Exercise 13.2

QUESTION 1

Express the following complex numbers in the standard form a+ib :

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

Sol :

$=1+2i+i+2i^2$

$=1+2i+i-2$ $[\because~(i^2=-1)]$

$=-1+3i$

(ii) $\dfrac{3+2i}{-2+i}$

Sol :

$=\dfrac{3+2i}{-2+i}\times\dfrac{-2-i}{-2-i}$

$=\dfrac{-6-3i-4i-2i^2}{4-i^2}$

$=\dfrac{-6-7i+2}{4+1}$ $[\because~(i^2=-1)]$

$=\dfrac{-4-7i}{5}$

$=\dfrac{-4}{5}-\dfrac{7i}{5}$

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

Sol :

$=\dfrac{1}{4+i^2+4i}$

$=\dfrac{1}{4-1+4i}$ $\left[\because~(i^2=-1)\right]$

$=\dfrac{1}{3+4i}$

$=\dfrac{1}{3+4i}\times\dfrac{3-4i}{3-4i}$

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

$=\dfrac{3-4i}{9+16}$ $[\because~(i^2=-1)]$

$=\dfrac{3}{25}-\dfrac{4i}{25}$

(iv) $\dfrac{1-i}{1+i}$

Sol :

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

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

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

$=\dfrac{1-1-2i}{2}$ $[\because~(i^2=-1)]$

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

$=-i$

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

Sol :

$=\dfrac{8+i^3+6i(2+i)}{2+3i}$

$=\dfrac{8+i^3+12i+6i^2}{2+3i}$

$=\dfrac{8-i+12i-6}{2+3i}$ $[\because~(i^2=-1)~(i^3=-i)]$

$=\dfrac{2+11i}{2+3i}\times\dfrac{2-3i}{2-3i}$

$=\dfrac{(2+11i)(2-3i)}{4+9}$

$=\dfrac{4-6i+22i-33i^2}{13}$

$=\dfrac{4+16i+33}{13}$ $[\because~(i^2=-1)]$

$=\dfrac{37+16i}{13}$

$=\dfrac{37}{13}+\dfrac{16i}{13}$

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

Sol :

$=\dfrac{(1+i)(1+\sqrt{3}i)}{1-i}\times \dfrac{1+i}{1+i}$

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

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

$=\dfrac{(1-1+2i)(1+\sqrt{3}i)}{1+1}$ $[\because~(i^2=-1)]$

$=\dfrac{2i+2\sqrt{3}i^2}{2}$

$=\dfrac{2i-2\sqrt{3}}{2}$

$=-\sqrt{3}+i$

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

Sol :

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

$=\dfrac{(2+3i)(4-5i)}{16-25i^2}$

$=\dfrac{(8-10i+12i-15i^2)}{16-25i^2}$

$=\dfrac{8+2i+15}{16+25}$ $[\because~(i^2=-1)]$

$=\dfrac{23+2i}{41}$

$=\dfrac{23}{41}+\dfrac{2i}{41}$

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

Sol :

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

$=\dfrac{1-i+3i-3i^2)}{1-i}$ $[\because~(i^3=-i)~(i^2=-1)]$

$=\dfrac{1-i+3i+3)}{1-i}\times \dfrac{1+i}{1+i}$

$=\dfrac{4+2i}{1-i}\times \dfrac{1+i}{1+i}$

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

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

$=\dfrac{4+6i-2}{1+1}$ $[\because~(i^2=-1)~(i^3=-i)]$

$=\dfrac{2+6i}{2}$

$=1+3i$

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

Sol :

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

$=\dfrac{1}{[1+8i^3+6i(1+2i)]}$

$=\dfrac{1}{1+8i^3+6i+12i^2}$

$=\dfrac{1}{1-8i+6i-12}$ $[\because~(i^2=-1)~(i^3=-i)]$

$=\dfrac{1}{-11-2i}\times \dfrac{-11+2i}{-11+2i}$

$=\dfrac{-11+2i}{(-11+2i)(-11-2i)}$

$=\dfrac{-11+2i}{(121-4i^2)}$

$=\dfrac{-11+2i}{(121+4}$

$=\dfrac{-11}{125}+\dfrac{2i}{125}$

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

Sol :

