\displaystyle \text{MULTIPLE CHOICE QUESTIONS (MCQs)}

\displaystyle \textbf{Question 1: }\text{The value of }(1+i)(1+i^2)(1+i^3)(1+i^4)\text{ is}
\displaystyle \text{(a) }2\qquad\text{(b) }0\qquad\text{(c) }1\qquad\text{(d) }i
\displaystyle \text{Answer:}
\displaystyle i^2=-1.
\displaystyle \therefore 1+i^2=1-1=0.
\displaystyle \therefore (1+i)(1+i^2)(1+i^3)(1+i^4)=0.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If }\frac{3+2i\sin\theta}{1-2i\sin\theta} \text{ is a real number and }0<\theta<2\pi,\text{ then }\theta=
\displaystyle \text{(a) }\pi\qquad\text{(b) }\frac{\pi}{2}\qquad \text{(c) }\frac{\pi}{3}\qquad\text{(d) }\frac{\pi}{6}
\displaystyle \text{Answer:}
\displaystyle \text{Let }s=\sin\theta.
\displaystyle \frac{3+2is}{1-2is} =\frac{(3+2is)(1+2is)}{1+4s^2}.
\displaystyle =\frac{3-4s^2+8is}{1+4s^2}.
\displaystyle \text{For the number to be real, its imaginary part must be zero.}
\displaystyle \therefore \frac{8s}{1+4s^2}=0 \Rightarrow s=0.
\displaystyle \therefore \sin\theta=0.
\displaystyle \text{Since }0<\theta<2\pi,\quad\theta=\pi.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If }(1+i)(1+2i)(1+3i)\cdots(1+ni)=a+ib,\text{ then}
\displaystyle 2\times5\times10\times\cdots\times(1+n^2)\text{ is equal to}
\displaystyle \text{(a) }\sqrt{a^2+b^2}\qquad \text{(b) }\sqrt{a^2-b^2}\qquad \text{(c) }a^2+b^2\qquad\text{(d) }a^2-b^2
\displaystyle \text{Answer:}
\displaystyle (1+i)(1+2i)(1+3i)\cdots(1+ni)=a+ib.
\displaystyle \text{Taking modulus squared on both sides,}
\displaystyle |1+i|^2|1+2i|^2|1+3i|^2\cdots|1+ni|^2=|a+ib|^2.
\displaystyle \therefore 2\times5\times10\times\cdots\times(1+n^2)=a^2+b^2.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }\sqrt{a+ib}=x+iy,\text{ then possible value of }\sqrt{a-ib}\text{ is}
\displaystyle \text{(a) }x^2+y^2\qquad \text{(b) }\sqrt{x^2+y^2}\qquad \text{(c) }x+iy\qquad\text{(d) }x-iy
\displaystyle \text{Answer:}
\displaystyle \sqrt{a+ib}=x+iy.
\displaystyle \therefore a+ib=(x+iy)^2.
\displaystyle \text{Taking conjugates,}\quad a-ib=(x-iy)^2.
\displaystyle \therefore \sqrt{a-ib}=\pm(x-iy).
\displaystyle \therefore x-iy\text{ is a possible value of }\sqrt{a-ib}.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }z=\cos\frac{\pi}{4} +i\sin\frac{\pi}{6},\text{ then}
\displaystyle \text{(a) }|z|=1,\ \arg(z)=\frac{\pi}{4}\qquad \text{(b) }|z|=1,\ \arg(z)=\frac{\pi}{6}
\displaystyle \text{(c) }|z|=\frac{\sqrt3}{2},\ \arg(z)=\frac{5\pi}{24}
\displaystyle \text{(d) }|z|=\frac{\sqrt3}{2},\ \arg(z)=\tan^{-1}\frac{1}{\sqrt2}
\displaystyle \text{Answer:}
\displaystyle z=\frac{1}{\sqrt2}+\frac{i}{2}.
\displaystyle |z|=\sqrt{\frac12+\frac14} =\sqrt{\frac34}=\frac{\sqrt3}{2}.
\displaystyle \tan(\arg z) =\frac{\text{Im}(z)}{\text{Re}(z)} =\frac{1/2}{1/\sqrt2}=\frac{1}{\sqrt2}.
\displaystyle \text{Since }z\text{ lies in the first quadrant,}
\displaystyle \arg(z)=\tan^{-1}\frac{1}{\sqrt2}.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The polar form of }(i^{25})^3\text{ is}
\displaystyle \text{(a) }\cos\frac{\pi}{2}+i\sin\frac{\pi}{2} \qquad\text{(b) }\cos\pi+i\sin\pi
\displaystyle \text{(c) }\cos\pi-i\sin\pi \qquad\text{(d) }\cos\frac{\pi}{2}-i\sin\frac{\pi}{2}
\displaystyle \text{Answer:}
\displaystyle i^{25}=i^{24}\cdot i=(i^4)^6i=i.
\displaystyle \therefore (i^{25})^3=i^3=-i.
\displaystyle -i=\cos\frac{\pi}{2}-i\sin\frac{\pi}{2}.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{If }i^2=-1,\text{ then the sum } i+i^2+i^3+\cdots\text{ up to }1000\text{ terms is equal to}
\displaystyle \text{(a) }1\qquad\text{(b) }-1\qquad \text{(c) }i\qquad\text{(d) }0
\displaystyle \text{Answer:}
\displaystyle i+i^2+i^3+i^4=i-1-i+1=0.
\displaystyle 1000=4\times250.
