KC Sinha Mathematics Solution Class 11 Chapter 10 समिश्र संख्याएँ (complex number) Exercise 10.1

Exercise 10.1

Question 1

Write the following as complex numbers
(i) $\sqrt{-27}$
Sol :
$=\sqrt{27} \mathrm{i}$
$=\sqrt{3 \times 3 \times 3} \mathrm{i}$
$=3 \sqrt{3} \mathrm{i}$

$=0+3 \sqrt{3} \mathrm{i}$

(ii) $\sqrt{-16}$
Sol :


(iii) $4-\sqrt{-5}$
Sol :
$4-\sqrt{5} i$



(iv) $-1-\sqrt{-5}$

(v) $1+\sqrt{-1}$

Question 2

Write the real and imaginary parts of the following complex numbers
(i) $2-i \sqrt{2}$
Sol :
Let
$z=2-i \cdot \sqrt{2}$

Re(z)=2 , In(z)=-√2

(ii) $-\frac{1}{5}+\frac{i}{5}$
(iii) $\frac{\sqrt{5}}{7} i$
(iv) $\sqrt{37}+\sqrt{-19}$
(v) $\frac{\sqrt{37}}{3}+\frac{3}{\sqrt{70}} i$

Question 3

Find the additive inverse of the following :
(i) -5+7i
Sol :
=5-7i

(ii) 4-3i
Sol :
-4+3i


(iii) 10
Sol :

Question 4

Find the sum of the following numbers $\frac{2}{3}+\frac{5}{3} i,-\frac{2}{3} i$ and $-\frac{5}{4}-i$
Sol :
$\left(\frac{2}{3}+\frac{5}{3} i\right)+\left(-\frac{2}{3} i\right)+\left(\frac{-5}{4}-i\right)$

$=\frac{2}{3}+\frac{5}{3} i-\frac{2}{3} i-\frac{5}{4}-i$

$=\frac{2}{3}-\frac{5}{4}+\left[\frac{5}{3}-\frac{2}{3}-1\right]i$

$=\frac{8-15}{12}+\left(\frac{5-2-3}{3}\right)i$

$=\frac{-7}{12}+0\times i$

$-\frac{7}{12}$

Question 5

Find the difference of the following complex numbers
(i) -3+2i and 13-i
Sol :
=(-3+2i)-(13-i)
=-3+2i-13+i
=-16+3i

(ii) 1-i and -1+6i
Sol :

Question 6

(1+i) and (3+i)
Sol :
Multiplication:
=(1+i)(3+i)
=3+i+3i+i²
=3+4i-1
=2+4i

Division :
$=\frac{1+i}{3+i}=\frac{1+i}{3+i}\times \frac{3-i}{3-i}$

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

$=\frac{3+2 i-(-1)}{9-(-1)}$

$=\frac{3+2 i+1}{9+1}=\frac{4+2 i}{10}$

$=\frac{4}{10}+\frac{20}{10}i$

$=\frac{2}{5}+\frac{i}{5}$

Question 7 

Find the multiplicative inverse of the following
(i) $2+\sqrt{3} i$
Sol :
$=\frac{1}{2+\sqrt{3}}$

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

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

$=\frac{2-\sqrt{3} i}{4-3(-1)}$

$=\frac{2-\sqrt{3} i}{7}$

$=\frac{2}{7}-\frac{\sqrt{3}}{7}i$


(ii) -3+4i

(iii) -i
Sol :
$=\frac{1}{-i}$

$=-\frac{1}{i} \times \frac{1}{i}$

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


(iv) 4-i3

(v) $(\sqrt{5}+i 3)$

(vi) 2-3i

Question 8

(i) Prove that $\operatorname{Re}\left(z_{1} z_{2}\right)=\operatorname{Re} z_{1} \operatorname{Re} z_{2}-\operatorname{Im} z_{1}, \operatorname{Im} z_{2}$
Sol :
$z_{1}=x_{1}+i_{y_{1}}$ and $z_{2}=x_{2}+i_{y_{2}}$

