\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,\qquad i^3=-i,\qquad i^4=1
\displaystyle (1+i)(1+i^2)(1+i^3)(1+i^4)
\displaystyle =(1+i)(1-1)(1-i)(1+1)
\displaystyle =0
\displaystyle \therefore \text{Correct option is (b).}
\\

\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 \frac{3+2i\sin\theta}{1-2i\sin\theta}\cdot\frac{1+2i\sin\theta}{1+2i\sin\theta}
\displaystyle =\frac{3-4\sin^2\theta+8i\sin\theta}{1+4\sin^2\theta}
\displaystyle \text{For the expression to be real, }8\sin\theta=0
\displaystyle \therefore \sin\theta=0
\displaystyle \text{Since }0<\theta<2\pi,\ \theta=\pi
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 3. }\text{If }(1+i)(1+2i)(1+3i)\cdots(1+ni)=a+ib,\text{ then }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)|^2=|a+ib|^2
\displaystyle |1+i|^2|1+2i|^2|1+3i|^2\cdots|1+ni|^2=a^2+b^2
\displaystyle 2\times5\times10\times\cdots\times(1+n^2)=a^2+b^2
\displaystyle \therefore \text{Correct option is (c).}
\\

\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 \therefore a-ib=(x-iy)^2
\displaystyle \therefore \sqrt{a-ib}=x-iy
\displaystyle \therefore \text{Correct option is (d).}
\\

\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}\qquad \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{\frac{1}{2}+\frac{1}{4}}=\frac{\sqrt3}{2}
\displaystyle \tan\arg(z)=\frac{\frac{1}{2}}{\frac{1}{\sqrt2}}=\frac{1}{\sqrt2}
\displaystyle \therefore \arg(z)=\tan^{-1}\frac{1}{\sqrt2}
\displaystyle \therefore \text{Correct option is (d).}
\\

\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})^3=i^{75}
\displaystyle i^{75}=i^{72}\cdot i^3=-i
\displaystyle -i=\cos\frac{\pi}{2}-i\sin\frac{\pi}{2}
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 7. }\text{If }i^2=-1,\text{ then the sum }i+i^2+i^3+\cdots\text{ upto }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\text{ terms form }250\text{ complete groups of }4\text{ terms.}
\displaystyle \therefore i+i^2+i^3+\cdots+i^{1000}=0
\displaystyle \therefore \text{Correct option is (d).}
\\

\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}
\displaystyle \arg(-2)=\pi,\qquad \arg(1+i\sqrt3)=\frac{\pi}{3}
\displaystyle \therefore \arg(z)=\pi-\frac{\pi}{3}=\frac{2\pi}{3}
\displaystyle \therefore \text{Correct option is (c).}
\\

\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 a=\cos\theta+i\sin\theta=e^{i\theta}
\displaystyle \frac{1+a}{1-a}=\frac{1+e^{i\theta}}{1-e^{i\theta}}
\displaystyle =\frac{2\cos\frac{\theta}{2}e^{i\theta/2}}{-2i\sin\frac{\theta}{2}e^{i\theta/2}}
\displaystyle =i\cot\frac{\theta}{2}
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 10. }\text{If }(1+i)(1+2i)(1+3i)\cdots(1+ni)=a+ib,\text{ then }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)|^2=|a+ib|^2
\displaystyle |1+i|^2|1+2i|^2|1+3i|^2\cdots|1+ni|^2=a^2+b^2
\displaystyle 2\cdot5\cdot10\cdot17\cdots(1+n^2)=a^2+b^2
\displaystyle \therefore \text{Correct option is (c).}
\\

\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}\qquad \text{(c) }\frac{(a^2-1)^2}{(4a^2-1)^2}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle x^2+y^2=\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 \text{Correct option is (a).}
\\

\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 1+i\text{ lies in the first quadrant.}
\displaystyle \therefore \arg(1+i)=\tan^{-1}1=\frac{\pi}{4}
\displaystyle \therefore \text{Correct option is (a).}
\\