\displaystyle \therefore i+i^2+i^3+\cdots+i^{1000}=250(0)=0.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If }z=\frac{-2}{1+i\sqrt3}, \text{ then the value of }\arg(z)\text{ is}
\displaystyle \text{(a) }\pi\qquad\text{(b) }\frac{\pi}{3}\qquad \text{(c) }\frac{2\pi}{3}\qquad\text{(d) }\frac{\pi}{4}
\displaystyle \text{Answer:}
\displaystyle z=\frac{-2}{1+i\sqrt3}\times \frac{1-i\sqrt3}{1-i\sqrt3}.
\displaystyle =\frac{-2(1-i\sqrt3)}{1+3} =-\frac12+\frac{\sqrt3}{2}i.
\displaystyle \text{Thus, }z\text{ lies in the second quadrant.}
\displaystyle \tan\alpha= \left|\frac{\sqrt3/2}{-1/2}\right|=\sqrt3 \Rightarrow\alpha=\frac{\pi}{3}.
\displaystyle \therefore \arg(z)=\pi-\frac{\pi}{3} =\frac{2\pi}{3}.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If }a=\cos\theta+i\sin\theta, \text{ then }\frac{1+a}{1-a}=
\displaystyle \text{(a) }\cot\frac{\theta}{2}\qquad \text{(b) }\cot\theta\qquad \text{(c) }i\cot\frac{\theta}{2}\qquad \text{(d) }i\tan\frac{\theta}{2}
\displaystyle \text{Answer:}
\displaystyle \frac{1+a}{1-a} =\frac{1+\cos\theta+i\sin\theta} {1-\cos\theta-i\sin\theta}.
\displaystyle 1+\cos\theta=2\cos^2\frac{\theta}{2}, \quad 1-\cos\theta=2\sin^2\frac{\theta}{2}.
\displaystyle \sin\theta=2\sin\frac{\theta}{2} \cos\frac{\theta}{2}.
\displaystyle \therefore \frac{1+a}{1-a} =i\cot\frac{\theta}{2}.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }(1+i)(1+2i)(1+3i)\cdots (1+ni)=a+ib,\text{ then}
\displaystyle 2\cdot5\cdot10\cdot17\cdots(1+n^2)=
\displaystyle \text{(a) }a-ib\qquad\text{(b) }a^2-b^2\qquad \text{(c) }a^2+b^2\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle (1+i)(1+2i)(1+3i)\cdots(1+ni)=a+ib.
\displaystyle \text{Taking modulus squared on both sides,}
\displaystyle |1+i|^2|1+2i|^2|1+3i|^2\cdots|1+ni|^2 =|a+ib|^2.
\displaystyle \therefore 2\cdot5\cdot10\cdot17\cdots(1+n^2) =a^2+b^2.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If } \frac{(a^2+1)^2}{2a-i}=x+iy,\text{ then }x^2+y^2\text{ is equal to}
\displaystyle \text{(a) }\frac{(a^2+1)^4}{4a^2+1}\qquad \text{(b) }\frac{(a+1)^2}{4a^2+1}
\displaystyle \text{(c) }\frac{(a^2-1)^2}{(4a^2-1)^2}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle x+iy=\frac{(a^2+1)^2}{2a-i}.
\displaystyle x^2+y^2=|x+iy|^2.
\displaystyle =\left|\frac{(a^2+1)^2}{2a-i}\right|^2.
\displaystyle =\frac{|(a^2+1)^2|^2}{|2a-i|^2}.
\displaystyle =\frac{(a^2+1)^4}{4a^2+1}.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{The principal value of the amplitude of }(1+i)\text{ is}
\displaystyle \text{(a) }\frac{\pi}{4}\qquad \text{(b) }\frac{\pi}{12}\qquad \text{(c) }\frac{3\pi}{4}\qquad\text{(d) }\pi
\displaystyle \text{Answer:}
\displaystyle z=1+i.
\displaystyle \tan\theta=\frac{\text{Im}(z)}{\text{Re}(z)}=1.
\displaystyle \text{Since }z\text{ lies in the first quadrant, } \theta=\frac{\pi}{4}.
\displaystyle \therefore \text{the principal amplitude is }\frac{\pi}{4}.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{The least positive integer }n \text{ such that }\left(\frac{2i}{1+i}\right)^n \text{ is a positive integer, is}
\displaystyle \text{(a) }16\qquad\text{(b) }8\qquad \text{(c) }4\qquad\text{(d) }2
\displaystyle \text{Answer:}
\displaystyle \frac{2i}{1+i} =\frac{2i(1-i)}{(1+i)(1-i)}=1+i.
\displaystyle 1+i=\sqrt2\left(\cos\frac{\pi}{4} +i\sin\frac{\pi}{4}\right).
\displaystyle \therefore (1+i)^n=(\sqrt2)^n \left(\cos\frac{n\pi}{4}+i\sin\frac{n\pi}{4}\right).
\displaystyle \text{For a positive integer, }\frac{n\pi}{4}=2k\pi.
\displaystyle \therefore n=8k.
\displaystyle \text{The least positive value is }n=8.
\displaystyle \text{Indeed, }(1+i)^8=16.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If }z\text{ is a non-zero complex number, then }\left|\frac{\bar z}{z\bar z}\right|^2 \text{ is equal to}
\displaystyle \text{(a) }\frac{|\bar z|}{|z|^3}\qquad \text{(b) }|z|\qquad\text{(c) }|\bar z|\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Since }z\neq0,\quad z\bar z=|z|^2.
\displaystyle \left|\frac{\bar z}{z\bar z}\right|^2 =\left|\frac{1}{z}\right|^2.
\displaystyle =\frac{1}{|z|^2}.
\displaystyle \text{Also, }\frac{|\bar z|}{|z|^3} =\frac{|z|}{|z|^3}=\frac{1}{|z|^2}.