$=\dfrac{3-4i}{4+4i-2i-2i^2}$

$=\dfrac{3-4i}{6+2i}$ $[\because~(i^2=1)]$

$=\dfrac{3-4i}{6+2i}\times \dfrac{6-2i}{6-2i}$

$=\dfrac{(3-4i)(6-2i)}{36-4i^2}$

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

$=\dfrac{18-30i-8}{36+4}$ $[\because~(i^2=-1)]$

$=\dfrac{10-30i}{40}$

$=\dfrac{1}{4}-\dfrac{3i}{4}$

(xi) $\left(\dfrac{1}{1-4i}-\dfrac{2}{1+i}\right)\left(\dfrac{3-4i}{5+i}\right)$

Sol :

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

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

$=\left(\dfrac{-1+9i}{1-3i+4}\right)\left(\dfrac{3-4i}{5+i}\right)$ $[\because~(i^2=-1)]$

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

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

$=\dfrac{-3+31i+36}{25+25i+3}$ $[\because~(i^2=-1)]$

$=\dfrac{33+31i}{28+25i}$

$=\dfrac{33+31i}{28+25i}\times\dfrac{28-25i}{28-25i}$

$=\dfrac{924+330i+868i+310i^2}{(28)^2-(25i)^2}$

$=\dfrac{924+1198i-310}{784+100}$ $[\because~(i^2=-1)]$

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

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

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

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

Sool :

$=\left(\dfrac{5+\sqrt{2}i}{1-\sqrt{2}i}\right)\times\left(\dfrac{1+\sqrt{2}i}{1+\sqrt{2}i}\right)$

$=\dfrac{(5+\sqrt{2}i)(1+\sqrt{2}i)}{1^2-(\sqrt{2}i)^2}$

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

$=\dfrac{5+6\sqrt{2}i-2}{1+2}$ $[\because~(i^2=-1)]$

$=\dfrac{3+6\sqrt{2}i}{3}$

$=\dfrac{3}{3}+\dfrac{6\sqrt{2}i}{3}$

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

QUESTION 2

Find the real values of x and y, if

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

Sol :

$=2x-3xi+2yi-3yi^2=4+i$

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

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

Comparing both the sides :

$2x+3y=4\dots(1)$

$-3x+2y=1\dots(2)$

Multiplying equation (1) by 3 and equation (2) by 2 :

$6x+9y=12\dots(3)$

$-6x+4y=2\dots(4)$

Adding equations (3) and (4) :

$13y=14$

$y=\dfrac{14}{13}$

Substituting the value of y in equation (1) :

$2x+3\times\dfrac{14}{13}=4$

$2x=4-\dfrac{42}{13}$

$2x=\dfrac{10}{13}$

$x=\dfrac{5}{13}$

$\therefore~x=\dfrac{5}{13}\text{ and } y=\dfrac{14}{13}$

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

Sol :

$=6x+3xi-4yi-2yi^2=10+10i$

$=6x+3xi-4yi+2y=10+10i$ $[\because~(i^2=-1)]$

$=6x+2y+i(3x-4y)=10+10i$

Comparing both sides:

$2x+3y=4\dots(1)$

$-3x+2y=1\dots(2)$

Multiplying equation (1) with 3 and equation (2) with 2 :

$6x+9y=12\dots(3)$

$-6x+4y=2\dots(4)$

Adding equations (3) and (4) :

$13y=14$

$y=\dfrac{14}{13}$

Substituting the value of y in equation (1):

$2x+3\times\dfrac{14}{13}=4$

$2x=4-\dfrac{42}{13}$

$2x=\dfrac{10}{13}$

$x=\dfrac{5}{13}$

$[\therefore~x=\dfrac{5}{13}\text{ and } y=\dfrac{14}{13}]$

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

Sol :

