Class 11: Complex Numbers – Exercise 13.2 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Express the following complex numbers in the form $a + ib$

i) $(1+i)(1+2i)$ ii) $\frac{3+2i}{-2+i}$ iii) $\frac{1}{(2+i)^2}$ iv) $\frac{1-i}{1+i}$

v) $\frac{(2+i)^3}{2+3i}$ vi) $\frac{(1+i)(1+\sqrt{3} i)}{1-i}$ vii) $\frac{2+3i}{4+5i}$ viii) $\frac{(1-i)^3}{1-i^3}$

ix) $(1+2i)^{-3}$ x) $\frac{3-4i}{(4+2i)(1+i)}$

xi) $\Big( \frac{1}{1-4i} - \frac{2}{1+i} \Big) \Big( \frac{3-4i}{5+i} \Big)$ xii) $\frac{5+\sqrt{2} i}{1 - \sqrt{2} i}$

Answer:

i) $(1+i)(1+2i)$

$= 1 + 2i + i + 2i^2$ $= 1 + 3i - 2$ $= - 1 + 3i$

ii) $\frac{3+2i}{-2+i}$

$=$ $\frac{3+2i}{-2+i}$ $\times$ $\frac{-2-i}{-2-i}$ $=$ $\frac{-6-3i-4i-2i^2}{4-i^2}$ $=$ $\frac{-6-7i+2}{4+1}$ $=$ $\frac{-4}{5}$ $- i \big($ $\frac{7}{5}$ $\big)$

iii) $\frac{1}{(2+i)^2}$

$=$ $\frac{1}{4+i^2 + 4i}$ $=$ $\frac{1}{3+4i}$ $\times$ $\frac{3-4i}{3-4i}$ $=$ $\frac{3-4i}{9-16i^2}$ $=$ $\frac{3}{25}$ $- i \big($ $\frac{4}{5}$ $\big)$

iv) $\frac{1-i}{1+i}$

$=$ $\frac{1-i}{1+i}$ $\times$ $\frac{1-i}{1-i}$ $=$ $\frac{1+i^2-2i}{1-i^2}$ $=$ $\frac{-2i}{2}$ $= -i$

v) $\frac{(2+i)^3}{2+3i}$

$=$ $\frac{( 4 + i^2 + 4 i )(2+i)}{2+3i}$ $=$ $\frac{8 + 2i^2 + 8i + 4i + i^3 + 4i^2}{2 + 3i}$ $=$ $\frac{2+11i}{2+3i}$ $\times$ $\frac{2-3i}{2-3i}$

$=$ $\frac{4 - 6i + 22 i - 33i^2}{4 - 9i^2}$ $=$ $\frac{37+ 16i}{13}$ $=$ $\frac{37}{13}$ $+ i \big($ $\frac{16}{13}$ $\big)$

vi) $\frac{(1+i)(1+\sqrt{3} i)}{1-i}$

$=$ $\frac{(1+i)(1+\sqrt{3} i)}{1-i}$ $\times$ $\frac{1+i}{1-i}$ $=$ $\frac{(1+i^2+2i)(1+\sqrt{3}i}{1-i^2}$

$=$ $\frac{2i ( 1 + \sqrt{3} i)}{2}$ $= i(1 + \sqrt{3}i)$ $= \sqrt{3} i^2 + i$ $= - \sqrt{3}+ i$

vii) $\frac{2+3i}{4+5i}$

$=$ $\frac{2+3i}{4+5i}$ $\times$ $\frac{4-5i}{4-5i}$ $=$ $\frac{8 - 10 i + 12 i - 15 i^2}{16 - 25 i^2}$ $=$ $\frac{23 + 2 i}{41}$ $=$ $\frac{23}{41}$ $+ i \big($ $\frac{2}{41}$ $\big)$

viii) $\frac{(1-i)^3}{1-i^3}$

$=$ $\frac{(1-i)^2 (1-i)}{1-i^3}$ $=$ $\frac{(1+i^2 - 2i)(1-i)}{1+i}$ $=$ $\frac{-2i(1-i)}{1+i}$ $=$ $\frac{-2(i - i^2)}{1+i}$ $=$ $\frac{-2(1+i)}{1+i}$ $= - 2$

ix) $(1+2i)^{-3}$

$=$ $\frac{1}{(1+2i)^2(1+2i)}$ $=$ $\frac{1}{(1+4i^2+4i)(1+2i)}$ $=$ $\frac{1}{(-3+4i)(1+2i)}$ $=$ $\frac{1}{-3+4i-6i+8^2}$

$=$ $\frac{1}{-11-2i}$ $=$ $\frac{1}{-11-2i}$ $\times$ $\frac{11-2i}{11-2i}$ $=$ $\frac{-11+2i}{121+4}$ $=$ $\frac{-11}{125}$ $+ i \big($ $\frac{2}{125}$ $\big)$

x) $\frac{3-4i}{(4+2i)(1+i)}$

$=$ $\frac{3-4i}{4 - 2i + 4i - 2i^2}$ $=$ $\frac{3-4i}{6+2i}$ $\times$ $\frac{6-2i}{6-2i}$ $=$ $\frac{18 - 24i - 6i + 8i^2}{36 + 4}$