\displaystyle \therefore \left|\frac{\bar z}{z\bar z}\right|^2 =\frac{|\bar z|}{|z|^3}.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{If }a=1+i,\text{ then }a^2\text{ equals}
\displaystyle \text{(a) }1-i\qquad\text{(b) }2i\qquad \text{(c) }(1+i)(1-i)\qquad\text{(d) }i-1
\displaystyle \text{Answer:}
\displaystyle a^2=(1+i)^2=1+2i+i^2.
\displaystyle =1+2i-1=2i.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{If }(x+iy)^{1/3}=a+ib, \text{ then }\frac{x}{a}+\frac{y}{b}=
\displaystyle \text{(a) }0\qquad\text{(b) }1\qquad \text{(c) }-1\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle (x+iy)^{1/3}=a+ib.
\displaystyle \therefore x+iy=(a+ib)^3.
\displaystyle =a^3+3a^2(ib)+3a(ib)^2+(ib)^3.
\displaystyle =(a^3-3ab^2)+i(3a^2b-b^3).
\displaystyle \therefore x=a(a^2-3b^2),\quad y=b(3a^2-b^2).
\displaystyle \therefore \frac{x}{a}+\frac{y}{b} =a^2-3b^2+3a^2-b^2.
\displaystyle =4(a^2-b^2).
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 17: }(\sqrt{-2})(\sqrt{-3}) \text{ is equal to}
\displaystyle \text{(a) }\sqrt6\qquad\text{(b) }-\sqrt6\qquad \text{(c) }i\sqrt6\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \sqrt{-2}=i\sqrt2,\qquad\sqrt{-3}=i\sqrt3.
\displaystyle \therefore (\sqrt{-2})(\sqrt{-3}) =(i\sqrt2)(i\sqrt3).
\displaystyle =i^2\sqrt6=-\sqrt6.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{The argument of } \frac{1-i\sqrt3}{1+i\sqrt3}\text{ is}
\displaystyle \text{(a) }60^\circ\qquad\text{(b) }120^\circ\qquad \text{(c) }210^\circ\qquad\text{(d) }240^\circ
\displaystyle \text{Answer:}
\displaystyle z=\frac{1-i\sqrt3}{1+i\sqrt3} \times\frac{1-i\sqrt3}{1-i\sqrt3}.
\displaystyle =\frac{(1-i\sqrt3)^2}{1+3} =-\frac12-\frac{\sqrt3}{2}i.
\displaystyle \text{Thus, }z\text{ lies in the third quadrant.}
\displaystyle \tan\alpha= \left|\frac{-\sqrt3/2}{-1/2}\right|=\sqrt3 \Rightarrow\alpha=60^\circ.
\displaystyle \therefore \arg(z)=180^\circ+60^\circ=240^\circ.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{If }z= \left(\frac{1+i}{1-i}\right),\text{ then }z^4\text{ equals}
\displaystyle \text{(a) }1\qquad\text{(b) }-1\qquad \text{(c) }0\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle z=\frac{1+i}{1-i} \times\frac{1+i}{1+i}.
\displaystyle =\frac{(1+i)^2}{1-i^2} =\frac{2i}{2}=i.
\displaystyle \therefore z^4=i^4=1.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{If }z= \frac{1+2i}{1-(1-i)^2},\text{ then }\arg(z)\text{ equals}
\displaystyle \text{(a) }0\qquad\text{(b) }\frac{\pi}{2}\qquad \text{(c) }\pi\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle (1-i)^2=1-2i+i^2=-2i.
\displaystyle \therefore 1-(1-i)^2=1+2i.
\displaystyle \therefore z=\frac{1+2i}{1+2i}=1.
\displaystyle \therefore \arg(z)=0.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{If }z= \frac{1}{(2+3i)^2},\text{ then }|z|=
\displaystyle \text{(a) }\frac{1}{13}\qquad \text{(b) }\frac{1}{5}\qquad \text{(c) }\frac{1}{12}\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle |z|=\left|\frac{1}{(2+3i)^2}\right| =\frac{1}{|2+3i|^2}.
\displaystyle |2+3i|^2=2^2+3^2=13.
\displaystyle \therefore |z|=\frac{1}{13}.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{If }z= \frac{1}{(1-i)(2+3i)},\text{ then }|z|=
\displaystyle \text{(a) }1\qquad \text{(b) }\frac{1}{\sqrt{26}}\qquad \text{(c) }\frac{5}{\sqrt{26}}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle |z|=\frac{1}{|1-i|\,|2+3i|}.
\displaystyle |1-i|=\sqrt{1^2+(-1)^2}=\sqrt2.
\displaystyle |2+3i|=\sqrt{2^2+3^2}=\sqrt{13}.
\displaystyle \therefore |z|=\frac{1}{\sqrt2\sqrt{13}} =\frac{1}{\sqrt{26}}.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{If }z=1-\cos\theta+i\sin\theta, \text{ then }|z|=
\displaystyle \text{(a) }2\sin\frac{\theta}{2}\qquad \text{(b) }2\cos\frac{\theta}{2}
\displaystyle \text{(c) }2\left|\sin\frac{\theta}{2}\right|\qquad \text{(d) }2\left|\cos\frac{\theta}{2}\right|
\displaystyle \text{Answer:}
\displaystyle |z|=\sqrt{(1-\cos\theta)^2+\sin^2\theta}.
\displaystyle =\sqrt{1-2\cos\theta+\cos^2\theta+\sin^2\theta}.
\displaystyle =\sqrt{2-2\cos\theta}.
\displaystyle =\sqrt{4\sin^2\frac{\theta}{2}} =2\left|\sin\frac{\theta}{2}\right|.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{If }x+iy=(1+i)(1+2i)(1+3i), \text{ then }x^2+y^2=
\displaystyle \text{(a) }0\qquad\text{(b) }1\qquad \text{(c) }100\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle x^2+y^2=|x+iy|^2.