$(3x-2yi)(4+i^2+4i)^2=10+10i)$

$(3x-2yi)(4+i^2+4i)=10+10i$ $[\because~(i^2=-1)]$

$(3x-2yi)(3+4i)=10+10i$

$9x+12xi-6yi-8yi^2=10+10i$ $[\because~(i^2=-1)]$

$9x+12xi-6yi+8y=10+10i$ $[\because~(i^2=-1)]$

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

Comparing both sides :

$9x+8y=10\dots(1)$

$12x-6y=10$

$6x-3y=5\dots(2)$

Multiplying equation (1) by 3 and equation (2) by 8 ,

$27x+24y=30\dots(3)$

$48x-24y=40\dots(4)$

Adding equations (3) and (4):

$75x=70$

$\therefore~x=\dfrac{14}{15}$

Substituting the value of x in equation (1) :

$9\times\dfrac{14}{15}+8y=10$

$\dfrac{126}{15}+8y=10$

$8y=10-\dfrac{126}{15}$

$8y=\dfrac{24}{15}$

$y=\dfrac{1}{5}$

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

Sol :

$\dfrac{x+xi-2i}{3+i}+\dfrac{2y-3yi+i}{3-i}=i$

$\dfrac{x+xi+2i}{3+i}+\dfrac{2y-3yi+i}{3-i}=i$ $[\because~(i^2=-1)]$

$\dfrac{(x+xi+2i)(3-i)+(2y-3yi+i)(3+i)}{(3+i)(3-i)}=i$

$\dfrac{3x+3xi-6i-xi-xi^2+2i^2+6y-9yi+3i+2yi-3yi^2+i^2}{10}=i$

$3x+3xi-6i-xi+x-2+6y-9yi+3i+2yi+3y-1=10i$ $[\because~(i^2=-1)]$

$4x+9y-3+2xi-7yi-3i=10i$

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

Comparing both the sides :

$4x+9y-3=0$

$4x+9y=3\dots(1)$

$2x-7y-3=10$

$2x-7y=13\dots(2)$

Multiplying equation (2) by 2 :

$4x-14y=26\dots(3)$

Subtracting equation (3) from 1 :

$\begin{matrix}4x+99y=\phantom{0}3\\4x-14y=26\\\underline{~-\phantom{4x}+\phantom{14y}\phantom{=}-}\\\underline{\qquad 23y=-23}\end{matrix}$

$[\therefore~y=-1]$

Substituting the value of y in equation (1) :

$4x-9=3$

$4x=3+9$

$x=3$

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

Sol :

$x+xi+yi+yi^2=2-5i$

$(x-y)+i(x+y)=2-5i$ $[\because~(i^2=-1)]$

Comparing both the sides

$x-y=2\dots(1)$

$x+y=-5\dots(2)$

Adding equations (1) and (2)

$2x=-3$

$x=\dfrac{-3}{2}$

Substituting the value of x in equation (1)

$\dfrac{-3}{2}-y=2$

$y=\dfrac{-3}{\phantom{-}2}-2$

$y=\dfrac{-7}{\phantom{-}2}$

QUESTION 3

Find the conjugates of the following complex numbers:

(i) $4-5i$

Sol :

$\text{ if }z=a+ib~,~\text{ then } \overline{z}=a-ib$

$\overline{z}=4+5i$

(ii) $\dfrac{1}{3+5i}$

Sol :

$=\left(\dfrac{1}{3+5i}\right)\times\left(\dfrac{3-5i}{3-5i}\right)$

$=\dfrac{3-5i}{9-25i^2}$

$=\dfrac{3-5i}{9+25}$ $[\because~(i^2=-1)]$

$=\dfrac{3}{34}-\dfrac{5i}{34}$

(iii) $\dfrac{1}{1+i}$

Sol :

