7); B(7, -7) represents the given complex numbers in the Argand diagram.

O is the mid-point of AB.
Co-ordinates of O are \(\left(\frac{7+7}{2}, \frac{7-7}{2}\right)\) = (7, 0)
Slope of \(\overleftrightarrow{\mathrm{AB}}=\frac{7+7}{7-7}=\frac{14}{0}\) = ∞
AB is parallel to Y-axis
PQ is perpendicular to AB
PQ is parallel to X-axis
Slope of PQ = 0
Equation of PQ is y – 0 = 0(x – 7)
i.e., y = 0

(ii) Find the equation of the straight line joining the point -9 + 6i, 11 – 4i in the Argand plane.
Solution:
Given points are -9 + 6i, 11 – 4i
Let A = (-9, 6); B = (11, -4)
Equation of the straight line AB is y – 6 = \(\frac{-4-6}{11+9}\) (x + 9)
⇒ y – 6 = \(\frac{-1}{2}\) (x + 9)
⇒ 2y – 12 = -x – 9
⇒ x + 2y – 3 = 0

Question 2.
If Z = x + iy and if the point P in the Argand plane represents Z, then describe geometrically the locus of z satisfying the equation.
(i) |z – 2 – 3i| = 5
(ii) 2|z – 2| = |z – 1|
(iii) Img z² = 4
(iv) \({Arg}\left(\frac{z-1}{z+1}\right)=\frac{\pi}{4}\)
Solution:
(i) z = x + iy and |z – 2 – 3i| = 5
|z – 2 – 3i| = 5
⇒ |x + iy – 2 – 3i| = 5
⇒ |(x – 2) + i(y – 3)| = 5
⇒ \(\sqrt{(x-2)^{2}+(y-3)^{2}}\) = 5
⇒ (x – 2)² + (y – 3)² = 25
⇒ x² – 4x + 4 + y² – 6y + 9 = 25
∴ Locus of P is x² + y² – 4x – 6y – 12 = 0

(ii) 2|z – 2| = |z – 1|
2|x + iy – 2| = |x + iy – 1|
⇒ 2|(x – 2) + iy| = |(x – 1) + iy|
⇒ \(2 \sqrt{(x-2)^{2}+y^{2}}=\sqrt{(x-1)^{2}+y^{2}}\)
Squaring both sides
⇒ 4[(x – 2)² + y²] = (x – 1)² + y²
⇒ 4(x² – 4x + 4 + y²) = x² – 2x + 1 + y²
⇒ 4x² + 4y² – 16x + 16 = x² + y² – 2x + 1
∴ Locus of P is 3x² + 3y² – 14x + 15 = 0

(iii) Img z² = 4
∵ z = x + iy
⇒ z² = (x + iy)²
⇒ z² = x² + i²y² + 2ixy
⇒ z² = (x² – y²) + i(2xy)
∴ Img (z²) = 2xy = 4
∴ The locus of P is xy = 2

(iv) \({Arg}\left(\frac{z-1}{z+1}\right)=\frac{\pi}{4}\)

Since \({Arg}\left(\frac{z-1}{z+1}\right)=\frac{\pi}{4}\)
∴ \(\frac{2 y}{x^{2}+y^{2}-1}=\tan \frac{\pi}{4}\)
⇒ 2y = x² + y² – 1
⇒ x² + y² – 2y – 1 = 0
∴ The locus of z is x² + y² – 2y – 1 = 0.

Question 3.
Show that the points in the Argand diagram represented by the complex numbers 2 + 2i, -2 – 2i, 2√3 + 2√3 i are the vertices of an equilateral triangle.
Solution:
A(2, 2), B(-2, -2), C(-2√3, 2√3) represent the given complex number in the Argand diagram.

AB² = (2 + 2)² + (2 + 2)²
= 16 + 16
= 32
BC² = (-2 + 2√3)² + (-2 – 2√3)²
= 4 + 12 – 8√3 + 4 + 12 + 8√3
= 32
AC² = (-2√3 – 2)² + (2√3 – 2)²
= 12 + 4 + 8√3 + 12 + 4 – 8√3
= 32
AB² = BC² = AC²
⇒ AB = BC = CA
∴ ∆ABC is an Equilateral triangle.

Question 4.
Find the eccentricity of the ellipse whose equation is |z – 4| + |z – \(\frac{12}{5}\)| = 10
Solution:
SP + S’P = 2a
S = (4, 0), S’ = (\(\frac{12}{5}\), 0)
2a = 10 ⇒ a = 5
SS’ = 2ae
⇒ 4 – \(\frac{12}{5}\) = 2 × 5e
⇒ \(\frac{8}{5}\) = 10e
⇒ e = \(\frac{4}{25}\)

II.