\displaystyle =|(1+i)(1+2i)(1+3i)|^2.
\displaystyle =|1+i|^2|1+2i|^2|1+3i|^2.
\displaystyle =(1^2+1^2)(1^2+2^2)(1^2+3^2).
\displaystyle =2\times5\times10=100.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{If }z= \frac{1}{1-\cos\theta-i\sin\theta},\text{ then }\text{Re}(z)=
\displaystyle \text{(a) }0\qquad\text{(b) }\frac12\qquad \text{(c) }\cot\frac{\theta}{2}\qquad \text{(d) }\frac12\cot\frac{\theta}{2}
\displaystyle \text{Answer:}
\displaystyle z=\frac{1}{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)^2+\sin^2\theta}.
\displaystyle =\frac{1-\cos\theta+i\sin\theta} {2(1-\cos\theta)}.
\displaystyle =\frac12+\frac{i}{2} \frac{\sin\theta}{1-\cos\theta}.
\displaystyle =\frac12+\frac{i}{2}\cot\frac{\theta}{2}.
\displaystyle \therefore \text{Re}(z)=\frac12.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{If }x+iy= \frac{3+5i}{7-6i},\text{ then }y=
\displaystyle \text{(a) }\frac{9}{85}\qquad \text{(b) }-\frac{9}{85}\qquad \text{(c) }\frac{53}{85}\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle x+iy=\frac{3+5i}{7-6i} \times\frac{7+6i}{7+6i}.
\displaystyle =\frac{(3+5i)(7+6i)}{7^2+6^2}.
\displaystyle =\frac{21+18i+35i+30i^2}{85}.
\displaystyle =\frac{-9+53i}{85} =-\frac{9}{85}+\frac{53}{85}i.
\displaystyle \therefore y=\frac{53}{85}.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{If } \frac{1-ix}{1+ix}=a+ib,\text{ then }a^2+b^2=
\displaystyle \text{(a) }1\qquad\text{(b) }-1\qquad \text{(c) }0\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle a^2+b^2=|a+ib|^2.
\displaystyle =\left|\frac{1-ix}{1+ix}\right|^2.
\displaystyle =\frac{|1-ix|^2}{|1+ix|^2}.
\displaystyle =\frac{1+x^2}{1+x^2}=1.
\displaystyle \therefore a^2+b^2=1.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{If }\theta\text{ is the amplitude of } \frac{a+ib}{a-ib},\text{ then }\tan\theta=
\displaystyle \text{(a) }\frac{2a}{a^2+b^2}\qquad \text{(b) }\frac{2ab}{a^2-b^2}
\displaystyle \text{(c) }\frac{a^2-b^2}{a^2+b^2}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \frac{a+ib}{a-ib} =\frac{(a+ib)^2}{(a-ib)(a+ib)}.
\displaystyle =\frac{a^2-b^2+2abi}{a^2+b^2}.
\displaystyle \therefore \tan\theta =\frac{2ab/(a^2+b^2)}{(a^2-b^2)/(a^2+b^2)}.
\displaystyle =\frac{2ab}{a^2-b^2}.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{If }z= \frac{1+7i}{(2-i)^2},\text{ then}
\displaystyle \text{(a) }|z|=2\qquad \text{(b) }|z|=\frac12\qquad \text{(c) amp}(z)=\frac{\pi}{4}\qquad \text{(d) amp}(z)=\frac{3\pi}{4}
\displaystyle \text{Answer:}
\displaystyle (2-i)^2=4-4i+i^2=3-4i.
\displaystyle z=\frac{1+7i}{3-4i} \times\frac{3+4i}{3+4i}.
\displaystyle =\frac{3+4i+21i+28i^2}{25} =\frac{-25+25i}{25}=-1+i.
\displaystyle \text{Thus, }z\text{ lies in the second quadrant.}
\displaystyle \tan\alpha=\left|\frac{1}{-1}\right|=1 \Rightarrow\alpha=\frac{\pi}{4}.
\displaystyle \therefore \text{amp}(z)=\pi-\frac{\pi}{4} =\frac{3\pi}{4}.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{The amplitude of }\frac{1}{i} \text{ is equal to}
\displaystyle \text{(a) }0\qquad\text{(b) }\frac{\pi}{2}\qquad \text{(c) }-\frac{\pi}{2}\qquad\text{(d) }\pi
\displaystyle \text{Answer:}
\displaystyle \frac{1}{i}=\frac{i}{i^2}=-i.
\displaystyle \therefore \text{the principal amplitude of }-i \text{ is }-\frac{\pi}{2}.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{The argument of } \frac{1-i}{1+i}\text{ is}
\displaystyle \text{(a) }-\frac{\pi}{2}\qquad \text{(b) }\frac{\pi}{2}\qquad \text{(c) }\frac{3\pi}{2}\qquad \text{(d) }\frac{5\pi}{2}
\displaystyle \text{Answer:}
\displaystyle \frac{1-i}{1+i} =\frac{(1-i)^2}{(1+i)(1-i)}.
\displaystyle =\frac{1-2i+i^2}{2} =\frac{-2i}{2}=-i.
\displaystyle \therefore \arg\left(\frac{1-i}{1+i}\right) =-\frac{\pi}{2}.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{The amplitude of } \frac{1+i\sqrt3}{\sqrt3+i}\text{ is}
\displaystyle \text{(a) }\frac{\pi}{3}\qquad \text{(b) }-\frac{\pi}{3}\qquad \text{(c) }\frac{\pi}{6}\qquad \text{(d) }-\frac{\pi}{6}
\displaystyle \text{Answer:}
\displaystyle \arg(1+i\sqrt3)=\frac{\pi}{3}, \qquad\arg(\sqrt3+i)=\frac{\pi}{6}.
