Class 11: Complex Numbers

$\displaystyle \textbf{Question 1: } \text{Express the following complex numbers in the form }a+ib:$
$\displaystyle \text{i) }(1+i)(1+2i)\hspace{1.0cm}\text{ii) }\frac{3+2i}{-2+i}\hspace{1.0cm}\text{iii) }\frac{1}{(2+i)^2}$
$\displaystyle \text{iv) }\frac{1-i}{1+i}\hspace{1.0cm}\text{v) }\frac{(2+i)^3}{2+3i}\hspace{1.0cm}\text{vi) }\frac{(1+i)(1+\sqrt{3}i)}{1-i}$
$\displaystyle \text{vii) }\frac{2+3i}{4+5i}\hspace{1.0cm}\text{viii) }\frac{(1-i)^3}{1-i^3}\hspace{1.0cm}\text{ix) }(1+2i)^{-3}$
$\displaystyle \text{x) }\frac{3-4i}{(4+2i)(1+i)}$
$\displaystyle \text{xi) }\left(\frac{1}{1-4i}-\frac{2}{1+i}\right)\left(\frac{3-4i}{5+i}\right)\hspace{1.0cm}\text{xii) }\frac{5+\sqrt{2}i}{1-\sqrt{2}i}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) }(1+i)(1+2i)$
$\displaystyle =1+2i+i+2i^2=1+3i-2=-1+3i$

$\displaystyle \text{ii) }\frac{3+2i}{-2+i}$
$\displaystyle =\frac{3+2i}{-2+i}\times\frac{-2-i}{-2-i}$
$\displaystyle =\frac{-6-3i-4i-2i^2}{4-i^2}$
$\displaystyle =\frac{-4-7i}{5}=-\frac{4}{5}-\frac{7}{5}i$

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

$\displaystyle \text{iv) }\frac{1-i}{1+i}$
$\displaystyle =\frac{1-i}{1+i}\times\frac{1-i}{1-i}$
$\displaystyle =\frac{(1-i)^2}{1-i^2}=\frac{1-2i+i^2}{2}$
$\displaystyle =\frac{-2i}{2}=-i$

$\displaystyle \text{v) }\frac{(2+i)^3}{2+3i}$
$\displaystyle =\frac{(2+i)^2(2+i)}{2+3i}$
$\displaystyle =\frac{(3+4i)(2+i)}{2+3i}=\frac{2+11i}{2+3i}$
$\displaystyle =\frac{2+11i}{2+3i}\times\frac{2-3i}{2-3i}$
$\displaystyle =\frac{4-6i+22i-33i^2}{4-9i^2}$
$\displaystyle =\frac{37+16i}{13}=\frac{37}{13}+\frac{16}{13}i$

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

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

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

$\displaystyle \text{ix) }(1+2i)^{-3}$
$\displaystyle =\frac{1}{(1+2i)^3}$
$\displaystyle =\frac{1}{(1+2i)^2(1+2i)}$
$\displaystyle =\frac{1}{(-3+4i)(1+2i)}$
$\displaystyle =\frac{1}{-3-6i+4i+8i^2}=\frac{1}{-11-2i}$
$\displaystyle =\frac{1}{-11-2i}\times\frac{-11+2i}{-11+2i}$
$\displaystyle =\frac{-11+2i}{121+4}=-\frac{11}{125}+\frac{2}{125}i$

$\displaystyle \text{x) }\frac{3-4i}{(4+2i)(1+i)}$
$\displaystyle =\frac{3-4i}{4+4i+2i+2i^2}=\frac{3-4i}{2+6i}$
$\displaystyle =\frac{3-4i}{2+6i}\times\frac{2-6i}{2-6i}$
$\displaystyle =\frac{6-18i-8i+24i^2}{4-36i^2}$
$\displaystyle =\frac{-18-26i}{40}=-\frac{9}{20}-\frac{13}{20}i$

$\displaystyle \text{xi) }\left(\frac{1}{1-4i}-\frac{2}{1+i}\right)\left(\frac{3-4i}{5+i}\right)$
$\displaystyle \frac{1}{1-4i}=\frac{1+4i}{17}$
$\displaystyle \frac{2}{1+i}=\frac{2(1-i)}{2}=1-i$
$\displaystyle \therefore \frac{1}{1-4i}-\frac{2}{1+i}$
$\displaystyle =\frac{1+4i}{17}-(1-i)=\frac{-16+21i}{17}$
$\displaystyle \frac{3-4i}{5+i}=\frac{(3-4i)(5-i)}{(5+i)(5-i)}$
$\displaystyle =\frac{15-3i-20i+4i^2}{26}=\frac{11-23i}{26}$
$\displaystyle \therefore \left(\frac{-16+21i}{17}\right)\left(\frac{11-23i}{26}\right)$
$\displaystyle =\frac{-176+368i+231i-483i^2}{442}$
$\displaystyle =\frac{307+599i}{442}=\frac{307}{442}+\frac{599}{442}i$

$\displaystyle \text{xii) }\frac{5+\sqrt{2}i}{1-\sqrt{2}i}$
$\displaystyle =\frac{5+\sqrt{2}i}{1-\sqrt{2}i}\times\frac{1+\sqrt{2}i}{1+\sqrt{2}i}$
$\displaystyle =\frac{5+5\sqrt{2}i+\sqrt{2}i+2i^2}{1-2i^2}$
$\displaystyle =\frac{3+6\sqrt{2}i}{3}=1+2\sqrt{2}i$