Question 1.
If \(\frac{z_{3}-z_{1}}{z_{2}-z_{1}}\) is a real number, show that the points represented by the complex numbers z₁, z₂, z₃ are collinear.
Solution:
Let z₁ = x₁ + iy₁; z₂ = x₂ + iy₂; z₃ = x₃ + iy₃

Given \(\frac{z_{3}-z_{1}}{z_{2}-z_{1}}\) is a real number.
Imaginary part = 0
⇒ (y₃ – y₁) (x₂ – x₁) – (x₃ – x₁) (y₂ – y₁) = 0
⇒ (y₃ – y₁) (x₂ – x₁) = (x₃ – x₁) (y₂ – y₁)
⇒ \(\frac{y_{3}-y_{1}}{x_{3}-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
The points A(x₁, y₁), B(x₂, y₂), C(x₃, y₃) represents the complex numbers z₁, z₂, z₃ respectively.
Slope of \(\stackrel{\leftrightarrow}{\mathrm{AC}}\) = Slope of \(\stackrel{\leftrightarrow}{\mathrm{AB}}\)
∴ A, B, C are collinear.

Question 2.
Show that the points in the Argand plane represented by the complex numbers 2 + i, 4 + 3i, 2 + 5i, 3i are the vertices of a square.
Solution:
A(2, 1), B(4, 3) C(2, 5), D(0, 3) represent the given complex number in the Argand plane

AB² = (2 – 4)² + (1 – 3)² = 4 + 4 = 8
BC² = (4 – 2)² + (3 – 5)² = 4 + 4 = 8
CD² = (2 – 0)² + (5 – 3)² = 4 + 4 = 8
DA² = (0 – 2)² + (3 – 1)² = 4 + 4 = 8
AB² = BC² = CD² = DA²
⇒ AB = BC = CD = DA ………(1)
AC² = (2 – 2)² + (1 – 5)² = 0 + 16 = 16
BD² = (4 – 0)² + (3 – 3)² = 16 + 0 = 16
AC² = BD²
⇒ AC = BD …….(2)
By (1), (2)
A, B, C, D are the vertices of a square.

Question 3.
Show that the points in the Argand plane represented by the complex numbers -2 + 7i, \(-\frac{3}{2}+\frac{1}{2} i\), 4 – 3i, \(\frac{7}{2}\)(1 + i) are the vertices of a rhombus.
Solution:
A(-2, 7), B(\(-\frac{3}{2}\), \(\frac{1}{2}\)), C(4, -3), D(\(\frac{7}{2}\), \(\frac{7}{2}\)) represents the given complex numbers in the Argand diagram.

∴ AB² = BC² = CD² = DA²
⇒ AB = BC = CD = DA ……….(1)
AC² = (-2 – 4)² + (7 + 3)²
= 36 + 100
= 136
BD² = \(\left(-\frac{3}{2}-\frac{7}{2}\right)^{2}+\left(\frac{1}{2}-\frac{7}{2}\right)^{2}\)
= 25 + 9
= 34
AC ≠ BD ……..(2)
∴ A, B, C, D are the vertices of a Rhombus.

Question 4.
Show that the points in the Argand diagram represented by the complex numbers z₁, z₂, z₃ are collinear, if and only if there exist three real numbers p, q, r not all zero, satisfying pz₁ + qz₂ + rz₃ = 0 and p + q + r = 0.
Solution:
pz₁ + qz₂ + rz₃ = 0
⇔ rz₃ = -pz₁ – qz₂
⇔ z₃ = \(\frac{-p z_{1}-q z_{2}}{r}\) [∵ r ≠ 0]
∵ p + q + r = 0
⇔ r = -p – q
⇔ z₃ = \(-\frac{\left(p z_{1}+q z_{2}\right)}{-(p+q)}\)
⇔ z₃ = \(\frac{p z_{1}+q z_{2}}{p+q}\)
⇔ z₃ divides the line segment joining z₁, z₂ in the ratio q : p
⇔ z₁, z₂, z₃ are collinear

Question 5.
The points P, Q denotes the complex numbers z₁, z₂ in the Argand diagram. O is the origin. If \(\bar{z}_{1} \bar{z}_{2}+\bar{z}_{2} \bar{z}_{1}=0\), show that POQ = 90°.
Solution:
Let z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂
Then P(x₁, y₁), Q(x₂, y₂), O(0, 0)
\(\overline{\mathbf{z}}_{1}\) = x1 – iy1, \(\overline{\mathbf{z}}_{2}\) = x₂ – iy₂
\(\bar{z}_{1} \bar{z}_{2}+\bar{z}_{2} \bar{z}_{1}\) = (x₁ + iy₁) (x₂ – iy₂) + (x₂ + iy₂) (x₁ – iy₁) = 0
⇒ x₁x₂ + y₁y₂ – ix₁y₂ + ix₂y₁ + x₁x₂ + y₁y₂ – ix₂y₁ + ix₁y₂ = 0
⇒ 2(x₁x₂ + y₁y₂) = 0
⇒ x₁x₂ + y₁y₂ = 0
⇒ x₁x₂ = -y₁y₂
⇒ \(\left(\frac{-x_{1}}{y_{1}}\right)\left(-\frac{x_{2}}{y_{2}}\right)=-1\)
Slope of OP × Slope of OQ = -1
∴ OP, OQ are perpendicular
⇒ ∠POQ = 90°