$=x_{1} x_{2}+i x_{1} y_{2}+i y_{1} x_{2}+i^{2} y_{1} y_{2}$

$z_1z_2 =x_1 x_{2} -y_{1} y_{2}+i\left(x_{1}, y_{2}+y_{1} x_{2}\right)$


$Re \left(z_1 z_{2}\right)=x_{1} x_{2}-y_{1} y_{2}$

$\operatorname{Re}(z_1z_2)=\operatorname{Re}\left(z_{1}\right) \operatorname{Re}\left(z_{2}\right)-\operatorname{Im}\left(z_{1}\right) \operatorname{Im}\left(z_{2}\right)$


(ii) $[\text { Let }] z_{1}=2-i, z_{2}=-2+i$ find
(a) $\operatorname{Re}\left(\frac{z_{1} z_{2}}{\overline{z_{1}}}\right)$
(b) $\operatorname{Im}\left(\frac{1}{z_{1} \overline{z_{2}}}\right)$
Sol :
$\bar{z}_{1}=\overline{2-i}$
=2+i

$\bar{z}_{2}=-\overline{2+i}$
=-2-i

(a) $\frac{z_{1} z_{2}}{\overline{z_{1}}}=\frac{(2-i)(-2+i)}{\overline{2-i}}$

$=\frac{-4+2 i+2 i-i^{2}}{2+i}$

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

$=\frac{-3+4 i}{2+i} \times \frac{2-i}{2-i}$

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

$=\frac{-6+11 i+4}{4+1}$

$=\frac{-2+11 i}{5}$

$\frac{z_{1} z_{2}}{\overline{z_{1}}}=-\frac{2}{5}+\frac{11}{5} i$

$\operatorname{Re}\left(\frac{z_1z_2}{\bar{z}_{1}}\right)=-\frac{2}{5}$


(b) $\frac{1}{z_1\bar{z}_{2}}=\frac{1}{(2-i)(-2-i)}$

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

$=\frac{1}{-4-1}=-\frac{1}{5}$

$\text {Im}\left(\frac{1}{z_{1} \overline{z_{2}}}\right)=0$

Question 9

Express the following in the form a+ib
(i) (3+2i)(3-2i)
Sol :
$=3^{2}-(2 i)^{2}$

$=9-4 i^{2}$

=9-4(-1)
=9+4
=13
=$13+0 \times i$


(ii) $(i-2)^{2}$
Sol :
$=i^{2}+2^{2}-2 \cdot i \cdot 2$

=-1+4-4i
=3-4i


(iii) $\frac{2-i}{4+3 i}$
Sol :
$=\frac{2-i}{4+3 i} \times \frac{4-3 i}{4-3i}$

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

$=\frac{8-10 i+3(-1)}{16-9(-1)}$

$=\frac{5-10 i}{25}$

$=\frac{5}{25}-\frac{10}{25}i$


(iv) $\frac{1+2 i+3 i^{2}}{1-2 i+3 i^{2}}$
Sol :
$\frac{1+2 i+3 i^{2}}{1-2 i+3 i^{2}}$

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

$=\frac{-2+2 i}{-2-2 i}$

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

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

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

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

$=\frac{-2i }{2}$

=-i
=0-i


(v) $\left(\frac{1+i}{1-i}\right)^{2}$
Sol :
$\left(\frac{1+1}{1-i}\right)^{2}$

$=\left(\frac{1+i}{1-i} \times \frac{1+j}{1+i}\right)^{2}$

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

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

$=\left(\frac{2i}{2}\right)^{2}$

$=i^{2}=-1$

=-1+0×i


(vi) $\left(\frac{1+2 i}{2+i}\right)^{2}$
Sol :

(vii) $\frac{6+3 i}{2-i}$
Sol :


(viii) $\frac{5+\sqrt{2} i}{1+\sqrt{2} i} Sol :$=\frac{5+\sqrt{2}+}{1-\sqrt{2} i} \times \frac{1+\sqrt{2} i}{1+\sqrt{2} i}=\frac{5+5 \sqrt{2} i+\sqrt{2} i+2 i^{2}}{1^{2}-(\sqrt{2 i})^{2}}=\frac{5+6 \sqrt{2} i+2(-1)}{1-2(-1)}=\frac{3+6 \sqrt{2} i}{3}=3\frac{(1+2 \sqrt{2} i)}{3}$