$\displaystyle \\$

$\displaystyle \textbf{Question 2: } \text{Find the real values of }x\text{ and }y,\text{ if:}$
$\displaystyle \text{i) }(x+iy)(2-3i)=4+i\hspace{1.0cm}\text{ii) }(3x-2iy)(2+i)^2=10(1+i)$
$\displaystyle \text{iii) }\frac{(1+i)x-2i}{3+i}-\frac{(2-3i)y+i}{3-i}=i$
$\displaystyle \text{iv) }(1+i)(x+iy)=2-5i$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) }(x+iy)(2-3i)=4+i$
$\displaystyle \Rightarrow 2x+2iy-3ix-3i^2y=4+i$
$\displaystyle \Rightarrow (2x+3y)+i(-3x+2y)=4+i$
$\displaystyle \text{Comparing the real and imaginary parts,}$
$\displaystyle 2x+3y=4\qquad\ldots\text{(i)}$
$\displaystyle -3x+2y=1\qquad\ldots\text{(ii)}$
$\displaystyle \text{Multiplying equation (i) by }3\text{ and equation (ii) by }2,$
$\displaystyle 6x+9y=12$
$\displaystyle -6x+4y=2$
$\displaystyle \text{Adding the equations,}$
$\displaystyle 13y=14\Rightarrow y=\frac{14}{13}$
$\displaystyle \text{Substituting }y=\frac{14}{13}\text{ in equation (i),}$
$\displaystyle 2x+3\left(\frac{14}{13}\right)=4$
$\displaystyle 2x=\frac{10}{13}\Rightarrow x=\frac{5}{13}$
$\displaystyle \therefore x=\frac{5}{13}\text{ and }y=\frac{14}{13}$

$\displaystyle \text{ii) }(3x-2iy)(2+i)^2=10(1+i)$
$\displaystyle \Rightarrow (3x-2iy)(4+4i+i^2)=10+10i$
$\displaystyle \Rightarrow (3x-2iy)(3+4i)=10+10i$
$\displaystyle \Rightarrow 9x+12ix-6iy-8i^2y=10+10i$
$\displaystyle \Rightarrow (9x+8y)+i(12x-6y)=10+10i$
$\displaystyle \text{Comparing the real and imaginary parts,}$
$\displaystyle 9x+8y=10\qquad\ldots\text{(i)}$
$\displaystyle 12x-6y=10\qquad\ldots\text{(ii)}$
$\displaystyle \text{Multiplying equation (i) by }6\text{ and equation (ii) by }8,$
$\displaystyle 54x+48y=60$
$\displaystyle 96x-48y=80$
$\displaystyle \text{Adding the equations,}$
$\displaystyle 150x=140\Rightarrow x=\frac{14}{15}$
$\displaystyle \text{Substituting }x=\frac{14}{15}\text{ in equation (i),}$
$\displaystyle 9\left(\frac{14}{15}\right)+8y=10$
$\displaystyle 8y=\frac{8}{5}\Rightarrow y=\frac{1}{5}$
$\displaystyle \therefore x=\frac{14}{15}\text{ and }y=\frac{1}{5}$

$\displaystyle \text{iii) }\frac{(1+i)x-2i}{3+i}-\frac{(2-3i)y+i}{3-i}=i$
$\displaystyle \Rightarrow \frac{[(1+i)x-2i](3-i)-[(2-3i)y+i](3+i)}{(3+i)(3-i)}=i$
$\displaystyle [(1+i)x-2i](3-i)=(4x-2)+i(2x-6)$
$\displaystyle [(2-3i)y+i](3+i)=(9y-1)+i(-7y+3)$
$\displaystyle \therefore \frac{(4x-9y-1)+i(2x+7y-9)}{10}=i$
$\displaystyle \Rightarrow (4x-9y-1)+i(2x+7y-9)=10i$
$\displaystyle \text{Comparing the real and imaginary parts,}$
$\displaystyle 4x-9y=1\qquad\ldots\text{(i)}$
$\displaystyle 2x+7y=19\qquad\ldots\text{(ii)}$
$\displaystyle \text{Multiplying equation (ii) by }2,$
$\displaystyle 4x+14y=38\qquad\ldots\text{(iii)}$
$\displaystyle \text{Subtracting equation (i) from equation (iii),}$
$\displaystyle 23y=37\Rightarrow y=\frac{37}{23}$
$\displaystyle \text{Substituting }y=\frac{37}{23}\text{ in equation (i),}$
$\displaystyle 4x-9\left(\frac{37}{23}\right)=1$
$\displaystyle 4x=\frac{356}{23}\Rightarrow x=\frac{89}{23}$
$\displaystyle \therefore x=\frac{89}{23}\text{ and }y=\frac{37}{23}$