Question 6.
The complex number z has argument θ, 0 < θ < \(\frac{\pi}{2}\) and satisfy the equation |z – 3i| = 3. Then prove that (cot θ – \(\frac{6}{z}\)) = i
Solution:
Let z = cos θ + i sin θ
Given |z – 3i| = 3.
⇒ |(cos θ + i sin θ) – 3i| = 3
⇒ |cos θ + i(sin θ – 3)| = 3
⇒ \(\sqrt{\cos ^{2} \theta+(\sin \theta-3)^{2}}\) = 3
⇒ cos²θ + sin²θ – 6 sin θ + 9 = 9
⇒ 1 – 6 sin θ = 0
⇒ 6 sin θ = 1
⇒ sin θ = \(\frac{1}{6}\)
since 0 < θ < \(\frac{\pi}{2}\)

∴ cos θ = \(\frac{\sqrt{35}}{6}\)
cot θ = √35
∴ cot θ – \(\frac{6}{z}\) = cot θ – \(\frac{6}{\cos \theta+i \sin \theta}\)
= cot θ – 6(cos θ – i sin θ)
= √35 – 6\(\left[\frac{\sqrt{35}}{6}-i \cdot \frac{1}{6}\right]\)
= √35 – √35 + i
= i
Hence cot θ – \(\frac{6}{z}\) = i

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Complex Numbers Solutions Exercise 1(c) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Complex Numbers Solutions Exercise 1(c)

I.

Question 1.
Express the following complex numbers in modulus-amplitude form.
(i) 1 – i
Solution:
1 – i
Let 1 – i = r(cos θ + i sin θ)
Equating real and imaginary parts
r cos θ = 1, r sin θ = -1
⇒ θ lies in IV quadrant
Squaring and adding
r² (cos²θ + sin²θ) = 1 + 1 = 2
⇒ r² = 2
⇒ r = √2
tan θ = – 1
⇒ θ = \(-\frac{\pi}{4}\)
∴ 1 – i = √2(cos(\(-\frac{\pi}{4}\)) + i sin(\(-\frac{\pi}{4}\)))

(ii) 1 + i√3
Solution:
1 + i√3 = r (cos θ + i sin θ)
Equating real and imaginary parts
r cos θ = 1 ………(1)
r sin θ = √3 ……..(2)
θ lies in the I quadrant
Squaring and adding (1) and (2)
r² (cos²θ + sin²θ) = 1 + 3
⇒ r² = 4
⇒ r = 2
Dividing (2) by (1)
\(\frac{r \sin \theta}{r \cos \theta}\) = √3
⇒ tan θ = √3
⇒ θ = \(\frac{\pi}{3}\)
∴ 1 + i√3 = 2(cos \(\frac{\pi}{3}\) + i sin \(\frac{\pi}{3}\))

(iii) -√3 + i
Solution:
-√3 + i = r(cos θ + i sin θ)
Equating real and imaginary parts
r cos θ = -√3 …….(1)
r sin θ = 1 …….(2)
⇒ θ lies in II quadrant
Squaring and adding (1), (2)
r² (cos²θ + sin²θ) = 3 + 1 = 4
⇒ r² = 4
⇒ r = 2
Dividing (2) by (1)
\(\frac{r \sin \theta}{r \cos \theta}=-\frac{1}{\sqrt{3}}\)
⇒ tan θ = \(-\frac{1}{\sqrt{3}}\) and θ lies in II quadrant
⇒ θ = 180° – 30° = 150° = \(\frac{5 \pi}{6}\)
∴ -√3 + i = 2(cos \(\frac{5 \pi}{6}\) + i sin \(\frac{5 \pi}{6}\))

(iv) -1 – i√3
Solution:
1 – i√3 = r (cos θ + i sin θ)
Equating real and imaginary parts
r cos θ = -1, r sin θ = -√3
Squaring and adding
r² (cos²θ + sin²θ) = 1 + 3
⇒ r² = 4
⇒ r = 2
∴ cos θ = \(-\frac{1}{2}\), sin θ = \(-\frac{\sqrt{3}}{2}\)
⇒ θ = -120° = \(-\frac{2 \pi}{3}\)
∴ -1 – i√3 = 2[cos(\(-\frac{2 \pi}{3}\)) + i sin(\(-\frac{2 \pi}{3}\))]

Question 2.
Simplify -2i(3 + i) (2 + 4i) (1 + i) and obtain the modulus of that complex number.
Solution:
-2i(3 + i) (2 + 4i) (1 + i)
= (-6i – 2i²) (2 + 2i + 4i + 4i²)
= (2 – 6i) (-2 + 6i)
= -4 + 12i – 36i² + 12i
= 32 + 24i
= 8(4 + 3i)
Modulus = |8(4 + 3i)|
= 8\(\sqrt{(4)^{2}+(3)^{2}}\)
= 8(5)
= 40

Question 3.
(i) If z ≠ 0, find Arg z + Arg \((\bar{z})\).
Solution:
If z = x + iy, then Arg(z) = \(\tan ^{-1}\left(\frac{y}{x}\right)\)
and \((\bar{z})\) = x – iy, then Arg (z) = \(\tan ^{-1}\left(\frac{-y}{x}\right)\) = –\(\tan ^{-1}\left(\frac{y}{x}\right)\)
∴ Arg (z) + Arg \((\bar{z})\) = \(\tan ^{-1}\left(\frac{y}{x}\right)\) + (-\(\tan ^{-1}\left(\frac{y}{x}\right)\)) = 0