$=\left(\dfrac{1}{1+i}\right)\times\left(\dfrac{1+i}{1+i}\right)$

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

$=\dfrac{1+i}{2}$ $[\because~(i^2=-1)]$

$=\dfrac{1}{2}+\dfrac{i}{5}$

(iv) $\dfrac{(3-i)^2}{2+i}$

Sol :

$=\dfrac{9+i^2-6i}{2+i}$

$=\dfrac{8-6i}{2+i}$ $[\because~(i^2=-1)]$

$=\left(\dfrac{8-6i}{2+i}\right)\times\left(\dfrac{2-i}{2-i}\right)$

$=\dfrac{(8-6i)(2-i)}{4-i^2}$

$=\dfrac{16-8i-12i+6i^2}{5}$ $[\because~(i^2=-1)]$

$=\dfrac{16-20i-6}{5}$

$=\dfrac{10-20i}{5}$

$=\dfrac{10}{5}-\dfrac{20i}{5}$

$=2+4i$

(v) $\dfrac{(1+i)(2+i)}{3+i}$

Sol :

$=\dfrac{2+i+2i+i^2}{3+i}$

$=\dfrac{2+i+2i-1}{3+i}$ $[\because~(i^2=-1)]$

$=\dfrac{1+3i}{3+i}\times\dfrac{3-i}{3-i}$

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

$=\dfrac{3-i+9i-3i^2}{9-i^2}$

$=\dfrac{3+8i+3}{9+1}$ $[\because~(i^2=-1)]$

$=\dfrac{6+8i}{10}$

$=\dfrac{3+4i}{5}$

(vi) $\dfrac{(3-2i)(2+3i)}{(1+2i)(2-i)}$

Sol :

$=\dfrac{6+9i-4i-6i^2}{2-i+4i-2i^2}$

$=\dfrac{6+5i-6i^2}{2+3i-2i^2}$

$=\dfrac{6+5i+6}{2+3i+2}$ $[\because~(i^2=-1)]$

$=\dfrac{12+5i}{4+3i}\times\dfrac{4-3i}{4-3i}$

$=\dfrac{(12+5i)(4-3i)}{16-9i^2}$

$=\dfrac{48+20i-36i-15i^2}{16+9}$

$=\dfrac{63-16i}{25}$ $[\because~(i^2=-1)]$

QUESTION 4

Find the multiplicative inverse of the following complex numbers:

(i) $1-i$

Sol :

Let $z=1-i$ then,

$\dfrac{1}{z}=\dfrac{1}{1-i}$

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

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

$=\dfrac{1+i}{1-i^2}$ $[\because~(i^2=-1)]$

$=\dfrac{1+i}{1+1}$

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

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

Sol :

$=1+3i^2+2\sqrt{3}i$

$=1-3+2\sqrt{3}i$ $[\because~(i^2=-1)]$

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

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

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

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

$={-2-2\sqrt{3}i}{4-12i^2}$

$={-2-2\sqrt{3}i}{4+12}$ $[\because~(i^2=-1)]$

$=\dfrac{-2}{16}-\dfrac{2\sqrt{3}i}{16}$

$=\dfrac{-1}{8}-\dfrac{\sqrt{3}i}{8}$

(iii) $4-3i$

Sol :

$z=4-3i$

$\dfrac{1}{z}={1}{4-3i}$

$={1}{4-3i}\times\dfrac{4+3i}{4+3i}$

$={4+3i}{16-9i^2}$

$={4+3i}{16+9}$ $[\because~(i^2=-1)]$

$={4}{25}+\dfrac{3i}{25}$

(iv) $\sqrt{5}+3i$

Sol :

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

$\dfrac{1}{z}={1}{\sqrt{5}+3i}$

$={1}{\sqrt{5}+3i}\times\dfrac{\sqrt{5}-3i}{\sqrt{5}-3i}$

$={\sqrt{5}-3i}{(\sqrt{5})^2-(3i)^2}$

$={\sqrt{5}-3i}{5-9i^2}$

$={\sqrt{5}}{14}-\dfrac{3i}{14}$ $[\because~(i^2=-1)$

QUESTION 5

If $z_{1}=2-i$ , $z_{2}=1+i$ , find $\left|\dfrac{z_{1}+z_{2}+1}{z_{1}-z_{2}+1}\right|$