$=$ $\frac{10-30i}{40}$ $=$ $\frac{1-3i}{4}$ $=$ $\frac{1}{4}$ $+ i \big($ $\frac{-3}{4}$ $\big)$

xi) $\Big($ $\frac{1}{1-4i}$ $-$ $\frac{2}{1+i}$ $\Big) \Big($ $\frac{3-4i}{5+i}$ $\Big)$

$= \Big($ $\frac{1+i-2+8i}{1-4i+i-4i^2}$ $\Big) \Big($ $\frac{3-4i}{5+i}$ $\Big)$ $= \Big($ $\frac{-1+9i}{5-3i}$ $\Big) \Big($ $\frac{3-4i}{5+i}$ $\Big)$ $=$ $\frac{-3+27i+4i-36i^2}{25-15i+5i-3i^2}$

$=$ $\frac{33+ 31i}{28-10i} \times \frac{28+10i}{28+10i}$ $=$ $\frac{924+330i+868i-310}{784+100}$ $=$ $\frac{614+1198i}{884}$ $=$ $\frac{614}{884}$ $+ i \big($ $\frac{1198}{884}$ $\big)$

xii) $\frac{5+\sqrt{2} i}{1 - \sqrt{2} i}$

$=$ $\frac{5+\sqrt{2} i}{1 - \sqrt{2} i}$ $\times$ $\frac{1 + \sqrt{2} i}{1 + \sqrt{2} i}$ $=$ $\frac{5 + \sqrt{2} i + 5 \sqrt{2} i + 2 i^2}{1+2}$

$=$ $\frac{3 + 6 \sqrt{2} i}{3}$ $= 1 + 2\sqrt{2} i$ $= 1 + i ( 2\sqrt{2})$

$\\$

Question 2: Find the real values of $x and$ latex y, if

i) $(x+iy)(2-3i) = 4+i$ ii) $(3x-2iy)(2+i)^2 = 10(1+i)$

iii) $\frac{(1+i)x- 2i}{3+i}$ $-$ $\frac{(2-3i)y+i}{3-i}$ $= i$ iv) $(1+i)(x+iy)= 2- 5i$

Answer:

i) $(x+iy)(2-3i) = 4+i$

$\Rightarrow 2x + i 2y - 3x i - 3y i^2 = 4 + i$

$\Rightarrow ( 2x + 3y) + i ( -3x + 2y) = 4 + i$

Therefore comparing, we get

$2x + 3y = 4$ … … … … … i)

$-3x + 2y = 1$ … … … … … ii)

Solving i) and ii), multiplying i) by $3$ and ii) by $2$ and adding we get

${ \hspace{1.0cm} 6x + 9y = 12} \\ \underline { (+) - 6x + 4 y = 2 } \\ { \hspace{2.0cm}13 y = 14}$

$\Rightarrow y =$ $\frac{14}{13}$

Now substituting in ii) we get

$2x + 3 \big($ $\frac{14}{13}$ $\big) = 4 \Rightarrow 2x =$ $\frac{10}{13}$ $\Rightarrow x =$ $\frac{10}{26}$

Hence $x =$ $\frac{10}{26}$ and $y =$ $\frac{14}{13}$

ii) $(3x-2iy)(2+i)^2 = 10(1+i)$

$\Rightarrow ( 3x - 2iy) ( 4 + i^2 + 4i) = 10 + 10 i$

$\Rightarrow ( 3x - i 2y) ( 3 + 4i) = 10 + 10i$

$\Rightarrow 9x - i 6 y + i 12 x + 8 y = 10 + 10 i$

$\Rightarrow ( 9x + 8 y) + i ( 12 x - 6y) = 10 + 10 i$

Therefore comparing, we get

$9x + 8y = 10$ … … … … … i)

$12 x - 6y = 10$ … … … … … ii)

Solving i) and ii), multiplying i) by $6$ and ii) by $8$ and adding we get

${ \hspace{1.0cm} 54x + 48 y = 60 } \\ \underline { (+) 96x - 48 y = 80 } \\ { \hspace{2.0cm} 150x = 140}$

$\Rightarrow x =$ $\frac{14}{15}$

Now substituting in i) we get

$9 \big($ $\frac{14}{15}$ $\big) + 8y = 10 \Rightarrow 8y =$ $\frac{8}{5}$ $\Rightarrow y =$ $\frac{1}{5}$

Hence $x =$ $\frac{14}{15}$ and $y =$ $\frac{1}{5}$

iii) $\frac{(1+i)x- 2i}{3+i}$ $-$ $\frac{(2-3i)y+i}{3-i}$ $= i$

$\Rightarrow$ $\frac{[ (1+i) x - 2 i ] (3-i)+[(2-3i)y + 1](3+i)}{9+1}$ $= i$

$\Rightarrow 3x + i 3x - 6i - i x - i^2 x + 2 i^2+ 6y - i 9y + 3 i + i 2y - i^2 3y + i^2 = 10i$