(ii) If z₁ = -1 and z₂ = -i, then find Arg(z₁z₂)
Solution:
z₁ = -1
⇒ z₁ = cos π + i sin π
⇒ Arg z₁ = π
z₂ = -i
⇒ z₂ = \(\cos \left(-\frac{\pi}{2}\right)+i \sin \left(-\frac{\pi}{2}\right)\)
⇒ Arg z₂ = \(-\frac{\pi}{2}\)
∴ Arg (z₁z₂) = Arg z₁ + Arg z₂
= π – \(\frac{\pi}{2}\)
= \(\frac{\pi}{2}\)

(iii) If z₁ = -1, z₂ = i then find Arg\(\left(\frac{\mathbf{z}_{1}}{\mathbf{z}_{2}}\right)\)
Solution:
z₁ = -1 = cos π + i sin π
⇒ Arg z₁ = π
z₂ = i = cos \(\frac{\pi}{2}\) + i sin \(\frac{\pi}{2}\)
⇒ Arg z₂ = \(\frac{\pi}{2}\)
∴ Arg\(\left(\frac{\mathbf{z}_{1}}{\mathbf{z}_{2}}\right)\) = Arg z₁ – Arg z₂
= π – \(\frac{\pi}{2}\)
= \(\frac{\pi}{2}\)

Question 4.
(i) If (cos 2α + i sin 2α) (cos 2β + i sin 2β) = cos θ + i sin θ, then find the value of θ.
Solution:
cos θ + i sin θ
= (cos 2α + i sin 2α) (cos 2β + i sin 2β)
= cos 2α . cos 2β + i sin 2α . cos 2β + i cos 2α sin 2β + i² sin 2α . sin 2β
= (cos 2α . cos 2β – sin 2α . sin 2β) + i(sin 2α cos 2β + cos 2α sin 2β)
= cos 2(α + β) + i sin 2(α + β)
∴ θ = 2(α + β)

(ii) If √3 + i = r (cos θ + i sin θ), then find the value of θ in radian measure.
Solution:
Given that
√3 + i = r(cos θ + i sin θ)
⇒ r cos θ = √3, r sin θ = 1
⇒ r² (cos²θ + sin²θ) = 3 + 1
⇒ r² = 4
⇒ r = 2
∴ cos θ = \(\frac{\sqrt{3}}{2}\) and sin θ = \(\frac{1}{2}\)
⇒ θ = \(\frac{\pi}{6}\)

(iii) If x + iy = cis α . cis β then find the value of x² + y².
Solution:
Given x + iy = cis α . cis β
⇒ x + iy = (cos α + i sin α) (cos β + i sin β)
⇒ x + iy = cos(α + β) + i sin(α + β)
Equating real and imaginary parts
x = cos(α + β) and y = sin(α + β)
∴ x² + y² = cos²(α + β) + sin²(α + β) = 1

(iv) If \(\frac{z_{2}}{z_{1}}\), (z₁ ≠ 0) is an imaginary number, then find the value of \(\left|\frac{2 z_{1}+z_{2}}{2 z_{1}-z_{2}}\right|\)
Solution:
\(\frac{z_{2}}{z_{1}}\), (z₁ ≠ 0) is purely imaginary.
We can suppose that \(\frac{z_{2}}{z_{1}}\) = iy, where y ∈ R – {0}

(v) If (√3 + i)¹⁰⁰ = 2⁹⁹ (a + ib), then show that a² + b² = 4.
Solution:

Question 5.
(i) If z = x + iy and |z| = 1, find the locus of z.
Solution:
Given z = x + iy
Also |z| = 1
⇒ |x + iy| = 1
⇒ \(\sqrt{x^{2}+y^{2}}=1\)
⇒ x² + y² = 1
∴ Locus of z is x² + y² = 1

(ii) If the amplitude of (z – 1) is \(\frac{\pi}{2}\), then find the locus of z.
Solution:
Given z = x + iy
z – 1 = x + iy – 1 = (x – 1) + iy
Since Amplitude (z – 1) is \(\frac{\pi}{2}\)
Real part of z – 1 is zero.
∴ x – 1 = 0
∴ Locus of z is x – 1 = 0

(iii) If the Arg \(\overline{\mathbf{z}}_{1}\) and Arg z₂ are \(\frac{\pi}{5}\) and \(\frac{\pi}{3}\) respectively, find (Arg z₁ + Arg z₂)
Solution:
Arg \(\overline{\mathbf{z}}_{1}\) = \(\frac{\pi}{5}\)
⇒ Arg \(\overline{\mathbf{z}}_{1}\) = -Arg z₁ = \(-\frac{\pi}{5}\)
Arg z₂ = \(\frac{\pi}{3}\)
∴ Arg z₁ + Arg z₂ = \(-\frac{\pi}{5}+\frac{\pi}{3}\) = \(\frac{2 \pi}{15}\)

(iv) If z = \(\frac{1+2 i}{1-(1-i)^{2}}\) then find Arg(z).
Solution:
Given z = \(\frac{1+2 i}{1-(1-i)^{2}}\)
= \(\frac{1+2 i}{1-(1-2 i+i)^{2}}\)
= \(\frac{1+2 i}{1+2 i}\)
= 1
= cos 0° + i sin 0°
∴ Arg (z) = 0°

II.