\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 \therefore \left(\frac{2i}{1+i}\right)^n=(1+i)^n
\displaystyle 1+i=\sqrt2\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)
\displaystyle \text{For a positive integer value, }n=8
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 14. }\text{If }z\text{ is a non-zero complex number, then }\frac{|\overline z|^2}{|z\overline z|}\text{ is equal to}
\displaystyle \text{(a) }\left|\frac{\overline z}{z}\right|\qquad \text{(b) }|z|\qquad \text{(c) }|\overline z|\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle |\overline z|=|z|
\displaystyle \therefore \frac{|\overline z|^2}{|z\overline z|}=\frac{|z|^2}{|z|\,|\overline z|}
\displaystyle =\frac{|z|^2}{|z|^2}=1
\displaystyle \left|\frac{\overline z}{z}\right|=\frac{|\overline z|}{|z|}=1
\displaystyle \therefore \text{Correct option is (a).}
\\

\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
\displaystyle =1+2i+i^2
\displaystyle =2i
\displaystyle \therefore \text{Correct option is (b).}
\\

\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=(a+ib)^3
\displaystyle =a^3+3a^2bi+3a(ib)^2+(ib)^3
\displaystyle =(a^3-3ab^2)+i(3a^2b-b^3)
\displaystyle \therefore x=a^3-3ab^2,\qquad y=3a^2b-b^3
\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{Correct option is (d) none of these.}
\\

\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\cdot i\sqrt3
\displaystyle =i^2\sqrt6=-\sqrt6
\displaystyle \therefore \text{Correct option is (b).}
\\

\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 \arg(1-i\sqrt3)=300^\circ
\displaystyle \arg(1+i\sqrt3)=60^\circ
\displaystyle \therefore \arg\left(\frac{1-i\sqrt3}{1+i\sqrt3}\right)=300^\circ-60^\circ=240^\circ
\displaystyle \therefore \text{Correct option is (d).}
\\

\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}\cdot\frac{1+i}{1+i}
\displaystyle =\frac{(1+i)^2}{1-i^2}=\frac{1+2i+i^2}{2}=i
\displaystyle \therefore z^4=i^4=1
\displaystyle \therefore \text{Correct option is (a).}
\\

\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 z=\frac{1+2i}{1+2i}=1
\displaystyle \therefore \arg(z)=0
\displaystyle \therefore \text{Correct option is (a).}
\\

\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|
\displaystyle =\frac{1}{|2+3i|^2}
\displaystyle =\frac{1}{2^2+3^2}=\frac{1}{13}
\displaystyle \therefore \text{Correct option is (a).}
\\

\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 =\frac{1}{\sqrt2\cdot\sqrt{13}}
\displaystyle =\frac{1}{\sqrt{26}}
\displaystyle \therefore \text{Correct option is (b).}
\\

\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}\qquad \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}}
\displaystyle =2\left|\sin\frac{\theta}{2}\right|
\displaystyle \therefore \text{Correct option is (c).}
\\

\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=|(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\cdot5\cdot10=100
\displaystyle \therefore \text{Correct option is (c).}
\\

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

\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 \frac{3+5i}{7-6i}=\frac{(3+5i)(7+6i)}{(7-6i)(7+6i)}
\displaystyle =\frac{21+18i+35i+30i^2}{49+36}
\displaystyle =\frac{-9+53i}{85}
\displaystyle \therefore y=\frac{53}{85}
\displaystyle \therefore \text{Correct option is (c).}
\\

\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=\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 \text{Correct option is (a).}
\\

\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}\qquad \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^2+b^2}
\displaystyle =\frac{a^2-b^2+2abi}{a^2+b^2}
\displaystyle \therefore \tan\theta=\frac{\frac{2ab}{a^2+b^2}}{\frac{a^2-b^2}{a^2+b^2}}
\displaystyle =\frac{2ab}{a^2-b^2}
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 29. }\text{If }z=\frac{1+7i}{(2-i)^2},\text{ then}
\displaystyle \text{(a) }|z|=2\qquad \text{(b) }|z|=\frac{1}{2}\qquad \text{(c) }\mathrm{amp}(z)=\frac{\pi}{4}\qquad \text{(d) }\mathrm{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}\cdot\frac{3+4i}{3+4i}
\displaystyle =\frac{3+4i+21i+28i^2}{25}
\displaystyle =\frac{-25+25i}{25}=-1+i
\displaystyle \therefore \mathrm{amp}(z)=\frac{3\pi}{4}
\displaystyle \therefore \text{Correct option is (d).}
\\