$\displaystyle \text{iv) }(1+i)(x+iy)=2-5i$
$\displaystyle \Rightarrow x+iy+ix+i^2y=2-5i$
$\displaystyle \Rightarrow (x-y)+i(x+y)=2-5i$
$\displaystyle \text{Comparing the real and imaginary parts,}$
$\displaystyle x-y=2\qquad\ldots\text{(i)}$
$\displaystyle x+y=-5\qquad\ldots\text{(ii)}$
$\displaystyle \text{Adding equations (i) and (ii),}$
$\displaystyle 2x=-3\Rightarrow x=-\frac{3}{2}$
$\displaystyle \text{Substituting }x=-\frac{3}{2}\text{ in equation (i),}$
$\displaystyle -\frac{3}{2}-y=2$
$\displaystyle -y=\frac{7}{2}\Rightarrow y=-\frac{7}{2}$
$\displaystyle \therefore x=-\frac{3}{2}\text{ and }y=-\frac{7}{2}$

$\displaystyle \\$

$\displaystyle \textbf{Question 3: } \text{Find the conjugates of the following complex numbers:}$
$\displaystyle \text{i) }4-5i\hspace{1.0cm}\text{ii) }\frac{1}{3+5i}\hspace{1.0cm}\text{iii) }\frac{1}{1+i}$
$\displaystyle \text{iv) }\frac{(3-i)^2}{2+i}\hspace{1.0cm}\text{v) }\frac{(1+i)(2+i)}{3+i}\hspace{1.0cm}\text{vi) }\frac{(3-2i)(2+3i)}{(1+2i)(2-i)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Note: If }z=x+iy,\text{ then the conjugate of }z\text{ is }\overline{z}=x-iy.$
$\displaystyle \text{i) }\text{If }z=4-5i,\text{ then }\overline{z}=4+5i.$

$\displaystyle \text{ii) }\text{If }z=\frac{1}{3+5i},$
$\displaystyle z=\frac{1}{3+5i}\times\frac{3-5i}{3-5i}=\frac{3-5i}{34}=\frac{3}{34}-\frac{5}{34}i.$
$\displaystyle \therefore \overline{z}=\frac{3}{34}+\frac{5}{34}i.$

$\displaystyle \text{iii) }\text{If }z=\frac{1}{1+i},$
$\displaystyle z=\frac{1}{1+i}\times\frac{1-i}{1-i}=\frac{1-i}{2}=\frac{1}{2}-\frac{1}{2}i.$
$\displaystyle \therefore \overline{z}=\frac{1}{2}+\frac{1}{2}i.$

$\displaystyle \text{iv) }\text{If }z=\frac{(3-i)^2}{2+i},$
$\displaystyle z=\frac{9+i^2-6i}{2+i}=\frac{8-6i}{2+i}\times\frac{2-i}{2-i}$
$\displaystyle =\frac{16-12i-8i+6i^2}{4-i^2}=\frac{10-20i}{5}=2-4i.$
$\displaystyle \therefore \overline{z}=2+4i.$

$\displaystyle \text{v) }\text{If }z=\frac{(1+i)(2+i)}{3+i},$
$\displaystyle z=\frac{2+2i+i+i^2}{3+i}=\frac{1+3i}{3+i}\times\frac{3-i}{3-i}$
$\displaystyle =\frac{3+9i-i-3i^2}{9-i^2}=\frac{6+8i}{10}=\frac{3}{5}+\frac{4}{5}i.$
$\displaystyle \therefore \overline{z}=\frac{3}{5}-\frac{4}{5}i.$

$\displaystyle \text{vi) }\text{If }z=\frac{(3-2i)(2+3i)}{(1+2i)(2-i)},$
$\displaystyle z=\frac{6-4i+9i-6i^2}{2+4i-i-2i^2}=\frac{12+5i}{4+3i}\times\frac{4-3i}{4-3i}$
$\displaystyle =\frac{48+20i-36i-15i^2}{16+9}=\frac{63-16i}{25}=\frac{63}{25}-\frac{16}{25}i.$
$\displaystyle \therefore \overline{z}=\frac{63}{25}+\frac{16}{25}i.$

$\displaystyle \\$

$\displaystyle \textbf{Question 4: } \text{Find the multiplicative inverse of the following complex numbers:}$
$\displaystyle \text{i) }1-i\hspace{1.0cm}\text{ii) }(1+i\sqrt{3})^2\hspace{1.0cm}\text{iii) }4-3i\hspace{1.0cm}\text{iv) }\sqrt{5}+3i$
$\displaystyle \text{Answer:}$
$\displaystyle \text{If }z=x+iy\text{ is a non-zero complex number, then its multiplicative inverse is }z^{-1}=\frac{1}{z}.$
$\displaystyle \text{i) }z=1-i$
$\displaystyle z^{-1}=\frac{1}{1-i}\times\frac{1+i}{1+i}$
$\displaystyle =\frac{1+i}{1-i^2}=\frac{1+i}{2}=\frac{1}{2}+\frac{1}{2}i$

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

$\displaystyle \text{iii) }z=4-3i$
$\displaystyle z^{-1}=\frac{1}{4-3i}\times\frac{4+3i}{4+3i}$
$\displaystyle =\frac{4+3i}{16+9}=\frac{4}{25}+\frac{3}{25}i$