Question 10

Simplify the following
(i) $2 i^{2}+6 i^{3}+3 i^{16}-6 i^{19}+4 i^{25}$
Sol :
$=2 i^{2}+6 i^{2}i+3\left(i^{2}\right)^{8}-6\left(i^{2}\right)^{9} \cdot i+4\left(i^{2}\right)^{12} \cdot i$

$=2(-1)+6(-1) i+3(-1)^{8}-6(-1)^{9} i+4(-1)^{12}i$

=-2-6i+3-6i+4i
= 1+4i


(ii) $1+i^{10}+i^{110}+i^{1000}$
Sol :
$=1+\left(i^{2}\right)^{5}+\left(i^{2}\right)^{55}+\left(i^{2}\right)^{500}$

$=1+(-1)^{5}+(-1)^{55}+(-1)^{500}$

=1-1-1+1
=0

(iii) $i^{n}+i^{n+1}+i^{n+2}+i^{n+3}$
Sol :
$=i^{n}+i^{n} \cdot i+i^{n} \cdot i^{2}+i^{n} \cdot i^{3}$

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

$=i^{n}+i^{n} \cdot i-i^{n}-i^{n}. i$

=0

(iv) $\left\{i^{17}-\left(\frac{1}{i}\right)^{34}\right\}^{2}$
Sol :
$=\left\{\left(i^{2}\right)^{8} \cdot i-\frac{1}{\left(i^{2}\right)^{17}}\right\}^{2}$

$=\left\{(-1)^{8} \cdot i-\frac{1}{(-1)^{17}}\right\}^{2}$

$=\left\{i-\frac{1}{-1}\right\}^{2}$

$=\{i+1\}^{2}$

$=i^{2}+1^{2}+2 \cdot i \cdot 1$

=-1+1+2i

=2i


(v) $(-i)^{4 n+3}$ where n is positive integer
Sol :
$=(-i)^{4 n} \cdot(-i)^{3}$

$=-i^{4 n} \cdot i^{3}$

$=-\left(i^{2}\right)^{2 n} \cdot i^{2} \cdot i$

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

$=-(1)(-1) i$


(vi) $\left(\frac{1+i}{1-i}\right)^{4 n+1}$ where n is a positive integer
Sol :
$=\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^{4 n+1}$

$=\left(\frac{1+i+i+i^{2}}{1^2-i^{2}}\right)^{4 n+1}$

$=\left(\frac{1+2 i-1}{1-(-1)}\right)^{4 n+1}$

$=\left(\frac{2 i}{2}\right)^{4 n+1}$

$=i^{4 n+1}$

$=i^{4 n} \cdot i$

$=\left(i^{2}\right)^{2 n} \cdot i$

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

= i


(vii) $(2 i)^{3}$
Sol :
=8 i^{3}=8(-i)=-8

(viii) $(8 i)\left(-\frac{1}{8} i\right)$

(ix) $(5 i)\left(-\frac{3}{5} i\right)$

(x) $(-5 i)\left(\frac{1}{8} i\right)$
Sol :
$=-\frac{5}{8} i^{2}$
$=-\frac{5}{8}(-1)=\frac{5}{8}$

(xi) $(-i)(2 i)\left(-\frac{1}{8} i\right)^{3}$
Sol:
$=-2 i^{2} \times\left(\frac{-1}{512} i^{3}\right)$

$\frac{1}{256}(-1)(-1)=\frac{i}{256$


(xii) $i^{-35}$
Sol :
$=\frac{1}{i^{35}}=\frac{1}{\left(i^{2}\right)^{17} \cdot i}$

$=\frac{\int}{(-1)^{17} i}$

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


$=\frac{-i}{-1}=i$

(xiii) $i^{-39}$

(xiv) $i^{9}+i^{19}$


(xv) $\left[\begin{array}{l}\left.i^{18}+\left(\frac{1}{i}\right)^{25}\right]^{3}\end{array}\right]$
Sol :
$=\left[\left(i^{2}\right)^{9}+\frac{1}{\left(i^{2}\right)^{12} \cdot i}\right]^{3}$