Sol :

$\left|\dfrac{z_{1}+z_{2}+1}{z_{1}-z_{2}+1}\right|=$ $\left|\dfrac{(2-i)+(1+i)+1}{(2-i)-(1+i)+1}\right|$

$=\left|\dfrac{4}{1-i}\right|$

$=\dfrac{4}{\left|1-i\right|}$

Also , $\left|1-i\right|=\sqrt{1^2+i^2}$ $\left[\because~(\left|a+ib\right|=\sqrt{a^2+b^2})\right]$

$=\sqrt{2}$

$\therefore \left|\dfrac{z_{1}+z_{2}+1}{z_{1}-z_{2}+1}\right|=\dfrac{4}{\sqrt{2}}$

QUESTION 6

If $z_1=2-i$ , $z_2=-2+i$ , find

(i) $Re\left(\dfrac{z_1z_2}{\overline{z_1}}\right)$

(ii) $Im\left(\dfrac{1}{z_1\overline{z_1}}\right)$

Sol :

(i)Given $z_1=2-i$ , $z_2=-2+i$ $\overline{z_1}=2+i$

$Re\left(\dfrac{z_1z_2}{\overline{z_1}}\right)=$ $\left(\dfrac{(2-i)(-2+i)}{2+i}\right)$

$=\left(\dfrac{-4+2i+2i-i^2}{2+i}\right)$

$=\left(\dfrac{-4+4i+1}{2+i}\right)$ $[\because~(i^2=-1)]$

$=\left(\dfrac{-3+4i}{2+i}\right)$

$=\left[\dfrac{-3+4i}{2+i}\times\left(\dfrac{2-i}{2+i}\right)\right]$

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

$=\left(\dfrac{-6+11i+4}{4+1}\right)$ $[\because~(i^2=-1)]$

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

$Re\left(\dfrac{z_1z_2}{\overline{z_1}}\right)=\dfrac{-2}{\phantom{-}5}$

(ii) Given $z_1=2-i$ , $z_2=-2+i$ $\overline{z_1}=2+i$

$Im\left(\dfrac{1}{z_1\overline{z_1}}\right)=$ $\dfrac{1}{(2-i)(2+i)}$

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

$=\dfrac{1}{5}$ [\because~(i^2=-1)]

$Im\left(\dfrac{1}{z_1\overline{z_1}}\right)=0$ Since no term containing i is present

QUESTION 7

Find the modulus of $\dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}$

Sol :

$=\dfrac{(1+i)(1+i)-(1-i)(1-i)}{(1-i)(1+i)}$

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

$=\dfrac{1+2i-1-(1-2i-1)}{1+1}$ $[\because~(i^2=-1)]$

$=\dfrac{2i+2i)}{2}$

$=2i$

$\therefore~{\left|2i\right|=\sqrt{0^2+2^2}}$

$=2$ $\left[\because~{\left|a+bi\right|=\sqrt{a^2+b^2}}\right]$

$\Rightarrow \left|\dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}\right|=2$

QUESTION 8

If $x+iy=\dfrac{a+ib}{a-ib}$ , prove that $x^2+y^2=1$

Sol :

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

Taking mod on the both sides:

$\left|x+iy\right|=\left|\dfrac{a+ib}{a-ib}\right|$

$\sqrt{x^2+y^2}=\dfrac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}}$

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

$x^2+y^2=1$ squaring both the sides

hence proved

QUESTION 9

Find the least positive integral value of n for which $\left(\dfrac{1+i}{1-i}\right)^n$

Sol :

$\left(=\dfrac{1+i}{1-i}\right)^n$

$=\left[\dfrac{1+i}{1-i}\times\dfrac{1+i}{1+i}\right]^n$

$=\left[\dfrac{1+i^2+2i}{1-i^2}\right]^n$ $[\because~(i^2=-1)]$

$=\left[\dfrac{1-1+2i}{1+1}\right]^n$

$=\left[\dfrac{2i}{2}\right]^n$

$=i^n$

For $=i^n$ to be real , the least positive value of n will be 2 .