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

$\displaystyle \\$

$\displaystyle \textbf{Question 5: } \text{If }z_1=2-i\text{ and }z_2=1+i,\text{ find }\left|\frac{z_1+z_2+1}{z_1-z_2+1}\right|.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{If }z=x+iy,\text{ then }|z|=\sqrt{x^2+y^2}.$
$\displaystyle z_1+z_2+1=(2-i)+(1+i)+1=4$
$\displaystyle z_1-z_2+1=(2-i)-(1+i)+1=2-2i$
$\displaystyle \therefore \frac{z_1+z_2+1}{z_1-z_2+1}=\frac{4}{2-2i}$
$\displaystyle =\frac{4}{2-2i}\times\frac{2+2i}{2+2i}$
$\displaystyle =\frac{8+8i}{4+4}=1+i$
$\displaystyle \therefore \left|\frac{z_1+z_2+1}{z_1-z_2+1}\right|=|1+i|$
$\displaystyle =\sqrt{1^2+1^2}=\sqrt{2}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: } \text{If }z_1=2-i\text{ and }z_2=-2+i,\text{ find:}$
$\displaystyle \text{i) }\text{Re}\left(\frac{z_1z_2}{z_1}\right)\hspace{1.0cm}\text{ii) }\text{Im}\left(\frac{1}{z_1z_2}\right)$
$\displaystyle \text{Answer:}$
$\displaystyle z_1=2-i,\qquad z_2=-2+i$
$\displaystyle \text{i) }\frac{z_1z_2}{z_1}=z_2,\text{ since }z_1\neq0$
$\displaystyle \therefore \frac{z_1z_2}{z_1}=-2+i$
$\displaystyle \therefore \text{Re}\left(\frac{z_1z_2}{z_1}\right)=-2$

$\displaystyle \text{ii) }z_1z_2=(2-i)(-2+i)$
$\displaystyle =-4+2i+2i-i^2=-3+4i$
$\displaystyle \therefore \frac{1}{z_1z_2}=\frac{1}{-3+4i}$
$\displaystyle =\frac{1}{-3+4i}\times\frac{-3-4i}{-3-4i}$
$\displaystyle =\frac{-3-4i}{9+16}=-\frac{3}{25}-\frac{4}{25}i$
$\displaystyle \therefore \text{Im}\left(\frac{1}{z_1z_2}\right)=-\frac{4}{25}$