$=\left[(-1)^{9}+\frac{1}{(-1)^{12} \cdot i}\right]^{3}$

$=\left[-1+\frac{1}{i}\right]^{3}$

$=\left[-1+\frac{1}{i} \times \frac{i}{i}\right]^{3}$

$=\left[-1+\frac{i}{i^{2}}\right]^{3}$

$=\left[\begin{array}{ll}-1 & -i\end{array}\right]^{3}$

$=[-1+(-i)]^{3}$

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

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

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

=-1-(-1)i-3i+3
=-1+i-3i+3
=2-2i


(xvi) $i^{6}+i^{8}$

(xvii) $i+i^{2}+i^{3}+i^{4}$

(xviii) $i^{12}+i^{13}+i^{14}+i^{15}$

(xix) $i^{4}+i^{8}+i^{12}+i^{16}$

Question 11

Write the following in the form a+ib
$\frac{1}{(2+i)^{2}}-\frac{1}{(2-i)^{2}}$
Sol :
$\frac{1}{(2+i)^{2}}-\frac{1}{(2-i)^{2}}=\frac{1}{2^{2}+i^{2}+2 \cdot 2} \cdot-\frac{1}{2^{2}+1^{2}-2 \cdot 2 \cdot i}$

$=\frac{1}{4-1+4 i}-\frac{1}{4-1-4 i}$

$=\frac{1}{3+4 i}-\frac{1}{3-4 i}$

$=\frac{(3-4 i)-(3+4 i)}{(3+4 i)(3-4 i)}$

$=\frac{3-4 i-3 -4 i}{3^{2}-(4 i)^{2}}$

$=\frac{-8 i}{9-16(-1)}$

$=-\frac{8 i}{25}$

$=0-\frac{8 i}{25}$

Question 12

Express the following in the form of a+ib
(i) $\left(\frac{1}{5}+i \frac{2}{5}\right)-\left(4+i \frac{5}{2}\right)$
Sol :
$=\frac{1}{5}+i \frac{2}{5}-4-i \frac{5}{2}$

$=\frac{1}{5}-4+i\left(\frac{2}{5}-\frac{5}{2}\right)$

$=\frac{1-20}{5}+i\left(\frac{4-25}{10}\right)$

$=-\frac{19}{5}-\frac{i \cdot 21}{10}$


(ii) (7-i2)-(4+i)+(-3+i5)
Sol :
=7-i.2-4-i-3+i.5
=0+i.2

(iii) $\left[\left(\frac{1}{3}+i \frac{7}{3}\right)+\left(4+i \frac{1}{3}\right)\right]-\left(-\frac{4}{3}+i\right)$
Sol :
$\frac{1}{3}+i \cdot \frac{7}{3}+4+i \frac{1}{3}+\frac{4}{3}-i$

$\frac{1}{3}+4+\frac{4}{3}+i \frac{7}{3}+i \frac{1}{3}-i$

$\frac{1+12+4}{3}+\frac{i.7+i.1-i.3}{3}$

$=\frac{17}{3}+\frac{5}{3} i$

(iv) $i^{3}+(6+i 3)-(20+i 5)+(14+i 3)$
(v) $(7+i 5)(7-i 5)$
(vi) $3 i^{3}\left(15 i^{6}\right)$
(vii) $\sqrt{3}+(\sqrt{3}-i 2)-(3-i 2)$

(viii) $(1+i)^{4}$
Sol :
$=\left[(1+i)^{2}\right]^{2}$

$=\left[i^{2}+i^{2}+2 \cdot 1 \cdot i\right]^{2}$


$=[1-1+2i]^{2}$

$=4 i^{2}=4(-1)$
=-4
=-4+0.i


(ix) $\left(\frac{1}{2}+i 2\right)$

(x) $\left(-2-i \frac{1}{3}\right)^{3}$
Sol :
$=\left[(-2)+\left(-\frac{1}{3} i\right)\right]^{3}$