Question 1.
Simplify the following complex numbers and find their modulus.
(i) \(\frac{(2+4 i)(-1+2 i)}{(-1-i)(3-i)}\)
(ii) \(\frac{(1+i)^{3}}{(2+i)(1+2 i)}\)
Solution:

Question 2.
(i) If (1 – i) (2 – i) ( 3 – i) …… (1 – ni) = x – iy then prove that 2 . 5 . 10 …….. (1 + n²) = x² + y².
Solution:
Given (1 – i) (2 – i) (3 – i) (1 – ni) = x – iy
|1 – i| |2 – i| |3 – i| …… |1 – ni| = |x – iy|
\(\sqrt{1+1} \sqrt{4+1} \sqrt{9+1} \ldots \sqrt{1+n^{2}}=\sqrt{x^{2}+y^{2}}\)
∴ 2 . 5 . 10 ……. (1 + n²) = x² + y²

(ii) If the real part of \(\frac{z+1}{z+i}\) is 1, then find the locus of z.
Solution:

(iii) If |z – 3 + i | = 4 determine the locus of z.
Solution:
Let z = x + iy
Given |z – 3 + i| = 4
⇒ |x + iy – 3 + i | = 4
⇒ (x – 3) + i(y + 1) = 4
⇒ \(\sqrt{(x-3)^{2}+(y+1)^{2}}\) = 4
⇒ (x – 3)² + (y + 1)² = 16
⇒ x² – 6x + 9 + y² + 2y + 1 = 16
⇒ x² + y² – 6x + 2y – 6 = 0
∴ The locus of z is x² + y² – 6x + 2y – 6 = 0

(iv) If |z + ai| = |z – ai| then find the locus of z.
Solution:
Let z = x + iy
Given |z + ai | = |z – ai|
⇒ |x + iy + ai| = |x + iy – ai|
⇒ |x + i(y + a)| = |x + i(y – a)|
⇒ \(\sqrt{x^{2}+(y+a)^{2}}=\sqrt{x^{2}+(y-a)^{2}}\)
⇒ x² + (y + a)² = x² + (y – a)²
⇒ (y + a)² – (y – a)² = 0
⇒ 4ay = 0
⇒ y = 0
∴ The locus of z is y = 0

Question 3.
If z = (x + iy) and if the point P in the Argand plane represents z, then describe geometrically the locus of P satisfying the equation.
(i) |2z – 3| = 7
Solution:
|2z – 3| = 7
Let z = x + iy
Given |2z – 3| = 7
⇒ |2(x + iy) – 3| = 7
⇒ |(2x – 3) + i(2y)| = 7
⇒ \(\sqrt{(2 x-3)^{2}+(2 y)^{2}}\) = 7
⇒ 4x² – 12x + 9 + 4y² = 49
⇒ 4x² + 4y² – 12x – 40 = 0
⇒ x² + y² – 3x – 10 = 0
∴ The locus of z is x² + y² – 3x – 10 = 0
This equation represents a circle.
Centre = (\(\frac{3}{2}\), 0)
Radius = \(\sqrt{\frac{9}{4}-(-10)}\) = \(\frac{7}{2}\) units

(ii) |z|² = 4 Re(z + 2)
Solution:
|z|² = 4 Re (z + 2)
Given |z|² = 4 Re (z + 2)
⇒ |x + iy|² = 4 Re (x + iy + z)
⇒ (\(\sqrt{x^{2}+y^{2}}\)) = 4(x + 2)
⇒ x² + y² = 4x + 8
∴ The locus of z is x² + y² – 4x – 8 = 0
This equation represents a circle. Center = (2, 0)
Radius = \(\sqrt{4-(-8)}\) = 2√3 units.

(iii) |z + i|² – |z – i|² = 2
Solution:
Given |z + i|² – |z –  i|² = 2
⇒ |x + iy + i|² – |x + iy – i|² = 2
⇒ |x + i(y + 1)|² – |x + i(y – 1)|² = 2
⇒ \(\left[\sqrt{x^{2}+(y+1)^{2}}\right]^{2}\) – \(\left[\sqrt{x^{2}+(y-1)^{2}}\right]^{2}\) = 2
⇒ x² + (y + 1 )² – [x² + (y – 1)²] = 2
⇒ x² + (y + 1)² – x² – (y – 1)² = 2
⇒ 4y = 2
⇒ 2y – 1 = 0
The locus of P is 2y – 1 = 0
This equation represents a straight line parallel to the x-axis.