$\displaystyle \\$

$\displaystyle \textbf{Question 7: } \text{Find the modulus of }\frac{1+i}{1-i}-\frac{1-i}{1+i}.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{1+i}{1-i}-\frac{1-i}{1+i}$
$\displaystyle =\frac{(1+i)^2-(1-i)^2}{(1-i)(1+i)}$
$\displaystyle =\frac{(1+2i+i^2)-(1-2i+i^2)}{1-i^2}$
$\displaystyle =\frac{2i+2i}{2}=2i$
$\displaystyle \therefore |2i|=\sqrt{0^2+2^2}=2$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: } \text{If }x+iy=\frac{a+ib}{a-ib},\text{ prove that }x^2+y^2=1.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }x+iy=\frac{a+ib}{a-ib},\text{ where }a^2+b^2\neq0.$
$\displaystyle \text{Taking the conjugate of both sides,}$
$\displaystyle \overline{x+iy}=\overline{\left(\frac{a+ib}{a-ib}\right)}$
$\displaystyle x-iy=\frac{\overline{a+ib}}{\overline{a-ib}}=\frac{a-ib}{a+ib}$
$\displaystyle \therefore (x+iy)(x-iy)=\frac{a+ib}{a-ib}\times\frac{a-ib}{a+ib}$
$\displaystyle x^2+y^2=\frac{(a+ib)(a-ib)}{(a-ib)(a+ib)}=1$
$\displaystyle \therefore x^2+y^2=1.$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: } \text{Find the least positive integral value of }n\text{ for which }\left(\frac{1+i}{1-i}\right)^n\text{ is real.}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{1+i}{1-i}=\frac{1+i}{1-i}\times\frac{1+i}{1+i}$
$\displaystyle =\frac{(1+i)^2}{1-i^2}=\frac{1+2i+i^2}{2}=i$
$\displaystyle \therefore \left(\frac{1+i}{1-i}\right)^n=i^n$
$\displaystyle \text{For }n=1,\qquad i^1=i,\text{ which is not real.}$
$\displaystyle \text{For }n=2,\qquad i^2=-1,\text{ which is real.}$
$\displaystyle \therefore \text{The least positive integral value of }n\text{ is }2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: } \text{Find the real values of }\theta\text{ for which the complex number}$
$\displaystyle \frac{1+i\cos\theta}{1-2i\cos\theta}\text{ is purely real.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }z=\frac{1+i\cos\theta}{1-2i\cos\theta}.$
$\displaystyle z=\frac{1+i\cos\theta}{1-2i\cos\theta}\times\frac{1+2i\cos\theta}{1+2i\cos\theta}$
$\displaystyle =\frac{(1+i\cos\theta)(1+2i\cos\theta)}{1+4\cos^2\theta}$
$\displaystyle =\frac{1+2i\cos\theta+i\cos\theta+2i^2\cos^2\theta}{1+4\cos^2\theta}$
$\displaystyle =\frac{1-2\cos^2\theta+3i\cos\theta}{1+4\cos^2\theta}$
$\displaystyle =\frac{1-2\cos^2\theta}{1+4\cos^2\theta}+i\left(\frac{3\cos\theta}{1+4\cos^2\theta}\right)$
$\displaystyle \text{For }z\text{ to be purely real, }\text{Im}(z)=0.$
$\displaystyle \therefore \frac{3\cos\theta}{1+4\cos^2\theta}=0$
$\displaystyle \text{Since }1+4\cos^2\theta>0,\text{ we get }\cos\theta=0.$
$\displaystyle \therefore \theta=\frac{(2n+1)\pi}{2},\qquad n\in\mathbb{Z}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: } \text{Find the smallest positive integer value of }n\text{ for which}$
$\displaystyle \frac{(1+i)^n}{(1-i)^{n-2}}\text{ is a real number.}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{(1+i)^n}{(1-i)^{n-2}}$
$\displaystyle =\frac{(1+i)^n}{(1-i)^n}\times(1-i)^2$
$\displaystyle =\left(\frac{1+i}{1-i}\right)^n(1-i)^2$
$\displaystyle =\left(\frac{1+i}{1-i}\times\frac{1+i}{1+i}\right)^n(1-2i+i^2)$
$\displaystyle =\left(\frac{1+2i+i^2}{1-i^2}\right)^n(-2i)$
$\displaystyle =\left(\frac{2i}{2}\right)^n(-2i)$
$\displaystyle =-2i^{n+1}$
$\displaystyle \text{For the expression to be real, }n+1\text{ must be even.}$
$\displaystyle \therefore n\text{ must be an odd positive integer.}$
$\displaystyle \text{The smallest positive odd integer is }n=1.$
$\displaystyle \text{Indeed, for }n=1,\qquad -2i^{1+1}=-2i^2=2,\text{ which is real.}$
$\displaystyle \therefore \text{The smallest positive integer value of }n\text{ is }1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: } \text{If }\left(\frac{1+i}{1-i}\right)^3-\left(\frac{1-i}{1+i}\right)^3=x+iy,\text{ find }(x,y).$
$\displaystyle \text{Answer:}$
$\displaystyle \left(\frac{1+i}{1-i}\right)^3-\left(\frac{1-i}{1+i}\right)^3=x+iy$
$\displaystyle \Rightarrow \left(\frac{1+i}{1-i}\times\frac{1+i}{1+i}\right)^3-\left(\frac{1-i}{1+i}\times\frac{1-i}{1-i}\right)^3=x+iy$
$\displaystyle \Rightarrow \left(\frac{(1+i)^2}{1-i^2}\right)^3-\left(\frac{(1-i)^2}{1-i^2}\right)^3=x+iy$
$\displaystyle \Rightarrow \left(\frac{2i}{2}\right)^3-\left(\frac{-2i}{2}\right)^3=x+iy$
$\displaystyle \Rightarrow i^3-(-i)^3=x+iy$
$\displaystyle \Rightarrow -i-i=x+iy$
$\displaystyle \Rightarrow -2i=x+iy$
$\displaystyle \text{Comparing the real and imaginary parts,}$
$\displaystyle x=0\text{ and }y=-2$
$\displaystyle \therefore (x,y)=(0,-2)$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: } \text{If }\frac{(1+i)^2}{2-i}=x+iy,\text{ find }x+y.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{(1+i)^2}{2-i}=x+iy$
$\displaystyle \Rightarrow \frac{2i}{2-i}\times\frac{2+i}{2+i}=x+iy$
$\displaystyle \Rightarrow \frac{4i+2i^2}{4+1}=x+iy$
$\displaystyle \Rightarrow \frac{-2+4i}{5}=x+iy$
$\displaystyle \text{Comparing the real and imaginary parts,}$
$\displaystyle x=-\frac{2}{5}\text{ and }y=\frac{4}{5}$
$\displaystyle \therefore x+y=-\frac{2}{5}+\frac{4}{5}=\frac{2}{5}$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: } \text{If }\left(\frac{1-i}{1+i}\right)^{100}=a+ib,\text{ find }(a,b).$
$\displaystyle \text{Answer:}$
$\displaystyle \left(\frac{1-i}{1+i}\right)^{100}=a+ib$
$\displaystyle \Rightarrow \left(\frac{1-i}{1+i}\times\frac{1-i}{1-i}\right)^{100}=a+ib$
$\displaystyle \Rightarrow \left(\frac{(1-i)^2}{1-i^2}\right)^{100}=a+ib$
$\displaystyle \Rightarrow \left(\frac{-2i}{2}\right)^{100}=a+ib$
$\displaystyle \Rightarrow (-i)^{100}=a+ib$
$\displaystyle \Rightarrow \left((-i)^4\right)^{25}=a+ib$
$\displaystyle \Rightarrow 1=a+ib$
$\displaystyle \text{Comparing the real and imaginary parts,}$
$\displaystyle a=1\text{ and }b=0$
$\displaystyle \therefore (a,b)=(1,0)$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: } \text{If }a=\cos\theta+i\sin\theta,\text{ find the value of }\frac{1+a}{1-a}.$
$\displaystyle \text{Answer:}$
$\displaystyle a=\cos\theta+i\sin\theta$
$\displaystyle \frac{1+a}{1-a}=\frac{(1+\cos\theta)+i\sin\theta}{(1-\cos\theta)-i\sin\theta}$
$\displaystyle =\frac{(1+\cos\theta)+i\sin\theta}{(1-\cos\theta)-i\sin\theta}\times\frac{(1-\cos\theta)+i\sin\theta}{(1-\cos\theta)+i\sin\theta}$
$\displaystyle =\frac{[(1+\cos\theta)+i\sin\theta][(1-\cos\theta)+i\sin\theta]}{(1-\cos\theta)^2+\sin^2\theta}$
$\displaystyle =\frac{1-\cos^2\theta-\sin^2\theta+2i\sin\theta}{1+\cos^2\theta-2\cos\theta+\sin^2\theta}$
$\displaystyle =\frac{2i\sin\theta}{2(1-\cos\theta)}$
$\displaystyle =\frac{i\sin\theta}{1-\cos\theta}$
$\displaystyle =i\left(\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}}\right)$
$\displaystyle =i\cot\frac{\theta}{2}$
$\displaystyle \therefore \frac{1+a}{1-a}=i\cot\frac{\theta}{2},\qquad \theta\neq2n\pi,\ n\in\mathbb{Z}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: } \text{Evaluate the following:}$
$\displaystyle \text{i) }2x^3+2x^2-7x+72,\text{ when }x=\frac{3-5i}{2}$
$\displaystyle \text{ii) }x^4-4x^3+4x^2+8x+44,\text{ when }x=3+2i$
$\displaystyle \text{iii) }x^4+4x^3+6x^2+4x+9,\text{ when }x=-1+i\sqrt{2}$
$\displaystyle \text{iv) }x^6+x^4+x^2+1,\text{ when }x=\frac{1+i}{\sqrt{2}}$
$\displaystyle \text{v) }2x^4+5x^3+7x^2-x+41,\text{ when }x=-2-\sqrt{3}i$
$\displaystyle \text{Answer:}$
$\displaystyle \text{i) }x=\frac{3-5i}{2}$
$\displaystyle \Rightarrow 2x-3=-5i$
$\displaystyle \Rightarrow (2x-3)^2=(-5i)^2$
$\displaystyle \Rightarrow 4x^2-12x+9=-25$
$\displaystyle \Rightarrow 2x^2-6x+17=0$
$\displaystyle \text{Now, }2x^3+2x^2-7x+72$
$\displaystyle =x(2x^2-6x+17)+8x^2-24x+72$
$\displaystyle =x(0)+4(2x^2-6x+17)+4$
$\displaystyle =0+4(0)+4=4$