$\Rightarrow 4x + i ( 2x) - 6i - 2 + 9y + i ( -7y) + 3i - 1 = 10i$

$\Rightarrow ( 4x + 9y - 3) + i ( 2x - 7y - 3) = 10i$

$\Rightarrow 4x + 9y - 3 = 0 \Rightarrow 4x + 9y = 3$ … … … … … i)

$\Rightarrow 2x - 7y - 3 = 10 \Rightarrow 2x - 7y = 13$ … … … … … ii)

Solving i) and ii), multiplying i) by $7$ and ii) by $9$ and adding we get

$46 x = 117 + 21 \Rightarrow x =$ $\frac{138}{46}$ $= 3$

Substituting in i) we get

$4 ( 3) + 9y = 3 \Rightarrow y = -1$

Hence $x = 3$ and $y = -1$

iv) $(1+i)(x+iy)= 2- 5i$

$x + ix + iy + i^2 y = 2 - 5i$

$(x-y) + i( x+y) = 2 - 5i$

Comapring we get

$x-y = 2$ … … … … … i)

$x+y = -5$ … … … … … ii)

Adding i) and ii) we get

$2x = - 3 \Rightarrow x =$ $\frac{-3}{2}$

Substituting in i) we get

$y = x - 2 =$ $\frac{-3}{2}$ $- 2 =$ $\frac{-7}{2}$

Hence $x =$ $\frac{-3}{2}$ and $y =$ $\frac{-7}{2}$

$\\$

Question 3: Find the conjugates of the following complex numbers:

i) $4-5i$ ii) $\frac{1}{3+5i}$ iii) $\frac{1}{1+i}$

iv) $\frac{(3-i)^2}{2+i}$ v) $\frac{(1+i)(2+i)}{3+i}$ vi) $\frac{(3-2i)(2+3i)}{(1+2i)(2-i)}$

Answer:

Note: if $z = x + iy$ , then conjugate of $z$ is $\overline{z} = x - iy$

i) If $z = 4-5i \Rightarrow \overline{z} = 4 + 5i$

ii) If $z =$ $\frac{1}{3+5i}$ $=$ $\frac{1}{3+5i}$ $\times$ $\frac{3 - 5i}{3 - 5i}$ $=$ $\frac{3 - 5i}{34}$ $=$ $\frac{3}{34}$ $-i$ $\frac{5}{34}$

Therefore $\overline{z} =$ $\frac{3}{34}$ $+ i$ $\frac{5}{34}$

iii) If $z =$ $\frac{1}{1+i}$ $=$ $\frac{1}{1+i}$ $\times$ $\frac{1-i}{1-i}$ $=$ $\frac{1-i}{2}$ $=$ $\frac{1}{2}$ $- i \big($ $\frac{1}{2}$ $\big)$

Therefore $\overline{z} =$ $\frac{1}{2}$ $+ i \big($ $\frac{1}{2}$ $\big)$

iv) If $z =$ $\frac{(3-i)^2}{2+i}$ $=$ $\frac{9+i^2-6i}{2+i}$ $=$ $\frac{8-6i}{2+i}$ $=$ $\frac{8-6i}{2+i}$ $\times$ $\frac{2-i}{2-i}$

$=$ $\frac{16-12i-8i-6}{4+1}$ $=$ $\frac{10-20i}{5}$ $= 2-4i$

Therefore $\overline{z} = 2+4i$

v) If $z =$ $\frac{(1+i)(2+i)}{3+i}$ $=$ $\frac{2+2i+i+i^2}{3+i}$ $=$ $\frac{1+3i}{3+i}$ $=$ $\frac{1+3i}{3+i}$ $\times$ $\frac{3-i}{3-i}$

$=$ $\frac{3+9i - i-3i^2}{9+1}$ $=$ $\frac{6+8i}{10}$ $=$ $\frac{3}{4}$ $+ i \big($ $\frac{4}{5}$ $\big)$

Therefore $\overline{z} =$ $\frac{3}{4}$ $- i \big($ $\frac{4}{5}$ $\big)$

vi) If $z =$ $\frac{(3-2i)(2+3i)}{(1+2i)(2-i)}$ $=$ $\frac{6-4i+9i-6i^2}{2+4i-i-2i^2}$ $=$ $\frac{12+5i}{4+3i}$ $\times$ $\frac{4-3i}{4-3i}$

$=$ $\frac{48+20i - 36i - 15i^2}{16+9}$ $=$ $\frac{63}{25}$ $- i \big($ $\frac{16}{25}$ $\big)$

Therefore $\overline{z} =$ $\frac{63}{25}$ $+ i \big($ $\frac{16}{25}$ $\big)$

$\\$

Question 4: Find the multiplicative inverse of the following complex numbers:

i) $1-i$ ii) $(1+i\sqrt{3})^2$ iii) $4-3i$ iv) $\sqrt{5}+3i$

Answer:

If $z = x + iy$ is the complex number, then multiplication inverse of $z$ is $z^{-1}$ or $\frac{1}{z}$

i) $z = 1-i$

$z^{-1} =$ $\frac{1}{1-i}$ $\times$ $\frac{1+i}{1+i}$ $=$ $\frac{1+i}{1-i^2}$ $=$ $\frac{1}{2}$ $+$ $\frac{1}{2}$ $i$

ii) $z = (1+i\sqrt{3})^2 = 1+3i^2+i (2\sqrt{3})= -2 +i (2\sqrt{3})$

$z^{-1} =$ $\frac{1}{z}$ $=$ $\frac{1}{-2+i (2\sqrt{3})}$ $\times$ $\frac{-2-i (2\sqrt{3}) }{-2-i (2\sqrt{3}) }$ $=$ $\frac{-2-i (2\sqrt{3}) }{4 + 12}$ $=$ $\frac{-1}{8}$ $-i$ $\frac{\sqrt{3}}{6}$

iii) $z = 4-3i$

$z^{-1} =$ $\frac{1}{z}$ $=$ $\frac{1}{4-3i}$ $\times$ $\frac{4+3i}{4+3i}$ $=$ $\frac{4+3i}{16+9}$ $=$ $\frac{4}{25}$ $+ i \big($ $\frac{3}{25}$ $\big)$

iv) $z = \sqrt{5}+3i$

$z^{-1} =$ $\frac{1}{z}$ $=$ $\frac{1}{\sqrt{5}+ 3i}$ $\times$ $\frac{\sqrt{5}- 3i}{\sqrt{5}-3i}$ $=$ $\frac{\sqrt{5}-3i}{5+9}$ $=$ $\frac{\sqrt{5}}{14}$ $-$ $\frac{3}{14}$ $i$

$\\$

Question 5: If $z_1 = 2 - i, z_2 = 1+i$ , find $\Big|$ $\frac{z_1+z_2+1}{z_1-z_2+1}$ $\Big|$

Answer:

If $z = x + iy$ , then $|z| = \sqrt{x^2+y^2}$

$z_1 = 2 - i$ $z_2 = 1 + i$

$\Rightarrow z_1 + z_2 + 1 = 2 - i + 1 + i + 1 = 4$

$\Rightarrow z_1 - z_2 + i = 2 - i - 1 - i + 1 = 1 - i$

$\Rightarrow$ $\frac{z_1 + z_2 + 1}{z_1 - z_2 + i}$ $=$ $\frac{4}{1-i}$ $\times \frac{1+i}{1+i}$ $=$ $\frac{4+4i}{2}$ $= 2+2i$

$\Rightarrow \Big|$ $\frac{z_1 + z_2 + 1}{z_1 - z_2 + i}$ $\Big| = \sqrt{2^2+2^2} = \sqrt{8} =2\sqrt{2}$

$\\$

Question 6: If $z_1 = 2-i, z_2 = -2+i$ , find i) $Re \Big($ $\frac{z_1z_2}{z_1}$ $\Big)$ ii) $Im \Big($ $\frac{1}{z_1z_2}$ $\Big)$

Answer:

$z_1 = 2-i, \hspace{1.0cm} z_2 = -2+i \hspace{1.0cm} \overline{z_1} = 2 + i$

i) $z_1z_2 = ( 2 - i) ( -2 + i) = -4 + 2i + 2i - i^2 = - 3 + 4i$

$\therefore$ $\frac{z_1z_2}{\overline{z_1}}$ $=$ $\frac{-3+4i}{2+i}$ $\times$ $\frac{2-i}{2-i}$ $=$ $\frac{-6+8i+3i-4i^2}{4+1}$ $=$ $\frac{-2+5i}{5}$ $=$ $\frac{-2}{5}$ $+ i$

$\therefore Re \big($ $\frac{z_1z_2}{\overline{z_1}}$ $\big) =$ $\frac{-2}{5}$

ii) $z_1 = 2 - i \hspace{1.0cm} \overline{z_1} = 2+i$

$\therefore z_1 \overline{z_1} = (2-i)(2+i) = 4 + 1 = 5$

$\therefore$ $\frac{1}{z_1 \overline{z_1}}$ $=$ $\frac{1}{5}$

$\therefore Im \big($ $\frac{1}{z_1 \overline{z_1}}$ $\big) = 0$

$\\$

Question 7: Find the modulus of $\frac{1+i}{1-i}$ $-$ $\frac{1-i}{1+i}$

Answer:

$\frac{1+i}{1-i}$ $-$ $\frac{1-i}{1+i}$ $=$ $\frac{(1+i^2+2i)-(1+i^2-2i)}{1-i^2}$ $=$ $\frac{2i+2i}{2}$ $= 2i$