\displaystyle \therefore \arg\left(\frac{1+i\sqrt3}{\sqrt3+i}\right) =\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{The value of } \frac{i^5+i^6+i^7+i^8+i^9}{1+i}\text{ is}
\displaystyle \text{(a) }\frac12(1+i)\qquad \text{(b) }\frac12(1-i)\qquad \text{(c) }1\qquad\text{(d) }\frac12
\displaystyle \text{Answer:}
\displaystyle i^5=i,\quad i^6=-1,\quad i^7=-i,\quad i^8=1,\quad i^9=i.
\displaystyle \therefore i^5+i^6+i^7+i^8+i^9=i.
\displaystyle \therefore \frac{i}{1+i} =\frac{i(1-i)}{(1+i)(1-i)}.
\displaystyle =\frac{i-i^2}{2}=\frac{1+i}{2}.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 34: }\frac{1+2i+3i^2} {1-2i+3i^2}\text{ equals}
\displaystyle \text{(a) }i\qquad\text{(b) }-1\qquad \text{(c) }-i\qquad\text{(d) }4
\displaystyle \text{Answer:}
\displaystyle \frac{1+2i+3i^2}{1-2i+3i^2} =\frac{1+2i-3}{1-2i-3}.
\displaystyle =\frac{-2+2i}{-2-2i} =\frac{1-i}{1+i}.
\displaystyle =\frac{(1-i)^2}{(1+i)(1-i)} =\frac{-2i}{2}=-i.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{The value of } \frac{i^{592}+i^{590}+i^{588}+i^{586}+i^{584}} {i^{582}+i^{580}+i^{578}+i^{576}+i^{574}}-1\text{ is}
\displaystyle \text{(a) }-1\qquad\text{(b) }-2\qquad \text{(c) }-3\qquad\text{(d) }-4
\displaystyle \text{Answer:}
\displaystyle i^{592}=1,\quad i^{590}=-1,\quad i^{588}=1,
\displaystyle i^{586}=-1,\quad i^{584}=1.
\displaystyle \therefore i^{592}+i^{590}+i^{588}+i^{586}+i^{584}=1.
\displaystyle i^{582}=-1,\quad i^{580}=1,\quad i^{578}=-1,
\displaystyle i^{576}=1,\quad i^{574}=-1.
\displaystyle \therefore i^{582}+i^{580}+i^{578}+i^{576}+i^{574}=-1.
\displaystyle \therefore \frac{1}{-1}-1=-1-1=-2.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{The value of }(1+i)^4+ (1-i)^4\text{ is}
\displaystyle \text{(a) }8\qquad\text{(b) }4\qquad \text{(c) }-8\qquad\text{(d) }-4
\displaystyle \text{Answer:}
\displaystyle (1+i)^2=1+2i+i^2=2i.
\displaystyle \therefore (1+i)^4=(2i)^2=-4.
\displaystyle (1-i)^2=1-2i+i^2=-2i.
\displaystyle \therefore (1-i)^4=(-2i)^2=-4.
\displaystyle \therefore (1+i)^4+(1-i)^4=-4-4=-8.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 37: }\text{If }z=a+ib\text{ lies in third quadrant, then }\frac{\bar z}{z}\text{ also lies in the third quadrant if}
\displaystyle \text{(a) }a>b>0\qquad\text{(b) }a<b<0\qquad \text{(c) }b<a<0\qquad\text{(d) }b>a>0
\displaystyle \text{Answer:}
\displaystyle z=a+ib\text{ lies in the third quadrant, so }a<0,\ b<0.
\displaystyle \frac{\bar z}{z}=\frac{a-ib}{a+ib} =\frac{(a-ib)^2}{a^2+b^2}.
\displaystyle =\frac{a^2-b^2}{a^2+b^2} -i\frac{2ab}{a^2+b^2}.
\displaystyle \text{For the third quadrant, }a^2-b^2<0.
\displaystyle \therefore |a|<|b|.
\displaystyle \text{Since }a<0,\ b<0,\text{ this gives }b<a<0.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{If }f(z)=\frac{7-z}{1-z^2}, \text{ where }z=1+2i,\text{ then }|f(z)|\text{ is}
\displaystyle \text{(a) }\frac{|z|}{2}\qquad \text{(b) }|z|\qquad\text{(c) }2|z|\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle z=1+2i\Rightarrow z^2=(1+2i)^2=-3+4i.
\displaystyle \therefore 7-z=7-(1+2i)=6-2i.
\displaystyle 1-z^2=1-(-3+4i)=4-4i.
\displaystyle \therefore f(z)=\frac{6-2i}{4-4i}.
\displaystyle =\frac{(6-2i)(4+4i)}{(4-4i)(4+4i)} =\frac{32+16i}{32}=1+\frac{i}{2}.
\displaystyle \therefore |f(z)|=\sqrt{1+\frac14} =\frac{\sqrt5}{2}.
\displaystyle \text{Also, }|z|=|1+2i|=\sqrt5.
\displaystyle \therefore |f(z)|=\frac{|z|}{2}.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 39: }\text{A real value of }x\text{ satisfies the equation }\frac{3-4ix}{3+4ix}=a-ib\ (a,b\in R),\text{ if }a^2+b^2=
\displaystyle \text{(a) }1\qquad\text{(b) }-1\qquad \text{(c) }2\qquad\text{(d) }-2
\displaystyle \text{Answer:}
\displaystyle \frac{3-4ix}{3+4ix}=a-ib.