As $i^2=-1$

QUESTION 10

Find the real values of $\theta$ for which the complex number $\dfrac{1+icos~\theta}{1-2icos~\theta}$ is purely real .

Sol :

$=\dfrac{1+icos~\theta}{1-2icos~\theta}\times\dfrac{1+2icos~\theta}{1+2icos~\theta}$

$=\dfrac{1+2icos~\theta+icos~\theta-2cos~\theta}{1+4icos^2~\theta}$

$=\dfrac{1-2cos~\theta +i3cos~\theta}{1+4cos^2~\theta}$

For it to be purely real , the imaginary part must be zero .

$3~cos~\theta=0$

This is true for odd multiples of $\dfrac{\pi}{2}$

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QUESTION 11

Find the smallest positive integer value of m for which $\dfrac{(1+i)^{n\phantom{-2}}}{(1-i)^{n-2}}$

Sol :

$=\dfrac{(1+i)^{m\phantom{-2}}}{(1-i)^{m-2}}$

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

$=\dfrac{(1+i)^{m}~(1-i)^{2}}{(1-i)^{m}}$

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

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

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

$=\left(\dfrac{1-1+2i}{1+1}\right)^m\times(1-1-2i)$ $[\because~(i^2=-1)]$

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

$=(i^m)\times(-2i)$

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

For this to be real , the smallest positive value of m will be 1

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

QUESTION 12

If $\left(\dfrac{1+i}{1-i}\right)^3-\left(\dfrac{1-i}{1+i}\right)^3=x+iy$ , find $(x,y)$

Sol :

$\left(\dfrac{1+i}{1-i}=\dfrac{1+i}{1-i}\times\dfrac{1+i}{1+i}\right)$

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

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

$=\dfrac{(1-1+2i}{1+1}$ $[\because~(i^2=-1)]$

$=\dfrac{2i}{2}$

$=i\dots(1)$

Also,

$\left(\dfrac{1-i}{1+i}=\dfrac{1-i}{1+i}\times\dfrac{1-i}{1-i}\right)$

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

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

$=\dfrac{1-1-2i}{1+1}$ $[\because~(i^2=-1)]$

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

$=-i\dots(2)$

It is given that ,

$\left(\dfrac{1+i}{1-i}\right)^3-\left(\dfrac{1-i}{1+i}\right)^3=x+iy$

$\Rightarrow \left(i\right)^3-\left(-i\right)^3=x+iy$ [from (1) and (2)]

$\Rightarrow i^3+i^3=x+iy$

$\Rightarrow 2i^3=x+iy$

$\Rightarrow 0-2i=x+iy$ $[\because~(i^3=-i)]$

$\Rightarrow x=0 \text{ and } y=-2$

Thus , $(x,y)=(0,-2)$

QUESTION 13

If $\dfrac{(1+i)^2}{2-i}=x+iy$ find $(x+y)$

Sol :

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

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

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

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

$=\dfrac{(1+i^2+2i)(2+i)}{(2^2-i^2)}$ $[\because~(i^2=-1)]$

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

$=\dfrac{4i+2i^2}{5}$

$=\dfrac{-2+4i}{5}$ $[\because~(i^2=-1)]$

$=\dfrac{-2}{\phantom{-}5}+\dfrac{4}{5i}\dots(1)$

It is given that

$\dfrac{(1+i)^2}{2-i}=x+iy$

$\dfrac{-2}{\phantom{-}5}+\dfrac{4}{5i}=x+iy$ [from (1)]