(iv) |z + 4i| + |z – 4i| = 10
Solution:
Given |z + 4i| + |z – 4i| = 10
⇒ |x + iy + 4i| + |x + iy – 4i| = 10
⇒ |x + i(y + 4)| + |x + i(y – 4)| = 10
⇒ \(\sqrt{x^{2}+(y+4)^{2}}+\sqrt{x^{2}+(y-4)^{2}}\)
⇒ \(\sqrt{x^{2}+(y+4)^{2}}\) = 10 – \(\sqrt{x^{2}+(y-4)^{2}}\)
⇒ x² + (y + 4)² = 100 + x² + (y – 4)² – 20 \(\sqrt{x^{2}+(y-4)^{2}}\)
⇒ 16y – 100 = – 20\(\sqrt{x^{2}+(y-4)^{2}}\)
⇒ 4y – 25 = -5\(\sqrt{x^{2}+(y-4)^{2}}\)
⇒ (4y – 25)² = 25[x² + (y – 4)²]
⇒ 16y² – 200y + 625 = 25x² + 25y² – 200y + 400
⇒ 25x² + 9y² = 225
∴ The locus of P is 25x² + 9y² = 225
⇒ \(\frac{x^{2}}{9}+\frac{y^{2}}{25}\) = 1
This equation represents an ellipse.
Center = (0, 0)
a² = 9, b² = 25
We know a² – b²(1 – e²)
⇒ 9 = 25(1 – e²)
⇒ 1 – e² = \(\frac{9}{25}\)
⇒ e = 1 – \(\frac{9}{25}\) = \(\frac{16}{25}\)
∴ Eccentricity = \(\frac{4}{5}\) and major axis parallel to y-axis.

Question 4.
(i) If z₁, z₂ are two non-zero complex numbers satisfying |z₁ + z₂| = |z₁| + |z₂|, show that Arg z₁ – Arg z₂ = 0.
Solution:

(ii) z = x + iy and the point P represents z in the Argand plane and \(\left|\frac{z-a}{z+\bar{a}}\right|\) = 1, Re(a) ≠ 0, then find the locus of P.
Solution:
Let z = x + iy and a = α + iβ
\(\frac{z-a}{z+\bar{a}}=\frac{(x+i y)-(\alpha+i \beta)}{(x+i y)+(\alpha-i \beta)}\)
\(\left|\frac{z-a}{z+\bar{a}}\right|\) = 1
⇒ \(\frac{|z-a|}{|z+\bar{a}|}\) = 1
⇒ |z – a| = |z + \(\overline{\mathrm{a}}\)|
⇒ |(x – α)| + i(y – β)| = |(x + α) + i(y – β)|
⇒ \(\sqrt{(x-\alpha)^{2}+(y-\beta)^{2}}=\sqrt{(x+\alpha)^{2}+(y-\beta)^{2}}\)
⇒ (x – α)² + (y – β)² = (x + α)² + (y – β)²
⇒ (x – α)² = (x + α)²
⇒ (x + α)² – (x – α)² = 0
⇒ 4αx = 0
Re(a) ≠ 0
⇒ a ≠ 0
⇒ x = 0
Locus of P is x = 0 i.e., Y-axis

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Complex Numbers Solutions Exercise 1(b) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Complex Numbers Solutions Exercise 1(b)

I.

Question 1.
Write the following complex numbers in the form A + iB.
(i) (2 – 3i) (3 + 4i)
(ii) (1 + 2i)³
(iii) \(\frac{a-i b}{a+i b}\)
(iv) \(\frac{4+3 i}{(2+3 i)(4-3 i)}\)
(v) (-√3 + √-2) (2√3 – i)
(vi) (-5i) \(\left(\frac{i}{8}\right)\)
(vii) (-i) 2i
(viii) i⁹
(ix) i^-19
(x) 3(7 + 7i) + i(7 + 7i)
(xi) \(\frac{2+5 i}{3-2 i}+\frac{2-5 i}{3+2 i}\)
Solution:
(i) (2 – 3i) (3 + 4i) = 6 + 8i – 9i – 12i²
= 6 – i + 12
= 18 – i
= 18 + i(-1)

(ii) (1 + 2i)³ = 1 + 3 . i² . 2i + 3 . 1 . 4i² + 8i³
= 1 + 6i – 12 – 8i
= -11 – 2i
= (-11) + i(-2)

(v) (-√3 + √-2) (2√3 – i) = (-√3 + i√2) (2√3 – i)
= -6 + i√3 + i2√6 + √2
= (-6 + √2) + i(√3 + 2√6)

(vi) (-5i) \(\left(\frac{i}{8}\right)\) = \(\frac{-5 i^{2}}{8}\)
= \(\frac{5}{8}\) + i(0)

(vii) (-i)(2i) = -2i²
= -2(-1)
= 2
= 2 + i(0)

(viii) i⁹ = i⁴ . i⁴ . i
= 1 . 1 . i
= i
= 0 + i(1)

(ix) i^-19 = \(\frac{1}{i^{19}}\)
= \(\frac{i}{i^{20}}\)
= \(\frac{i}{\left(i^{4}\right)^{5}}\)
= \(\frac{\mathrm{i}}{\mathrm{i}^{5}}\)
= i
= 0 + i(1)

(x) 3(7 + 7i) + i(7 + 7i)
= 21 + 21i + 7i – 7
= 14 + 28i

Question 2.
Write the conjugate of the following complex numbers.
(i) 3 + 4i
(ii) (15 + 3i) – (4 – 20i)
(iii) (2 + 5i) (-4 + 6i)
(iv) \(\frac{5 i}{7+i}\)
Solution:
(i) Let z = 3 + 4i
\(\bar{z}\) = 3 – 4i

(ii) Let z = (15 + 3i) – (4 – 20i)
= 15 + 3i – 4 + 20i
= 11 + 23i
\(\bar{z}\) = 11 – 23i

(iii) Let z = (2 + 5i) (-4 + 6i)
= -8 + 12i – 20i – 30
= -38 – 8i
\(\bar{z}\) = -38 + 8i