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

$=-8-\frac{1}{27} i^{3}-\frac{12}{3} i-\frac{6}{9} i^{2}$

$=-8-\frac{1}{27}(-i)-4 i-\frac{2}{3}(-1)$

$=-8+\frac{1}{27} i-4 i+\frac{2}{3}$

$=-8+\frac{2}{3}+\frac{1}{27} i-4 i$

$=\frac{-24+2}{3}+\frac{1 i-108 i}{27}$

$=\frac{-22}{3}-\frac{107 i}{27}$

Question 13

Find the following as a single complex number x+iy:

(i) $(\sqrt{6}+i 5)\left(\sqrt{6}-i \frac{1}{5}\right)$
Sol :
$=6-i \frac{\sqrt{6}}{5}+i\cdot 5 \sqrt{6}-i^{2}$

$=6-(-1)+i 5 \sqrt{1}-i \frac{\sqrt{6}}{5}$

$=7+\frac{i 25 \sqrt{6}-i \sqrt{6}}{5}$

$=7+i \frac{24 \sqrt{6}}{5}$


(ii) (5+i9)+(-3+i4)
Sol :
$=\frac{5+9 i}{-3+4 i} \times \frac{-3-4 i}{-3-4 i}$

$=\frac{-15-20i-27 i-36 i^{2}}{(-3)^{2}-(4i)^{2}}$

$=\frac{-15-47 i-36(-1)}{9-16(-1)}$

$=\frac{-15-47 i+36}{9+16}$

$=\frac{21-47 i}{25}$

$=\frac{21}{25}-\frac{47}{25} i$


(iv) $\left[\left(\sqrt{5}+\frac{i}{2}\right)(\sqrt{5}-i 2)\right]+(6+i 5)$
Sol :
$=\left[\begin{array}{ll}5-2 \sqrt{5}\end{array}i+\frac{\sqrt{5}}{2}i-i^{2}\right] \div(6+i 5)$

$=\left[\begin{array}{ll}5 +\frac{\sqrt{3}}{2} i-2 \sqrt{5} i+1\end{array}\right] \div(6+i 5)$

$=\left[6+\frac{\sqrt{5} i-4 \sqrt{5}i}{2}\right] \div(6+5 i)$

$=\left[\begin{array}{ll}6 -\frac{3 \sqrt{5}i}{2}\end{array}\right] \div(6+5 i)$

$=\left(\frac{12-3 \sqrt{5} i}{2}\right) \div(6+5 i)$

$=\frac{12-3 \sqrt{5} i}{2(6+5 i)}$

$=\frac{12-3 \sqrt{5}i}{2(6+5 i)} \times \frac{6-5 i}{6-5 i}$

$=\frac{72-60 i-18 \sqrt{5}i+15 \sqrt{5} i^{2}}{2\left[6^{2}-(5 i)^{2}\right]}$

$=\frac{(72-15 \sqrt{5})-i(60-18 \sqrt{5})}{2(36+25)}$

$=\frac{(72-15\sqrt5)-i (60+18 \sqrt{5})}{122}$


(v) $\frac{[(\sqrt{2}+i \sqrt{3})+(\sqrt{2}-i \sqrt{3})]}{[(\sqrt{3}+i \sqrt{2})+(\sqrt{3}-i \sqrt{2})]}$
Sol :
$=\frac{\sqrt{2}+i \sqrt{3}+\sqrt{2}-i\sqrt{3}}{\sqrt{3}+i\sqrt{2}+\sqrt{3}-i \sqrt{2}}$

$=\frac{2\sqrt2}{2 \sqrt{3}}$

$=\sqrt{\frac{2}{3}}+0\times i$


(vi) $\frac{(3+i \sqrt{5})(3-i \sqrt{5})}{(\sqrt{3}+i \sqrt{2})-(\sqrt{3}-i \sqrt{2})}$
Sol :
$=\frac{(3)^{2}-(i \sqrt{5})^{2}}{\sqrt{3}+i\sqrt{2}-\sqrt3+i\cdot \sqrt2}$