$x=-\dfrac{2}{5}\text{ and } y=\dfrac{4}{5}$

$\therefore~(x+y)=-\dfrac{2}{5}+\dfrac{4}{5}$

$Thus (x+y)=\dfrac{2}{5}$

QUESTION 14

If $\left(\dfrac{1-i}{1+i}\right)^100=a+ib$ find (a+b)

Sol :

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

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

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

$=\dfrac{1-1-2i}{1+1}$ $[\because~(i^2=-1)]$

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

$=-i\dots(1)$

It is given that

$\left(\dfrac{1-i}{1+i}\right)^100=a+ib$

$\Rightarrow \left(-i\right)^100=a+ib$ [from (i)]

$\Rightarrow i^100=a+ib$ $[\because~(-1)^n=1 \text{ if n is even }]$

$\Rightarrow i^{4\times 25}=a+ib$

$\Rightarrow 1+0i=a+ib$ $[\because~(i4=1)\text{ and } 1^n=1]$

$a=1\text{ and } b=0$

Thus , $(a+b)=(1,0)$

QUESTION 15

If $a=cos~\theta+isin~\theta$ , find the value of $\dfrac{1+a}{1-a}$

Sol :

$\dfrac{1+a}{1-a}=\dfrac{1+a}{1-a}\times\dfrac{1+a}{1+a}$

$=\dfrac{(1+a)^2}{1^2-a^2}$

$=\dfrac{1+a^2+2a}{1-a^2}$

on putting value of $a=cos~\theta+isin~\theta$

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

$=\dfrac{1+cos^2~\theta+i^2~sin^2~\theta+2i~cos~\theta ~sin~\theta+2cos~\theta+2i~sin~\theta}{1-cos^2~\theta-i^2sin^2~\theta-2icos~\theta~sin~\theta}$

$=\dfrac{1+cos^2~\theta-~sin^2~\theta+2i~cos~\theta ~sin~\theta+2cos~\theta+2i~sin~\theta}{1-cos^2~\theta+sin^2~\theta-2icos~\theta~sin~\theta}$ $[\because~(i^2=-1)]$

QUESTION 16

Evaluate the following:

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

Sol :

$x=\dfrac{3-5i}{2}$

$\Rightarrow x^2=\left(\dfrac{3-5i}{2}\right)^2$ [squaring both the sides]

$x^2=\dfrac{9+25i^2-30i}{4}$

$x^2=\dfrac{-16-30i}{4}$ $[\because~(i^2=-1)]$

$\Rightarrow x^3=x^2\times x^1$

$\Rightarrow x^3=\dfrac{-16-30i}{4}\times\dfrac{3-5i}{2}$ [putting value of $x^2\text{ and } x$ ]

$\Rightarrow x^3=\dfrac{-48+80i-90i+150i^2}{8}$

$\Rightarrow x^3=\dfrac{-198-10i}{8}$

$\therefore 2x^3+2x^2-7x+72$

$\therefore 2\left(\dfrac{-198-10i}{8}\right)+2\left(\dfrac{-16-30i}{4}\right)-7\left(\dfrac{3-5i}{2}\right)+72$

$\therefore \left(\dfrac{-198-10i}{4}\right)+\left(\dfrac{-16-30i}{2}\right)-\left(\dfrac{21-35i}{2}\right)+72$

$=\dfrac{-198-10i-32-60i-42+70i+288}{4}$

$=\dfrac{16}{4}$

= 4

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

Sol :

$x=3+2i$

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

$\Rightarrow x^2=9+4i^2+12i$

$\Rightarrow x^2=9-4+12i$ $[\because~(i^2=-1)]$

$\Rightarrow x^2=5+12i$

$\Rightarrow x^3=x^2\times x$

putting the values of $x^2 \text{ and } x$

$\Rightarrow x^3=(5+12i)\times(3+2i)$

$\Rightarrow x^3=(15+10i+36i-24)$

$\Rightarrow x^3=-9+46i$

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

$\Rightarrow x^4=(5+12i)^2$

$\Rightarrow x^4=25+144i^2+120i$

$\Rightarrow x^4=-119+120i$

$\therefore x^4+4x^3+4x^2+8x+44$

$=-119+120i-4(-9+46i)+4(5+12i)+8(3+2i)+44$

$=-119+120i+36-184i+20+48i+24+16i+44$

= 5

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

Sol :