$|2i| = \sqrt{2^2} = 2$

$\\$

Question 8: If $x+iy =$ $\frac{a+ib}{a-ib}$ , prove that $x^2 + y^2 = 1$

Answer:

$x+iy =$ $\frac{a+ib}{a-ib}$

$\overline{x+iy} = \Big($ $\overline{\frac{a+ib}{a-ib}}$ $\Big) =$ $\frac{\overline{a+ib}}{\overline{a-ib}}$

$x-iy =$ $\frac{a-ib}{a+ib}$

$\therefore ( x +iy)(x-iy) =$ $\frac{a+ib}{a-ib}$ $\times$ $\frac{a-ib}{a+ib}$

$\Rightarrow x^2 + y^2 =$ $\frac{a^2 +b^2}{a^2 +b^2}$ $= 1$

Hence proved.

Therefore the general solution is given by $\theta = 2 n \pi \pm$ $\frac{\pi}{2}$ $, n \in Z$

$\\$

Question 9: Find the least positive integral value of $n$ for which $\Big($ $\frac{1+i}{1-i}$ $\Big)^n$ is real.

Answer:

For $n = 1$

$\Big($ $\frac{1+i}{1-i}$ $\Big)^1 =$ $\frac{1+i^2+2i}{2}$ $= i$ which is not real.

For $n = 2$

$\Big($ $\frac{1+i}{1-i}$ $\Big)^2 =$ $\frac{1+i^2+2i}{1+i^2-2i}$ $=$ $\frac{2i}{-2i}$ $= -1$ which is real.

Therefore the least positive integral value of $n = 2$

$\\$

Question 10: Find the real values of $\theta$ for which the complex number $\frac{1+ i \cos \theta}{1 - 2 i \cos \theta}$ is purely real.

Answer:

Let $z =$ $\frac{1+ i \cos \theta}{1 - 2 i \cos \theta}$

$=$ $\frac{1+ i \cos \theta}{1 - 2 i \cos \theta}$ $\times$ $\frac{1 + 2 i \cos \theta}{1 + 2 i \cos \theta}$

$=$ $\frac{1+i \cos \theta + 2 i \cos \theta + 2 i \cos \theta + 2 i^2 \cos^2 \theta}{1+4 \cos^2 \theta}$

$=$ $\frac{(1-2 \cos^2 \theta) + (i ( 3 \cos \theta)}{1+4 \cos^2 \theta}$

For $z$ to be purely Real, $Im (z) = 0$

$\therefore$ $\frac{3 \cos \theta}{1+4 \cos^2 \theta}$ $= 0$

$\Rightarrow \cos \theta = 0$

$\Rightarrow \theta =$ $\frac{\pi}{2}$

$\\$

Question 11: Find the smallest positive integer value of $n$ for which $\frac{(1+i)^n}{(1-i)^{n-2}}$ is a real number.

Answer:

$\frac{(1+i)^n}{(1-i)^{n-2}}$

$=$ $\frac{ (1+i)^n }{ (1-i)^{n-2} }$ $\times (1-i)^2$

$= \Big($ $\frac{1+i}{1-i}$ $\times$ $\frac{1+i}{1+i}$ $\Big)^m \times ( 1+i^2 - 2i)$

$= \Big($ $\frac{1+i^2+ 2i}{1-i^2}$ $\Big)^m \times ( 1-1 - 2i)$

$= \Big($ $\frac{1-1+ 2i}{1+1}$ $\Big)^m \times ( - 2i)$

$= -2i (i^m)$

$= - 2 i^{m+1}$

For this to be real , the smallest positive value of $n$ will be $1$ .

Thus $i^{1+1} = i^2 = - 1$ , which is real.

$\\$

Question 12: If $\Big($ $\frac{1+i}{1-i}$ $\Big)^3 - \Big($ $\frac{1-i}{1+i}$ $\Big)^3 = x+iy$ , find $( x, y)$

Answer:

$\Big($ $\frac{1+i}{1-i}$ $\Big)^3 - \Big($ $\frac{1+i}{1-i}$ $\Big)^3 = x+iy$

$\Rightarrow \Big($ $\frac{1+i}{1-i}$ $\times$ $\frac{1+i}{1+i}$ $\Big)^3 - \Big($ $\frac{1+i}{1-i}$ $\times$ $\frac{1-i}{1-i}$ $\Big)^3 = x+iy$

$\Rightarrow \Big($ $\frac{2i}{2}$ $\Big)^3 - \Big($ $\frac{-2i}{2}$ $\Big)^3 = x+iy$

$\Rightarrow i^3 - ( -i)^3 = x + i y$

$\Rightarrow -i - i = x + iy$

$\Rightarrow -2i = x + iy$

$\Rightarrow x = 0$ and $y = - 2$

$\\$

Question 13: If $\frac{(1+i)^2}{2-i}$ $= x+iy$ , find $x+y$

Answer:

$\frac{(1+i)^2}{2-i}$ $= x + i y$

$\Rightarrow$ $\frac{2i}{2-i}$ $\times$ $\frac{2+i}{2+i}$ $= x+iy$

$\Rightarrow$ $\frac{4i + 2i^2}{4+1}$ $= x+iy$

$\Rightarrow$ $\frac{-2+4i}{5}$ $= x+iy$

$\Rightarrow x =$ $\frac{-2}{5}$ $, \ \ \ y =$ $\frac{4}{5}$

$\therefore x + y =$ $\frac{-2}{5}$ $+$ $\frac{4}{5}$ $=$ $\frac{2}{5}$

$\\$

Question 14: If $\Big($ $\frac{1-i}{1+ i}$ $\Big)^{100} = a+ib$ , find $( a, b)$

Answer:

$\Big($ $\frac{1-i}{1+ i}$ $\Big)^{100} = a+ib$

$\Rightarrow \Big($ $\frac{1-i}{1+ i}$ $\times$ $\frac{1-i}{1-i}$ $\Big)^{100} = a+ib$

$\Rightarrow \Big($ $\frac{-2i}{2}$ $\Big)^{100} = a+ib$

$\Rightarrow ( -i )^{100} = a+ib$

$\Rightarrow 1 = a + ib$

$\therefore a = 1 , b = 0$

$\therefore (a, b) = ( 1, 0)$

$\\$

Question 15: If $a = \cos \theta + i \sin \theta$ , find the value of $\frac{1+a}{1-a}$

Answer:

$a = \cos \theta + i \sin \theta$

$\frac{1+a}{1-a}$ $=$ $\frac{(1+\cos \theta) + i (\sin \theta) }{(1-\cos \theta) - i (\sin \theta) }$

$=$ $\frac{(1+\cos \theta) + i (\sin \theta) }{(1-\cos \theta) - i (\sin \theta) }$ $\times$ $\frac{(1-\cos \theta) + i (\sin \theta)}{(1-\cos \theta) + i (\sin \theta)}$

$=$ $\frac{(1+i \sin \theta)^2 - \cos^2 \theta}{(1-\cos \theta)^2 + \sin^2 \theta}$

$=$ $\frac{1 - \sin^2 \theta + 2 i \sin \theta- \cos^2 \theta}{1+ \cos^2 \theta - 2 \cos \theta + \sin^2 \theta}$

$=$ $\frac{2 i \sin \theta }{2 ( 1 - \cos \theta)}$

$=$ $\frac{i \sin \theta}{1 - \cos \theta}$

$= i$ $\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin^2 \frac{\theta}{2}}$

$= i \cot^2$ $\frac{\theta}{2}$

$\\$

Question 16:

i) $2x^3+2x^2-7x+72$ , when $x =$ $\frac{3-5i}{2}$

ii) $x^4-4x^3+4x^2+8x+44$ , when $x= 3 + 2i$

iii) $x^4+4x^3+6x^2+4x+9$ , when $x = - 1 +i \sqrt{2}$

iv) $x^6 + x^4 + x^2+1$ , when $x =$ $\frac{1+i}{\sqrt{2}}$

v) $2x^4 + 5x^3 + 7x^2 - x +41$ , when $x = - 2 - \sqrt{3}i$

Answer:

i) $x =$ $\frac{3-5i}{2}$

$\Rightarrow 2x = 3 - 5i$

$\Rightarrow ( 2x - 3)^2 = ( -5 i)^2$

$\Rightarrow 4x^2 + 9 - 12 x = - 25$

$\Rightarrow 4x^2 - 12 x + 34 = 0$

$\Rightarrow 2x^2- 6x + 17 = 0$

$\Rightarrow 2x^3 + 2 x^2 - 7x + 72 = 0$

Now $2x^3+2x^2-7x+72$

$= x ( 2x^2 - 6x + 17) + 6x^2 - 17x + 2x^2 - 7x + 72$

$= x(0) + 8x^2 - 24x + 72$

$= 4 ( 2x^2 - 6x +17) + 4$

$= 4(0) + 4 = 4$

ii) $x = 3 + 2i$

$\Rightarrow (x-3)^2 = (2i)^2$

$\Rightarrow x^2 + 9 - 6x = - 4$

$\Rightarrow x^2 - 6x + 13 = 0$

Now, $x^4-4x^3+4x^2+8x+44$

$= x^2( x^2 - 6x + 13) + 6x^2 - 13 x^2 - 4x^3 + 4x^2 + 8x + 144$

$= x^2 (0) + 2x^3 - 9x^2 + 8x + 44$

$= 2x ( x^2 - 6x + 13) + 12x^2 - 26 x - 9x^2 +8x+ 44$

$= 2x (0) + 3x^2 - 18x +44$

$= 3 ( x^2 - 6x + 13) + 5$

$= 3 ( 0) + 5 = 5$

iii) $x = - 1 + i \sqrt{2}$

$\Rightarrow ( x + 1) ^2 = ( i \sqrt{2})^2$

$\Rightarrow x^2 + 1 + 2 x = - 2$

$\Rightarrow x^2 + 2x + 3 = 0$

Now $x^4+4x^3+6x^2+4x+9$

$= x^2 ( x^2 + 2x + 3) + 2x^3 + 3x^2 + 4x + 9$

$= x (0) + 2x ( x^2 + 2x + 3) - ( x^2 - 2x + 9)$

$= 2x (0) - ( x^2 + 2x + 3) + 3 + 9$

$= 12$

iv) $x=$ $\frac{1+i}{\sqrt{2}}$

$(\sqrt{2} x)^2 = ( 1 + i)^2$

$\Rightarrow 2x^2 = 2i$

$\Rightarrow x^2 = i$

$\Rightarrow x^4 = - 1$

$\Rightarrow x^4 + 1 = 0$

Now, $x^6 + x^4 + x^2+1$

$= x^2 ( x^4 + 1) + ( x^4 +1 )$

$= x^2 ( 0) + (0) = 0$

v) $x = - 2 - \sqrt{3} i$

$x^2 = ( -2 - \sqrt{3} i)^2 = 1 + 4 \sqrt{3} i$

$x^3 = ( 1+4\sqrt{3} i)( - 2 - \sqrt{3} i) = 10 - 9 \sqrt{3} i$

$x^4 = ( 10 - 9 \sqrt{3} i) ( - 2 - \sqrt{3} i) = - 20 + 18 \sqrt{3} i - 10 \sqrt{3} i + 27 i^2 = -47 + 8 \sqrt{3} i$

Now, $2x^4 + 5x^3 + 7x^2 - x +41$

$= 2 ( -47 + 8\sqrt{3} i) + 5 ( 10 - 9 \sqrt{3} i) + 7 ( 1 + 4 \sqrt{3} i) - ( -2 - \sqrt{3} i) + 41$

$= - 94 + 16 \sqrt{3} i + 50 - 45 \sqrt{3} i + 7 + 28 \sqrt{3} i + 2 + \sqrt{3} i + 41$

$= ( - 94 + 50 + 7 + 2 + 41) + ( 16 \sqrt{3} i - 45 \sqrt{3} i + 28 \sqrt{3} i + \sqrt{3} i)$

$= 6+0 =6$

$\\$

Question 17: For a positive integer, find the value of $(1-i)^n \Big( 1 -$ $\frac{1}{i}$ $\Big)^n$

Answer:

$(1-i)^n \big( 1 -$ $\frac{1}{i}$ $\big)^n$ $= \Big[ (1-i) \big( 1 -$ $\frac{1}{i}$ $\big) \Big]^n$ $= \Big[$ $\frac{(1-i)(i-1)}{i}$ $\Big]^n$ $= \Big[$ $\frac{i+1-1+i}{i}$ $\Big]^n$ $= 2^n$

$\\$

Question 18: If $(1 +i) z = (1-i) \overline{z}$ , then show that $z = -i \overline{z}$

Answer:

$(1+i) z = ( 1- i) \overline{z}$

$\Rightarrow z = \Big($ $\frac{1-i}{1+i}$ $\Big) \overline{z}$

$\Rightarrow z = \Big($ $\frac{1-i}{1+i}$ $\times$ $\frac{1-i}{1-i}$ $\Big) \overline{z}$

$\Rightarrow z = \Big($ $\frac{-2i}{2}$ $\Big) \overline{z}$

$\Rightarrow z = -i \overline{z}$

Hence proved.

$\\$

Question 19: Solve the system of equations $Re(z^2) = 0 , |z|= 2$ .

Answer:

$Re(z^2) = 0 , |z| = 2$

Let $z = x + iy$

$z^2 = ( x + iy)^2 = ( x^2 - y^2)+ i ( 2xy)$

$Re(z^2) = 0 \Rightarrow x^2 - y^2 = 0$ … … … … … i)

$|z| = 2$

$\Rightarrow \sqrt{x^2+y^2} = 2$

$\Rightarrow x^2 + y^2 = 4$ … … … … … ii)

Adding i) and ii) we get

$2x^2 = 4 \Rightarrow x^2 = 2 \Rightarrow x = \pm \sqrt{2}$

$\therefore y = \pm \sqrt{2}$

Hence $x + iy = \pm \sqrt{2} \pm i \sqrt{2}$

For positive sign for $x : z = \sqrt{2} \pm i \sqrt{2} = \sqrt{2} ( 1 \pm i)$

For negative sign for $x : z = - \sqrt{2} \pm i \sqrt{2} = \sqrt{2} ( - 1 \pm i )$

$\\$

Question 20: If $\frac{z-1}{z+1}$ is purely imaginary number $(z \neq -1)$ , find the value of $|z|$

Answer:

Let $z = x + iy$

$\frac{z-1}{z+1}$ $=$ $\frac{x + iy-1}{x + iy+1}$ $=$ $\frac{(x-1) + iy}{(x +1) + iy}$