$\displaystyle \text{ii) }x=3+2i$
$\displaystyle \Rightarrow x-3=2i$
$\displaystyle \Rightarrow (x-3)^2=(2i)^2$
$\displaystyle \Rightarrow x^2-6x+9=-4$
$\displaystyle \Rightarrow x^2-6x+13=0$
$\displaystyle \text{Now, }x^4-4x^3+4x^2+8x+44$
$\displaystyle =x^2(x^2-6x+13)+2x^3-9x^2+8x+44$
$\displaystyle =2x(x^2-6x+13)+3x^2-18x+44$
$\displaystyle =2x(0)+3(x^2-6x+13)+5$
$\displaystyle =0+3(0)+5=5$

$\displaystyle \text{iii) }x=-1+i\sqrt{2}$
$\displaystyle \Rightarrow x+1=i\sqrt{2}$
$\displaystyle \Rightarrow (x+1)^2=(i\sqrt{2})^2$
$\displaystyle \Rightarrow x^2+2x+1=-2$
$\displaystyle \Rightarrow x^2+2x+3=0$
$\displaystyle \text{Now, }x^4+4x^3+6x^2+4x+9$
$\displaystyle =(x^2+2x)^2+2x^2+4x+9$
$\displaystyle =(-3)^2+2(x^2+2x)+9$
$\displaystyle =9+2(-3)+9=12$

$\displaystyle \text{iv) }x=\frac{1+i}{\sqrt{2}}$
$\displaystyle \Rightarrow x^2=\frac{(1+i)^2}{2}=\frac{2i}{2}=i$
$\displaystyle \Rightarrow x^4=i^2=-1$
$\displaystyle \text{Now, }x^6+x^4+x^2+1$
$\displaystyle =x^2(x^4+1)+(x^4+1)$
$\displaystyle =(x^2+1)(x^4+1)$
$\displaystyle =(x^2+1)(-1+1)=0$