$=\frac{9-(-1) 5}{i 2\sqrt{2}}$

$=\frac{14}{2 \sqrt{2} i}$

$=\frac{7}{\sqrt{2} i} \times \frac{\sqrt{2} i}{\sqrt{2} i}$

$=\frac{7 \sqrt{2} i}{2(-1)}$

$=-\frac{7 \sqrt{2}i}{2}$

$=0-7 \sqrt{2}$


(vii) $\left(\frac{1}{1-4 i}-\frac{2}{1+i}\right)\left(\frac{3-4 i}{5+i}\right)$
Sol :
$=\left(\frac{(1+i)-2(1-4 i)}{(1-4 i)(1+i)}\right)\left(\frac{3-4 i}{5+i}\right)$

$=\left(\frac{1+i-2+8 i}{1+i-4 i-4 i^{2}}\right)\left(\frac{3-4 i}{5+i}\right)$

$=\left(\frac{-1+9 i}{1-3 i+4}\right)\left(\frac{3-4 i}{5+i}\right)$

$=\left(\frac{-1+9 i}{5-3 i}\right)\left(\frac{3-4 i}{5+i}\right)$

$=\frac{-3+4 i+27 i-36 i^{2}}{25+5 i-15 i-3 i^{2}}$

$=\frac{-3+31 i+36}{25-10 i+3}$

$=\frac{33+31 i}{28-10 i} \times \frac{28+10 i}{28+10 i}$

$=\frac{924+330i+868 i+310 i^{2}}{(28)^{2}-(10 i)^{2}}$

$=\frac{924+1198 i-310}{784+100}$

$=\frac{614+1198 i}{884}$

$=\frac{2(307+599i)}{884}$

$=\frac{307}{442}+\frac{599}{442}i$

Question 14

If $\left(\frac{1+i}{1-i}\right)^{m}=1$ then find the least positive integral value of m
Sol :
$\left(\frac{1+i}{1-i}\right)^{m}=1$

$\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^{m}=1$

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

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

$\left(\frac{2 i}{2}\right)^{m}=1$

$i^{m}=1$

$i^{m}=(-1)^{2}$

$i^{m} =\left(i^{2}\right)^{2}$


$i^{m} =i^{4} \Rightarrow m=4$

Question 15

Find x and y
(i) (x+iy)+(7-5i)=9+4i
Sol :
x+iy=9+4i-7+5i
x+iy=2+9i

By equating real and imaginary part

x=2 , y=9


(ii) (x+iy)(2-3i)=4+i
Sol :
$x+iy=\frac{4+i}{2-3i}$

$x+i y=\frac{4+1}{2-3 i} \times \frac{2+3 i}{2+3 i}$

$=\frac{8+12 i+2 i+3 i^{2}}{(2)^{2}-(3 i)^{2}}$

$x+iy=\frac{8+14 i-3}{4-9(-1)}$

$x+i y=\frac{5+14 i}{13}$

$x+i y=\frac{5}{13}+\frac{14}{13}i$

By equating real and imaginary part

$x=\frac{5}{13}$ $y=\frac{14}{13}$


(iii) $\left(\frac{3}{\sqrt{5}} x-5\right)+i 2 \sqrt{5} y=\sqrt{2}$
Sol :
$\left(\frac{3}{\sqrt{5}} x-5\right)+i 2 \sqrt{5} y=\sqrt{2}+i \cdot 0$

By equating real and imaginary part

$\frac{3}{\sqrt{5}} x-5=\sqrt{2}$

$\frac{3}{\sqrt{5}} x=5+\sqrt{2}$

$x=\frac{\sqrt{5}(5+\sqrt{2})}{3}$

and

$2 \sqrt{5} y=0$

y=0


(iv) 4x+i(3x-y)=3-i6
Sol :
By equating real and imaginary parts

4x=3

$x=\frac{3}{4}$

and

3x-y=-6

$3\left(\frac{3}{4}\right)-y=-6$

$\frac{9}{4}-y=-6$

$\frac{9}{4}+6=y$

$\frac{9+24}{4}=y$

$\frac{33}{4}=y$


(v) (3y-2)+i(7-2x)=0
Sol :
By equating real and imaginary parts
3y-2=0

3y=2

$y=\frac{2}{3}$

and

7-2x=0

-2x=-7

$x=\frac{7}{2}$

Question 16

Simplify :
$\frac{20}{\sqrt{3}-\sqrt{-2}}+\frac{30}{3 \sqrt{-2}-2 \sqrt{3}}-\frac{14}{2 \sqrt{3}-\sqrt{-2}}$