$=$ $\frac{(x-1) + iy}{(x +1) + iy}$ $\times$ $\frac{(x+1) - iy}{(x +1) - iy}$

$=$ $\frac{[(x-1)+iy][(x+1)-iy]}{(x+1)^2 + y^2}$

$=$ $\frac{x^2 -1 + i ( yx+y) - i ( xy - y) + y^2}{(x+1)^2 + y^2}$

$=$ $\frac{(x^2 + y^2 -1) + i ( yx + y - xy + y)}{(x+1)^2 + y^2}$

$=$ $\frac{(x^2 + y^2 -1) }{(x+1)^2 + y^2}$ $+ i$ $\frac{ 2y}{(x+1)^2 + y^2}$

If it is purely imaginary, $Re(z) = 0$

$\Rightarrow x^2 + y^2 - 1 = 0$

$\Rightarrow x^2 + y^2 = 1$

$\Rightarrow \sqrt{x^2 + y^2} = 1$

$\Rightarrow |z| = 1$

$\\$

Question 21: If $z_1$ is a complex number other than $- 1$ such that $|z_1|= 1$ and $z_2 =$ $\frac{z_1-1}{z_1+1}$ then show that the real parts of $z_2$ is zero.

Answer:

Let $z_1 = x_1 + i y_2$ and $z_2 = x_ 2 + i y_2$

$|z| = 1 \Rightarrow {x_1}^2 + {y_1}^2 = 1$ … … … … … i)

$z_2 =$ $\frac{z_1 - 1}{z_1 + 1}$

$\Rightarrow z_2 =$ $\frac{(x_1-1)+iy_1}{(x_1+1)+iy_1}$

$=$ $\frac{(x_1-1)+iy_1}{(x_1+1)+iy_1}$ $\times$ $\frac{(x_1+1)- iy_1}{(x_1+1)-iy_1}$

$=$ $\frac{ ({x_1}^2-1) + i ( y_1x_1+y_1) - i ( y_1x_1-y_1)+{y_1}^2}{(x_1+1)^2+{y_1}^2}$

$=$ $\frac{( {x_1}^2+{y_1}^2-1)+i ( 2y_1)}{(x_1+1)^2+{y_1}^2}$

$=0 + i$ $\frac{2y_1}{(x_1+1)^2+{y_1}^2}$

Since there is no real part, $z_2$ is purely an imaginary numbers

$\Rightarrow x_2 = 0$

$\\$

Question 22: If $|z+1|= z + 2 (1+i)$ , find $z$

Answer:

Let $z = x + i y$

$|z+1| = z + 2 ( 1 + i)$

$\Rightarrow |x+1+iy| = ( x+iy) + 2 ( 1+i)$

$\Rightarrow \sqrt{(x+1)^2 + y^2} = (x+2) + i ( y + 2)$

Comparing, $y + 2 = 0 \Rightarrow y = - 2$

$\therefore \sqrt{ (x+1)^2+y^2 } = (x+2)$

$\Rightarrow (x+1)^2 + 4 = ( x+2)^2$

$\Rightarrow x^2 + 1 + 2x + 4 = ( x+2)^2$

$\Rightarrow 5-4 = 2x$

$\Rightarrow x =$ $\frac{1}{2}$

$\therefore z =$ $\frac{1}{2}$ $+ i (-2)$

$\\$

Question 23: Solve the equation $|z| = z+1+2i$

Answer:

Let $z = x + iy$

$|z| = z + 1 + 2i$

$|x+iy| = x + iy + 1 + 2i$

$\sqrt{x^2 + y^2} = (x+1) + i ( y+2)$

Comparing, $y+2 = 0 \Rightarrow y = -2$

and $\sqrt{x^2 + y^2} = (x+1)$

$\Rightarrow x^2 + y^2 = x^2 + 1 + 2x$

$\Rightarrow 2x = y^2 - 1 = ( -2)^2 - 1 = 3$

$\Rightarrow x =$ $\frac{3}{2}$

$\therefore z =$ $\frac{3}{2}$ $- 2i$

$\\$

Question 24: What is the smallest positive integer $n$ for which $(1+i)^{2n}= ( 1- i)^{2n}$ ?

Answer:

$(1+i)^{2n}= ( 1- i)^{2n}$

$\Rightarrow [(1+i)^{2}]^n= [( 1- i)^{2}]^n$

$\Rightarrow (2i)^n = ( -2i)^n$

$\Rightarrow (2i)^n = ( -1)^n (2i)^n$

$\Rightarrow ( -1 )^n = 1$

$\therefore n$ is a multiple of $2$ .

Hence the smallest positive number $n$ for which $(1+i)^{2n} = (1-i)^{2n}$ is $2$

$\\$

Question 25: If $z_1, z_2, z_3$ , are complex numbers such that, $|z_1|= |z_2|=|z_3|= \Big|$ $\frac{1}{z_1}$ $+$ $\frac{1}{z_2}$ $+$ $\frac{1}{z_3}$ $\Big| = 1$ , then find the value of $|z_1+z_2+z_3|$ .