Question 3.
Simplify
(i) i² + i⁴ + i⁶ + …….. + (2n + 1) terms
(ii) i¹⁸ – 3 . i⁷ + i² (1 + i⁴) (-i)²⁶
Solution:
(i) i² + i⁴ + i⁶ + …….. + (2n + 1) terms
= -1 + 1 – 1 + (2n + 1) terms
= -1

(ii) i¹⁸ – 3i² + i² (1 + i⁴) (-i)²⁶
= i¹⁶ . i² – 3 . i⁴ . i³ + i² (1 + 1) i²⁴ . i²
= 1 . (-1) – 3 . 1 . (-i) + (-1) (2) (1) (-1)
= -1 + 3i + 2
= 1 + 3i

Question 4.
Find a square root for the following complex numbers.
(i) 7 + 24i
(ii) -8 – 6i
(iii) (3 + 4i)
(iv) (-47 + i8√3)
Solution:

Question 5.
Find the multiplicative inverse of the following complex numbers.
(i) √5 + 3i
(ii) -i
(iii) i^-35
Solution:
(i) √5 + 3i
The multiplicative inverse of x + iy is \(\frac{x-i y}{x^{2}+y^{2}}\)

II.

Question 1.
(i) If (a + ib)² = x + iy, find x² + y²
Solution:
(a + ib)² = x + iy
⇒ a² + i(2ab) – b² = x + iy
⇒ (a² – b²) + i(2ab) = x + iy
⇒ (a² – b²) + i(2ab) = x + iy
Equating real and imaginary parts on both sides, we have
x = a² – b² and y = 2ab
x² + y² = (a² – b²)² + (2ab)²
= a⁴ – 2a²b² + b⁴ + 4a²b²
= a⁴ + 2a²b² + b⁴
= (a² + b²)²

(ii) If x + iy = \(\frac{3}{2+\cos \theta+i \sin \theta}\) then show that x² + y² = 4x – 3
Solution:
x + iy = \(\frac{3}{2+\cos \theta+i \sin \theta}\) \(\frac{(2+\cos \theta)-i \sin \theta}{(2+\cos \theta)-i \sin \theta}\)

(iii) If x + iy = \(\frac{1}{1+\cos \theta+i \sin \theta}\), show that 4x² – 1 = 0.
Solution:

Equating real parts on both sides, we have
x = \(\frac{1}{2}\)
⇒ 2x = 1
⇒ 4x² = 1
⇒ 4x² – 1= 0

(iv) If u + iv = \(\frac{2+i}{z+3}\) and z = x + iy, find u, v.
Solution:

Question 2.
(i) If z = 3 – 5i Show that z³ – 10z² + 58z – 136 = 0.
Solution:
Given z = 3 – 5i
⇒ z – 3 = -5i
⇒ (z – 3)² = 25i²
⇒ z² – 6z + 9 = -25
⇒ z² – 6z + 34 = 0
∴ z³ – 10z² + 58z – 136 = z(z² – 6z + 34) – 4z² + 24z – 136
= z(0) – 4(z² – 6z + 34)
= 0 – 4(0)
= 0
∴ z³ – 10z² + 58z – 136 = 0

(ii) If z = 2 – i√7 , then show that 3z³ – 4z² + z + 88 = 0.
Solution:
Given z = 2 – i√7
⇒ z – 2 = -i√7
⇒ (z – 2)² = (-i√7)²
⇒ z² – 4z + 4 = 7i²
⇒ z² – 4z + 4 = -7
⇒ z² – 4z + 11 = 0
∴ 3z³ – 4z² + z + 88 = 3z(z² – 4z + 11) + 8z² – 32z + 88
= 3z(0) + 8(z² – 4z + 11)
= 0 + 8(0)
= 0
∴ 3z³ – 4z² + z + 88 =0

(iii) Show that \(\frac{2-i}{(1-2 i)^{2}}\) and \(\left(\frac{-2-11 i}{25}\right)\) conjugate to each other.
Solution:

Question 3.
(i) If (x – iy)^1/3 = a – ib then show that \(\frac{x}{a}+\frac{y}{b}\) = 4(a² – b²)
Solution:
Given (x – iy)^1/3 = (a – ib)
⇒ x – iy = (a – ib)³
⇒ x – iy = a³ – 3a²ib + 3ai²b² – i³b³
⇒ x – iy = (a³ – 3ab²) – i(3a²b – b³)
Equating real and imaginary parts
x = a³ – 3ab²
⇒ \(\frac{x}{a}\) = a² – 3b²
y = 3a²b – b³
⇒ \(\frac{y}{b}\) = 3a² – b²
∴ \(\frac{x}{a}+\frac{y}{b}\) = a² – 3b² + 3a² – b²
= 4a² – 4b²
= 4(a² – b²)
∴ \(\frac{x}{a}+\frac{y}{b}\) = 4(a² – b²)

(ii) Write \(\left(\frac{a+i b}{a-i b}\right)^{2}-\left(\frac{a-i b}{a+i b}\right)^{2}\) in the form of x + iy.
Solution:

(iii) If x and y are real numbers such that \(\frac{(1+i) x-2 i}{3+i}+\frac{(2-3 i) y+i}{3-1}=i\), then determine the values of x and y.
Solution:

Question 4.
(i) Find the least positive integer n, satisfying \(\left(\frac{1+i}{1-i}\right)^{n}\) = 1
Solution:

(ii) If \(\left(\frac{1+i}{1-i}\right)^{3}-\left(\frac{1-i}{1+i}\right)^{3}\) = x + iy, find x and y.
Solution:

(iii) Find real values of ‘θ’ in order that \(\frac{3+2 i \sin \theta}{1-2 i \sin \theta}\) is a
(a) real numbers
(b) Purely imaginary number
Solution:

(iv) Find the real values of x and y if \(\frac{x-1}{3+i}+\frac{y-1}{3-i}=i\)
Solution:
Given \(\frac{x-1}{3+i}+\frac{y-1}{3-i}=i\)
⇒ \(\frac{(x-1)(3-i)+(y-1)(3+i)}{9-i^{2}}=i\)
⇒ 3x – xi – 3 + i + 3y – iy – 3 – i = 10i
⇒ (3x + 3y – 6) + i(-x + y) = 0 + 10i
Now equating real and imaginary parts
3x + 3y – 6 = 0
⇒ x + y – 2 = 0 ……..(1)
-x + y = 10
⇒ x – y + 10 = 0 ………(2)
(1) + (2) ⇒ 2x + 8 = 0
⇒ x = -4
From (1),
-4 + y – 2 = 0
⇒ y = 6
∴ x = -4, y = 6

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Complex Numbers Solutions Exercise 1(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Complex Numbers Solutions Exercise 1(a)

I.

Question 1.
If z₁ = (2, -1), z₂ = (6, 3), find z₁ – z₂.
Solution:
z₁ = (2, -1), z₂ = (6, 3)
∴ z₁ – z₂ = (2 – 6, -1 – 3) = (-4, -4)

Question 2.
If z₁ = (3, 5) and z₂ = (2, 6), find z₁ . z₂
Solution:
Given z₁ = (3, 5) = 3 + 5i
and z₂ = (2, 6) = 2 + 6i
z₁ . z₂ = (3 + 5i) . (2 + 6i)
= 6 + 10i + 18i + 30i²
= 6 + 28i + 30(-1) [since i² = -1]
= -24 + 28i
= (-24, 28)

Question 3.
Write the additive inverse of the following complex numbers.
(i) (√3, 5)
(ii) (-6, 5) + (10, -4)
(iii) (2, 1) (-4, 6)
Solution:
The additive inverse of (a, b) is (-a, -b)
(i) The additive inverse of (√3, 5) is (-√3, -5)
(ii) (-6, 5) + (10, -4)
= (-6 + 10, 5 + (-4))
= (4, 1)
∴ The additive inverse of (4, 1) is (-4, -1)
(iii) (2, 1) . (-4, 6)
= ((2 × -4 – 1 × 6), (1 × -4 + 2 × 6))
= (-8 – 6, -4 + 12)
= (-14, 8)
∴ The additive inverse of (-14, 8) is (14, -8)

II.

Question 1.
If z₁ = (6, 3); z₂ = (2, -1), find z₁/z₂.
Solution:
Given z₁ = (6, 3) = 6 + 3i
and z₂ = (2, -1) = 2 – i

Question 2.
If z = (cos θ, sin θ), find (z – \(\frac{1}{z}\))
Solution:
Given z = (cos θ, sin θ) = cos θ + i sin θ
⇒ \(\frac{1}{z}\) = cos θ – i sin θ
∴ z – \(\frac{1}{z}\) = (cos θ + i sin θ) – (cos θ – i sin θ)
= 2 i sin θ
= 0 + i (2 sin θ)
= (0, 2 sin θ)

Question 3.
Write the multiplicative inverse of the following complex numbers.
(i) (3, 4)
(ii) (sin θ, cos θ)
(iii) (7, 24)
(iv) (-2, 1)
Solution:
The multiplicative inverse of the complex number (a, b) is \(\left(\frac{a}{a^{2}+b^{2}}, \frac{-b}{a^{2}+b^{2}}\right)\)
(i) Multiplicative inverse of (3, 4) = \(\left(\frac{3}{3^{2}+4^{2}} \cdot \frac{-4}{3^{2}+4^{2}}\right)\) = \(\left(\frac{3}{25}, \frac{-4}{25}\right)\)
(ii) Multiplicative inverse of (sin θ, cos θ) = \(\left(\frac{\sin \theta}{\sin ^{2} \theta+\cos ^{2} \theta}, \frac{-\cos \theta}{\sin ^{2} \theta+\cos ^{2} \theta}\right)\) = (sin θ, -cos θ)
(iii) Multiplicative inverse of (7, 24) = \(\left(\frac{7}{7^{2}+24^{2}}, \frac{-24}{7^{2}+24^{2}}\right)\) = \(\left(\frac{7}{625}, \frac{-24}{625}\right)\)
(iv) Multiplicative inverse of (-2, 1) = \(\left(\frac{-2}{(-2)^{2}+(1)^{2}}, \frac{-1}{(-2)^{2}+(1)^{2}}\right)\) = \(\left(-\frac{2}{5},-\frac{1}{5}\right)\)