\displaystyle \text{Taking modulus on both sides,}
\displaystyle |a-ib|=\left|\frac{3-4ix}{3+4ix}\right| =\frac{|3-4ix|}{|3+4ix|}.
\displaystyle =\frac{\sqrt{9+16x^2}}{\sqrt{9+16x^2}}=1.
\displaystyle \therefore \sqrt{a^2+b^2}=1.
\displaystyle \therefore a^2+b^2=1.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 40: }\text{The complex number }z\text{ which satisfies the condition }\left|\frac{i+z}{i-z}\right|=1\text{ lies on}
\displaystyle \text{(a) circle }x^2+y^2=1\qquad \text{(b) the x-axis}\qquad\text{(c) the y-axis}
\displaystyle \text{(d) the line }x+y=1
\displaystyle \text{Answer:}
\displaystyle \left|\frac{i+z}{i-z}\right|=1 \Rightarrow |z+i|=|z-i|.
\displaystyle \text{Thus, }z\text{ is equidistant from the points } -i\text{ and }i.
\displaystyle \therefore z\text{ lies on the perpendicular bisector of the segment joining }-i\text{ and }i.
\displaystyle \therefore z\text{ lies on the x-axis.}
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 41: }\text{If }z\text{ is a complex number, then}
\displaystyle \text{(a) }|z|^2>|\bar z|^2\qquad \text{(b) }|z|^2=|\bar z|^2
\displaystyle \text{(c) }|z|^2<|\bar z|^2\qquad \text{(d) }|z|^2\geq|\bar z|^2
\displaystyle \text{Answer:}
\displaystyle \text{Let }z=x+iy.\text{ Then }\bar z=x-iy.
\displaystyle |z|=\sqrt{x^2+y^2},\qquad |\bar z|=\sqrt{x^2+y^2}.
\displaystyle \therefore |z|=|\bar z|.
\displaystyle \therefore |z|^2=|\bar z|^2.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 42: }\text{Which of the following is correct for any two complex numbers }z_1\text{ and }z_2?
\displaystyle \text{(a) }|z_1z_2|=|z_1||z_2|\qquad \text{(b) }\arg(z_1z_2)=\arg(z_1)\arg(z_2)
\displaystyle \text{(c) }|z_1+z_2|=|z_1|+|z_2|\qquad \text{(d) }|z_1+z_2|\geq|z_1|+|z_2|
\displaystyle \text{Answer:}
\displaystyle \text{For any two complex numbers, } |z_1z_2|=|z_1||z_2|.
\displaystyle \text{Also, }|z_1+z_2|\leq|z_1|+|z_2| \text{ by the triangle inequality.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 43: }\text{If the complex number } z=x+iy\text{ satisfies }|z+1|=1,\text{ then }z\text{ lies on}
\displaystyle \text{(a) x-axis}\qquad \text{(b) circle with centre }(-1,0)\text{ and radius }1
\displaystyle \text{(c) y-axis}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle |z+1|=1.
\displaystyle |x+1+iy|=1.
\displaystyle \Rightarrow (x+1)^2+y^2=1.
\displaystyle \text{This is a circle with centre }(-1,0) \text{ and radius }1.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \text{VERY SHORT ANSWER QUESTIONS}

\displaystyle \textbf{Question 1: }\text{Write the values of the square root of }i.
\displaystyle \text{Answer:}
\displaystyle \text{Let }\sqrt{i}=x+iy.
\displaystyle \therefore i=(x+iy)^2=x^2-y^2+2xyi.
\displaystyle \therefore x^2-y^2=0,\qquad 2xy=1.
\displaystyle \therefore x=y,\qquad 2x^2=1.
\displaystyle \therefore x=y=\pm\frac{1}{\sqrt2},\text{ with the same sign.}
\displaystyle \therefore \sqrt{i}=\pm\frac{1+i}{\sqrt2}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Write the values of the square root of }-i.
\displaystyle \text{Answer:}
\displaystyle \text{Let }\sqrt{-i}=x+iy.
\displaystyle \therefore -i=(x+iy)^2=x^2-y^2+2xyi.
\displaystyle \therefore x^2-y^2=0,\qquad 2xy=-1.
\displaystyle \therefore x=-y,\qquad 2x^2=1.
\displaystyle \therefore x=\pm\frac{1}{\sqrt2},\qquad y=\mp\frac{1}{\sqrt2}.
\displaystyle \therefore \sqrt{-i}=\pm\frac{1-i}{\sqrt2}.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If }x+iy=\sqrt{\frac{a+ib}{c+id}},\text{ then write the value of}
\displaystyle (x^2+y^2)^2.
\displaystyle \text{Answer:}
\displaystyle x+iy=\sqrt{\frac{a+ib}{c+id}}.
\displaystyle \therefore (x+iy)^2=\frac{a+ib}{c+id}.
\displaystyle \therefore |x+iy|^4=\left|\frac{a+ib}{c+id}\right|^2.
\displaystyle \therefore (x^2+y^2)^2 =\frac{|a+ib|^2}{|c+id|^2}.
\displaystyle \therefore (x^2+y^2)^2=\frac{a^2+b^2}{c^2+d^2}.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }\pi<\theta<2\pi\text{ and }z=1+\cos\theta+i\sin\theta,
\displaystyle \text{then write the value of }|z|.
\displaystyle \text{Answer:}
\displaystyle |z|=\sqrt{(1+\cos\theta)^2+\sin^2\theta}.
\displaystyle =\sqrt{2+2\cos\theta}.
\displaystyle =\sqrt{4\cos^2\frac{\theta}{2}} =2\left|\cos\frac{\theta}{2}\right|.