$\displaystyle \text{v) }x=-2-\sqrt{3}i$
$\displaystyle \Rightarrow x+2=-\sqrt{3}i$
$\displaystyle \Rightarrow (x+2)^2=(-\sqrt{3}i)^2$
$\displaystyle \Rightarrow x^2+4x+4=-3$
$\displaystyle \Rightarrow x^2+4x+7=0$
$\displaystyle \text{Now, }2x^4+5x^3+7x^2-x+41$
$\displaystyle =2x^2(x^2+4x+7)-3x^3-7x^2-x+41$
$\displaystyle =-3x(x^2+4x+7)+5x^2+20x+41$
$\displaystyle =5(x^2+4x+7)+6$
$\displaystyle =5(0)+6=6$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: } \text{For a positive integer }n,\text{ find the value of }(1-i)^n\left(1-\frac{1}{i}\right)^n.$
$\displaystyle \text{Answer:}$
$\displaystyle (1-i)^n\left(1-\frac{1}{i}\right)^n$
$\displaystyle =\left[(1-i)\left(1-\frac{1}{i}\right)\right]^n$
$\displaystyle =\left[\frac{(1-i)(i-1)}{i}\right]^n$
$\displaystyle =\left[\frac{i-1-i^2+i}{i}\right]^n$
$\displaystyle =\left[\frac{2i}{i}\right]^n$
$\displaystyle =2^n$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: } \text{If }(1+i)z=(1-i)\overline{z},\text{ show that }z=-i\overline{z}.$
$\displaystyle \text{Answer:}$
$\displaystyle (1+i)z=(1-i)\overline{z}$
$\displaystyle \Rightarrow z=\left(\frac{1-i}{1+i}\right)\overline{z}$
$\displaystyle \Rightarrow z=\left(\frac{1-i}{1+i}\times\frac{1-i}{1-i}\right)\overline{z}$
$\displaystyle \Rightarrow z=\left(\frac{(1-i)^2}{1-i^2}\right)\overline{z}$
$\displaystyle \Rightarrow z=\left(\frac{-2i}{2}\right)\overline{z}$
$\displaystyle \Rightarrow z=-i\overline{z}$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: } \text{Solve the system of equations }\mathrm{Re}(z^2)=0,\ |z|=2.$
$\displaystyle \text{Answer:}$
$\displaystyle \mathrm{Re}(z^2)=0,\qquad |z|=2$
$\displaystyle \text{Let }z=x+iy.$
$\displaystyle z^2=(x+iy)^2=(x^2-y^2)+2xy\,i$
$\displaystyle \mathrm{Re}(z^2)=0$
$\displaystyle \Rightarrow x^2-y^2=0\qquad\ldots\text{(i)}$
$\displaystyle |z|=2$
$\displaystyle \Rightarrow \sqrt{x^2+y^2}=2$
$\displaystyle \Rightarrow x^2+y^2=4\qquad\ldots\text{(ii)}$
$\displaystyle \text{Adding (i) and (ii),}$
$\displaystyle 2x^2=4$
$\displaystyle \Rightarrow x^2=2$
$\displaystyle \Rightarrow x=\pm\sqrt2$
$\displaystyle \text{From (i), }x^2=y^2\Rightarrow y=\pm x.$
$\displaystyle \therefore y=\pm\sqrt2$
$\displaystyle \therefore z=x+iy=\pm\sqrt2\pm i\sqrt2$
$\displaystyle \text{Hence the four solutions are }$
$\displaystyle z=\sqrt2(1+i),\ \sqrt2(1-i),\ \sqrt2(-1+i),\ \sqrt2(-1-i).$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: } \text{If }\frac{z-1}{z+1}\text{ is a purely imaginary number, where }z\neq-1,\text{ find }|z|.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }z=x+iy.$
$\displaystyle \frac{z-1}{z+1}=\frac{x+iy-1}{x+iy+1}$
$\displaystyle =\frac{(x-1)+iy}{(x+1)+iy}$
$\displaystyle =\frac{(x-1)+iy}{(x+1)+iy}\times\frac{(x+1)-iy}{(x+1)-iy}$
$\displaystyle =\frac{[(x-1)+iy][(x+1)-iy]}{(x+1)^2+y^2}$
$\displaystyle =\frac{x^2-1+y^2+2iy}{(x+1)^2+y^2}$
$\displaystyle =\frac{x^2+y^2-1}{(x+1)^2+y^2}+i\left(\frac{2y}{(x+1)^2+y^2}\right)$
$\displaystyle \text{Since }\frac{z-1}{z+1}\text{ is purely imaginary,}$
$\displaystyle \mathrm{Re}\left(\frac{z-1}{z+1}\right)=0.$
$\displaystyle \therefore \frac{x^2+y^2-1}{(x+1)^2+y^2}=0$
$\displaystyle \text{Since }z\neq-1,\ (x+1)^2+y^2\neq0.$
$\displaystyle \therefore x^2+y^2-1=0$
$\displaystyle \Rightarrow x^2+y^2=1$
$\displaystyle \Rightarrow |z|=\sqrt{x^2+y^2}=1$
$\displaystyle \therefore |z|=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: } \text{If }z_1\text{ is a complex number other than }-1\text{ such that }|z_1|=1$
$\displaystyle \text{and }z_2=\frac{z_1-1}{z_1+1},\text{ show that the real part of }z_2\text{ is zero.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }z_1=x_1+iy_1\text{ and }z_2=x_2+iy_2.$
$\displaystyle |z_1|=1$
$\displaystyle \Rightarrow \sqrt{x_1^2+y_1^2}=1$
$\displaystyle \Rightarrow x_1^2+y_1^2=1\qquad\ldots\text{(i)}$
$\displaystyle z_2=\frac{z_1-1}{z_1+1}$
$\displaystyle =\frac{(x_1-1)+iy_1}{(x_1+1)+iy_1}$
$\displaystyle =\frac{(x_1-1)+iy_1}{(x_1+1)+iy_1}\times\frac{(x_1+1)-iy_1}{(x_1+1)-iy_1}$
$\displaystyle =\frac{[(x_1-1)+iy_1][(x_1+1)-iy_1]}{(x_1+1)^2+y_1^2}$
$\displaystyle =\frac{x_1^2+y_1^2-1+2iy_1}{(x_1+1)^2+y_1^2}$
$\displaystyle =\frac{x_1^2+y_1^2-1}{(x_1+1)^2+y_1^2}+i\left(\frac{2y_1}{(x_1+1)^2+y_1^2}\right)$
$\displaystyle \text{Using }x_1^2+y_1^2=1,$
$\displaystyle z_2=0+i\left(\frac{2y_1}{(x_1+1)^2+y_1^2}\right)$
$\displaystyle \therefore \mathrm{Re}(z_2)=0.$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: } \text{If }|z+1|=z+2(1+i),\text{ find }z.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }z=x+iy.$
$\displaystyle |z+1|=z+2(1+i)$
$\displaystyle \Rightarrow |x+1+iy|=(x+iy)+2(1+i)$
$\displaystyle \Rightarrow \sqrt{(x+1)^2+y^2}=(x+2)+i(y+2)$
$\displaystyle \text{Since }|z+1|\text{ is a real number,}$
$\displaystyle y+2=0$
$\displaystyle \Rightarrow y=-2$
$\displaystyle \therefore \sqrt{(x+1)^2+4}=x+2$
$\displaystyle \Rightarrow (x+1)^2+4=(x+2)^2$
$\displaystyle \Rightarrow x^2+2x+5=x^2+4x+4$
$\displaystyle \Rightarrow 2x=1$
$\displaystyle \Rightarrow x=\frac{1}{2}$
$\displaystyle \text{Since }x+2=\frac{5}{2}>0,\text{ the solution is valid.}$
$\displaystyle \therefore z=\frac{1}{2}-2i.$
$\displaystyle \\$