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

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

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

$\Rightarrow x^2=1-2-2\sqrt{2}i$ $[\because~(i^2=-1)]$

$\Rightarrow x^2=-1-2\sqrt{2}i$

$\Rightarrow x^3=x^2\times x$

[putting the values of $x^2\text{ and } x$ ]

$\Rightarrow x^3=\left(-1-2\sqrt{2}i\right)\times \left(-1+i\sqrt{2}\right)$

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

$\Rightarrow x^3=5+\sqrt{2}i$

$\Rightarrow x^4=(x^2)^2$ [putting the value of $x^2$ ]

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

$\Rightarrow x^4=1+8i^2+4\sqrt{2}i$

$\Rightarrow x^4=-7+4\sqrt{2}i$

$\therefore x^4+4x^+6x^2+4x+9$

$-7+4\sqrt{2}i+20+4\sqrt{2}i-6-12\sqrt{2}i-4+4\sqrt{2}i+9$

= 12

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

Sol :

$x=\dfrac{1+i}{\sqrt{2}}$

$\Rightarrow x^2=\left(\dfrac{1+i}{\sqrt{2}}\right)^2$ [squaring both the sides]

$\Rightarrow x^2=\dfrac{1+i^2+2i}{2}$

$\Rightarrow x^2=\dfrac{1-1+2i}{2}$ $[\because~(i^2=-1)]$

$\Rightarrow x^2=i$

$\Rightarrow x^6=(x^2)^3$

$\Rightarrow x^6=i^3$ [putting value of $x^2$ ]

$\Rightarrow x^6=-i$

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

$\Rightarrow x^4=(i)^2$ [putting the value of $x^{2}$ ]

$\Rightarrow x^4=i^2$

$\Rightarrow x^4=-1$ $[\because~(i^2=-1)]$

$\therefore x^6+x^4+x^2+1$

$=-i-1+i+1$

= 0

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

Sol :

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

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

$\Rightarrow x^2=4+3i^2+4\sqrt{3}i$

$\Rightarrow x^2=4-3+4\sqrt{3}i$ $[\because~(i^2=-1)]$

$\Rightarrow x^2=1+4\sqrt{3}i$

$\Rightarrow x^3=x^2\times x$

putting the value of $x^2\text{ and } x$

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

$\Rightarrow x^3=-2-\sqrt{3}i-8\sqrt{3}i-12i^2$

$\Rightarrow x^3=-2-9\sqrt{3}i+12$ $[\because~(i^2=-1)]$

$\Rightarrow x^3=10-9\sqrt{3}i$

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

on putting the value of $x^2$

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

$\Rightarrow x^4=1+48i^2+8\sqrt{3}i$

$\Rightarrow x^4=1-48+8\sqrt{3}i$ $[\because~(i^2=-1)]$

$\Rightarrow x^4=-47+8\sqrt{3}i$

$\therefore~2x^4+5x^3+7x^2-x+41$

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

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

= 6

QUESTION 17

For a positive integer n , find the value of $(1-i)^n(1-\dfrac{1}{i})^n$

Sol :

$(1-i)^n(1-\dfrac{1}{i})^n$

$=(1-i)^n(1-\dfrac{i^4}{i})^n$ $[\because~(i^4=1)]$

$=(1-i)^n(1-i^3)^n$

$=(1-i)^n(1+i)^n$ $[\because~(i^3=-i)]$

$=[(1-i)(1+i)]^n$

$=[1-i^2]^n$

$=[1+1]^n$ $[\because~(i^2=-1)]$

$= 2^n$

Thus , the value of $(1-i)^n(1-\dfrac{1}{i})^n$ is $= 2^n$

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A quality solution

Rd sharma solution of chapter 13 complex number class 11

Exercise 13.2

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