\displaystyle \text{Since }\pi<\theta<2\pi,\quad \frac{\pi}{2}<\frac{\theta}{2}<\pi.
\displaystyle \therefore \cos\frac{\theta}{2}<0.
\displaystyle \therefore |z|=-2\cos\frac{\theta}{2}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }n\text{ is any positive integer, write the value of}
\displaystyle \frac{i^{4n+1}-i^{4n-1}}{2}.
\displaystyle \text{Answer:}
\displaystyle i^{4n+1}=i^{4n}\cdot i=i.
\displaystyle i^{4n-1}=i^{4n}\cdot i^{-1}=-i.
\displaystyle \therefore \frac{i^{4n+1}-i^{4n-1}}{2} =\frac{i-(-i)}{2}=i.
\displaystyle \therefore \text{the required value is }i.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Write the value of } \frac{i^{592}+i^{590}+i^{588}+i^{586}+i^{584}} {i^{582}+i^{580}+i^{578}+i^{576}+i^{574}}.
\displaystyle \text{Answer:}
\displaystyle \frac{i^{592}+i^{590}+i^{588}+i^{586}+i^{584}} {i^{582}+i^{580}+i^{578}+i^{576}+i^{574}}
\displaystyle =\frac{i^{584}(i^8+i^6+i^4+i^2+1)} {i^{574}(i^8+i^6+i^4+i^2+1)}.
\displaystyle =i^{10}=i^8i^2=-1.
\displaystyle \therefore \text{the required value is }-1.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Write }1-i\text{ in polar form.}
\displaystyle \text{Answer:}
\displaystyle z=1-i.
\displaystyle |z|=\sqrt{1^2+(-1)^2}=\sqrt2.
\displaystyle \tan\alpha=\left|\frac{-1}{1}\right|=1 \Rightarrow\alpha=\frac{\pi}{4}.
\displaystyle \text{Since }z\text{ lies in the fourth quadrant, } \arg(z)=-\frac{\pi}{4}.
\displaystyle \therefore 1-i=\sqrt2\left( \cos\left(-\frac{\pi}{4}\right)+i\sin\left(-\frac{\pi}{4}\right)\right).
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Write }-1+i\sqrt3\text{ in polar form.}
\displaystyle \text{Answer:}
\displaystyle z=-1+i\sqrt3.
\displaystyle |z|=\sqrt{(-1)^2+(\sqrt3)^2}=2.
\displaystyle \tan\alpha=\left|\frac{\sqrt3}{-1}\right|=\sqrt3 \Rightarrow\alpha=\frac{\pi}{3}.
\displaystyle \text{Since }z\text{ lies in the second quadrant, } \arg(z)=\pi-\frac{\pi}{3}=\frac{2\pi}{3}.
\displaystyle \therefore -1+i\sqrt3 =2\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right).
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Write the argument of }-i.
\displaystyle \text{Answer:}
\displaystyle -i=\cos\left(-\frac{\pi}{2}\right) +i\sin\left(-\frac{\pi}{2}\right).
\displaystyle \therefore \arg(-i)=-\frac{\pi}{2}.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Write the least positive integral value of }n\text{ for which}
\displaystyle \left(\frac{1+i}{1-i}\right)^n\text{ is real.}
\displaystyle \text{Answer:}
\displaystyle \frac{1+i}{1-i} =\frac{(1+i)^2}{(1-i)(1+i)} =\frac{1+2i+i^2}{2}=i.
\displaystyle \therefore \left(\frac{1+i}{1-i}\right)^n=i^n.
\displaystyle i^n\text{ is real when }n\text{ is even.}
\displaystyle \therefore \text{the least positive integral value of }n\text{ is }2.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Find the principal argument of }(1+i\sqrt3)^2.
\displaystyle \text{Answer:}
\displaystyle 1+i\sqrt3=2\left(\cos\frac{\pi}{3} +i\sin\frac{\pi}{3}\right).
\displaystyle \therefore (1+i\sqrt3)^2 =4\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right).
\displaystyle \therefore \text{the principal argument is }\frac{2\pi}{3}.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Find }z,\text{ if }|z|=4\text{ and }\arg(z)=\frac{5\pi}{6}.
\displaystyle \text{Answer:}
\displaystyle z=|z|\left(\cos(\arg z)+i\sin(\arg z)\right).
\displaystyle \therefore z=4\left(\cos\frac{5\pi}{6} +i\sin\frac{5\pi}{6}\right).
\displaystyle =4\left(-\frac{\sqrt3}{2}+\frac{i}{2}\right).
\displaystyle \therefore z=-2\sqrt3+2i.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }|z-5i|=|z+5i|,\text{ then find the locus of }z.
\displaystyle \text{Answer:}
\displaystyle \text{Let }z=x+iy.
\displaystyle |x+i(y-5)|=|x+i(y+5)|.
\displaystyle \therefore x^2+(y-5)^2=x^2+(y+5)^2.
\displaystyle \therefore y^2-10y+25=y^2+10y+25.
\displaystyle \therefore 20y=0\Rightarrow y=0.
\displaystyle \therefore \text{the locus of }z\text{ is the real axis.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If }\frac{(a^2+1)^2}{2a-i}=x+iy,\text{ find the value of}
\displaystyle x^2+y^2.
\displaystyle \text{Answer:}
\displaystyle x^2+y^2=|x+iy|^2.
\displaystyle =\left|\frac{(a^2+1)^2}{2a-i}\right|^2.
\displaystyle =\frac{(a^2+1)^4}{|2a-i|^2}.
\displaystyle =\frac{(a^2+1)^4}{4a^2+1}.
\displaystyle \therefore x^2+y^2=\frac{(a^2+1)^4}{4a^2+1}.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Write the value of }\sqrt{-25}\times\sqrt{-9}.