$\displaystyle \textbf{Question 23: } \text{Solve the equation }|z|=z+1+2i.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }z=x+iy.$
$\displaystyle |z|=z+1+2i$
$\displaystyle \Rightarrow |x+iy|=x+iy+1+2i$
$\displaystyle \Rightarrow \sqrt{x^2+y^2}=(x+1)+i(y+2)$
$\displaystyle \text{Since }|z|\text{ is a real number,}$
$\displaystyle y+2=0$
$\displaystyle \Rightarrow y=-2$
$\displaystyle \therefore \sqrt{x^2+y^2}=x+1$
$\displaystyle \Rightarrow x^2+y^2=(x+1)^2$
$\displaystyle \Rightarrow x^2+4=x^2+2x+1$
$\displaystyle \Rightarrow 2x=3$
$\displaystyle \Rightarrow x=\frac{3}{2}$
$\displaystyle \text{Since }x+1=\frac{5}{2}>0,\text{ the solution is valid.}$
$\displaystyle \therefore z=\frac{3}{2}-2i.$
$\displaystyle \\$

$\displaystyle \textbf{Question 24: } \text{What is the smallest positive integer }n\text{ for which }(1+i)^{2n}=(1-i)^{2n}\ ?$
$\displaystyle \text{Answer:}$
$\displaystyle (1+i)^{2n}=(1-i)^{2n}$
$\displaystyle \Rightarrow \left((1+i)^2\right)^n=\left((1-i)^2\right)^n$
$\displaystyle \Rightarrow (2i)^n=(-2i)^n$
$\displaystyle \Rightarrow (2i)^n=(-1)^n(2i)^n$
$\displaystyle \Rightarrow (-1)^n=1,\qquad \text{since }(2i)^n\neq0$
$\displaystyle \therefore n\text{ is an even positive integer.}$
$\displaystyle \therefore \text{The smallest positive integer }n\text{ is }2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 25: } \text{If }z_1,z_2,z_3\text{ are complex numbers such that}$
$\displaystyle |z_1|=|z_2|=|z_3|=\left|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}\right|=1,\text{ find }|z_1+z_2+z_3|.$
$\displaystyle \text{Answer:}$
$\displaystyle |z_1|=|z_2|=|z_3|=1$
$\displaystyle \therefore z_1\overline{z_1}=z_2\overline{z_2}=z_3\overline{z_3}=1$
$\displaystyle \Rightarrow \frac{1}{z_1}=\overline{z_1},\qquad \frac{1}{z_2}=\overline{z_2},\qquad \frac{1}{z_3}=\overline{z_3}$
$\displaystyle \therefore \frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}=\overline{z_1}+\overline{z_2}+\overline{z_3}$
$\displaystyle =\overline{z_1+z_2+z_3}$
$\displaystyle \therefore \left|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}\right|=\left|\overline{z_1+z_2+z_3}\right|$
$\displaystyle =|z_1+z_2+z_3|$
$\displaystyle \text{But }\left|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}\right|=1$
$\displaystyle \therefore |z_1+z_2+z_3|=1$
$\displaystyle \\$

$\displaystyle \textbf{Question 26: } \text{Find the number of solutions of }z^2+|z|^2=0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }z=x+iy.$
$\displaystyle \therefore |z|=\sqrt{x^2+y^2}$
$\displaystyle z^2+|z|^2=0$
$\displaystyle \Rightarrow (x+iy)^2+\left(\sqrt{x^2+y^2}\right)^2=0$
$\displaystyle \Rightarrow (x^2-y^2)+2ixy+x^2+y^2=0$
$\displaystyle \Rightarrow 2x^2+2ixy=0$
$\displaystyle \text{Equating the real and imaginary parts,}$
$\displaystyle 2x^2=0,\qquad 2xy=0$
$\displaystyle \Rightarrow x=0$
$\displaystyle \therefore z=iy,\qquad y\in\mathbb{R}$
$\displaystyle \text{Indeed, }(iy)^2+|iy|^2=-y^2+y^2=0.$
$\displaystyle \therefore \text{There are infinitely many solutions, namely }z=iy,\ y\in\mathbb{R}.$
$\displaystyle \\$