Sol :
$=\frac{20}{\sqrt{3}-\sqrt{2}i}-\frac{30}{2 \sqrt{3}-3 \sqrt{2} i}-\frac{14}{2 \sqrt{3}-\sqrt{2} i}$

$=\frac{20}{\sqrt{3}-\sqrt{2} i} \times \frac{\sqrt{3}+\sqrt{2} i}{\sqrt{3}+\sqrt{2} i}-\frac{30}{2 \sqrt{3}-3 \sqrt{2} i} \times \frac{2 \sqrt{3}+3 \sqrt{2} i}{2 \sqrt{3}+3 \sqrt{2}i}-\frac{14}{2 \sqrt{3}-\sqrt{2} i}\times\frac{2 \sqrt{3}+\sqrt{2}i}{2\sqrt3+\sqrt{2}i}$

$=\frac{20(\sqrt{3}+\sqrt{2} i)}{(\sqrt{3})^{2}-(\sqrt{2} i)^{2}}-\frac{30(2 \sqrt{3}+3 \sqrt{2} i)}{(2 \sqrt{3})^{2}-(3 \sqrt{2} i)^{2}}-\frac{14(2 \sqrt{3}+\sqrt{2} i)}{(2 \sqrt{3})^{2}-(\sqrt{2} i)^{2}}$

$=\frac{20(\sqrt{3}+\sqrt{2} i)}{3-2(-1)}-\frac{30(2 \sqrt{3}+3 \sqrt{2} \cdot i)}{12-18(-1)}-\frac{14(2 \sqrt{3}+\sqrt{2} i)}{12-2(-1)}$

$=\frac{20(\sqrt{3}+\sqrt{2} i)}{5}-\frac{30(2 \sqrt{3}+3 \sqrt{2} i)}{30}-\frac{14(2\sqrt3+\sqrt2 i)}{14}$

$=4 \sqrt{3}+4 \sqrt{2} i-2 \sqrt{3}-3 \sqrt{2} i-2 \sqrt{3}-\sqrt{2}i$

$=4 \sqrt{3}-4 \sqrt{3}+4 \sqrt{2}i-4\sqrt2 i$

Question 17

Sol :
$\frac{1-i x}{1+i x}=a-i b$..(i)

By conjugating both sides

$\frac{1+i x}{1-i x}=a+i b$..(ii)

Multiplying (ii) with (i) , we get

$\frac{1-ix}{1+i x} \times \frac{1+ix}{1-i x} \quad=(a-i b)(a+i b)$

$1=a^{2}-(i b)^{2}$

$1=a^{2}-(i) b^{2}$

$1=a^{2}+b^{2}$

Question 18

Sol :
Let $z=\frac{3+2 i \sin \theta}{1-2i \sin \theta}$

$z=\frac{3+2 i \sin \theta}{1-2i \sin \theta} \times \frac{1+2 i\sin \theta}{1+2i \sin \theta}$

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

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

$z=\frac{3-4 \sin ^{2} \theta+8i \sin \theta}{1+4 \sin ^{2} \theta}$

$z=\frac{3-4 \sin ^{2} \theta}{1+4 \sin ^{2} \theta}+\frac{i \cdot 8 \sin \theta}{1+4 \sin ^{2} \theta}$

Re(z)=0

$\frac{3-4 \sin ^{2} \theta}{1+4 \sin ^{2} \theta}=0$

$3-4 \sin ^{2} \theta=0$

$-4 \sin ^{2} \theta=-3$

$\sin ^{2} \theta=\frac{3}{4}$

$\sin ^{2} \theta=\left(\frac{\sqrt{3}}{2}\right)^{2}$

[if $\sin ^{2} \theta=\sin ^{2} \alpha$

then $\theta=n \pi \pm \alpha, n \in z$]

$\theta=n \pi \pm \frac{\pi}{3}, n \in z$