\displaystyle \text{Answer:}
\displaystyle \sqrt{-25}=5i,\qquad\sqrt{-9}=3i.
\displaystyle \therefore \sqrt{-25}\times\sqrt{-9}=(5i)(3i)=15i^2.
\displaystyle =-15.
\displaystyle \therefore \text{the required value is }-15.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Write the sum of the series }i+i^2+i^3+\cdots
\displaystyle \text{up to }1000\text{ terms.}
\displaystyle \text{Answer:}
\displaystyle i+i^2+i^3+i^4=i-1-i+1=0.
\displaystyle \text{The powers of }i\text{ repeat in cycles of }4.
\displaystyle 1000=4\times250.
\displaystyle \therefore i+i^2+i^3+\cdots+i^{1000}=250(0)=0.
\displaystyle \therefore \text{the required sum is }0.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Write the value of }\arg(z)+\arg(\overline{z}).
\displaystyle \text{Answer:}
\displaystyle \arg(\overline{z})=-\arg(z).
\displaystyle \therefore \arg(z)+\arg(\overline{z}) =\arg(z)-\arg(z)=0.
\displaystyle \therefore \text{the required value is }0.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{If }|z+4|\leq3,\text{ then find the greatest and least values of}
\displaystyle |z+1|.
\displaystyle \text{Answer:}
\displaystyle |z+4|\leq3\text{ represents a disc with centre }-4\text{ and radius }3.
\displaystyle |z+1|\text{ represents the distance of }z\text{ from the point }-1.
\displaystyle \text{The distance between the points }-4\text{ and }-1\text{ is }3.
\displaystyle \therefore \text{least value of }|z+1|=3-3=0.
\displaystyle \therefore \text{greatest value of }|z+1|=3+3=6.
\displaystyle \therefore \text{the greatest and least values are }6\text{ and }0,\text{ respectively.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{For any two complex numbers }z_1\text{ and }z_2\text{ and any two}
\displaystyle \text{real numbers }a,b,\text{ find the value of}
\displaystyle |az_1-bz_2|^2+|az_2+bz_1|^2.
\displaystyle \text{Answer:}
\displaystyle |az_1-bz_2|^2 =a^2|z_1|^2+b^2|z_2|^2-2ab\,\text{Re}(z_1\overline{z_2}).
\displaystyle |az_2+bz_1|^2 =a^2|z_2|^2+b^2|z_1|^2+2ab\,\text{Re}(z_1\overline{z_2}).
\displaystyle \text{Adding the two expressions, the cross terms cancel.}
\displaystyle \therefore |az_1-bz_2|^2+|az_2+bz_1|^2
\displaystyle =(a^2+b^2)(|z_1|^2+|z_2|^2).
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Write the conjugate of } \frac{2-i}{(1-2i)^2}.
\displaystyle \text{Answer:}
\displaystyle (1-2i)^2=1-4i+4i^2=-3-4i.
\displaystyle \therefore \frac{2-i}{(1-2i)^2} =\frac{2-i}{-3-4i}.
\displaystyle =\frac{(2-i)(-3+4i)}{(-3-4i)(-3+4i)}.
\displaystyle =\frac{-6+8i+3i-4i^2}{9+16}.
\displaystyle =\frac{-2+11i}{25} =-\frac{2}{25}+\frac{11}{25}i.
\displaystyle \therefore \overline{\left(\frac{2-i}{(1-2i)^2}\right)} =-\frac{2}{25}-\frac{11}{25}i.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{If }n\in N,\text{ then find the value of } i^n+i^{n+1}+i^{n+2}+i^{n+3}.
\displaystyle \text{Answer:}
\displaystyle i^n+i^{n+1}+i^{n+2}+i^{n+3} =i^n(1+i+i^2+i^3).
\displaystyle =i^n(1+i-1-i)=i^n(0)=0.
\displaystyle \therefore i^n+i^{n+1}+i^{n+2}+i^{n+3}=0.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Find the real value of }a\text{ for which } 3i^3-2ai^2+(1-a)i+5\text{ is real.}
\displaystyle \text{Answer:}
\displaystyle 3i^3-2ai^2+(1-a)i+5
\displaystyle =-3i+2a+(1-a)i+5.
\displaystyle =(2a+5)-(a+2)i.
\displaystyle \text{For the expression to be real, its imaginary part must be zero.}
\displaystyle \therefore -(a+2)=0.
\displaystyle \therefore a=-2.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{If }|z|=2\text{ and }\arg(z)=\frac{\pi}{4}, \text{ find }z.
\displaystyle \text{Answer:}
\displaystyle z=|z|\left(\cos(\arg z)+i\sin(\arg z)\right).
\displaystyle =2\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right).
\displaystyle =2\left(\frac{1}{\sqrt2}+\frac{i}{\sqrt2}\right).
\displaystyle \therefore z=\sqrt2+i\sqrt2.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{Write the argument of } (1+i\sqrt3)(1+i)(\cos\theta+i\sin\theta).
\displaystyle \text{Answer:}
\displaystyle \arg(1+i\sqrt3)=\frac{\pi}{3}.
\displaystyle \arg(1+i)=\frac{\pi}{4}.
\displaystyle \arg(\cos\theta+i\sin\theta)=\theta.
\displaystyle \therefore \arg\left[(1+i\sqrt3)(1+i) (\cos\theta+i\sin\theta)\right]
\displaystyle =\frac{\pi}{3}+\frac{\pi}{4}+\theta.
\displaystyle =\frac{7\pi}{12}+\theta.
\displaystyle \therefore \text{the required argument is } \frac{7\pi}{12}+\theta.
\displaystyle \\

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