\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}=-i
\displaystyle \therefore \mathrm{amp}\left(\frac{1}{i}\right)=-\frac{\pi}{2}
\displaystyle \therefore \text{Correct option is (c).}
\\

\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}=-i
\displaystyle \therefore \arg\left(\frac{1-i}{1+i}\right)=-\frac{\pi}{2}
\displaystyle \therefore \text{Correct option is (a).}
\\

\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}
\displaystyle \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{Correct option is (c).}
\\

\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) }\frac{1}{2}(1+i)\qquad \text{(b) }\frac{1}{2}(1-i)\qquad \text{(c) }1\qquad \text{(d) }\frac{1}{2}
\displaystyle \text{Answer:}
\displaystyle i^5+i^6+i^7+i^8+i^9=i-1-i+1+i=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{Correct option is (a).}
\\

\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)}
\displaystyle =\frac{1-2i+i^2}{2}=-i
\displaystyle \therefore \text{Correct option is (c).}
\\

\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}+i^{590}+i^{588}+i^{586}+i^{584}=1-1+1-1+1=1
\displaystyle i^{582}+i^{580}+i^{578}+i^{576}+i^{574}=-1+1-1+1-1=-1
\displaystyle \therefore \frac{1}{-1}-1=-2
\displaystyle \therefore \text{Correct option is (b).}
\\

\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=-8
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 37. }\text{If }z=a+ib\text{ lies in third quadrant, then }\frac{\overline 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 \frac{\overline z}{z}=\frac{a-ib}{a+ib}\cdot\frac{a-ib}{a-ib}
\displaystyle =\frac{(a-ib)^2}{a^2+b^2}
\displaystyle =\frac{a^2-b^2-2abi}{a^2+b^2}
\displaystyle \text{For third quadrant, real and imaginary parts must be negative.}
\displaystyle \frac{a^2-b^2}{a^2+b^2}<0\Rightarrow a^2<b^2\Rightarrow |a|<|b|
\displaystyle \frac{-2ab}{a^2+b^2}<0\Rightarrow ab>0
\displaystyle \text{Since }z\text{ lies in third quadrant, }a<0,\ b<0
\displaystyle \text{and }|a|<|b|\Rightarrow a>b
\displaystyle \therefore b<a<0
\displaystyle \therefore \text{Correct option is (c).}
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\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
\displaystyle z^2=(1+2i)^2=-3+4i
\displaystyle f(z)=\frac{7-(1+2i)}{1-(-3+4i)}
\displaystyle =\frac{6-2i}{4-4i}
\displaystyle =1+\frac{i}{2}
\displaystyle \therefore |f(z)|=\sqrt{1+\frac{1}{4}}=\frac{\sqrt5}{2}
\displaystyle |z|=\sqrt{1^2+2^2}=\sqrt5
\displaystyle \therefore |f(z)|=\frac{|z|}{2}
\displaystyle \therefore \text{Correct option is (a).}
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\displaystyle \textbf{Question 39. }\text{A real value of }x\text{ satisfies }\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 a^2+b^2=|a-ib|^2
\displaystyle =\left|\frac{3-4ix}{3+4ix}\right|^2
\displaystyle =\frac{|3-4ix|^2}{|3+4ix|^2}
\displaystyle =\frac{9+16x^2}{9+16x^2}=1
\displaystyle \therefore \text{Correct option is (a).}
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\displaystyle \textbf{Question 40. }\text{The complex number }z\text{ which satisfies }\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\text{-axis}\qquad \text{(c) the }y\text{-axis}\qquad \text{(d) the line }x+y=1
\displaystyle \text{Answer:}
\displaystyle \left|\frac{i+z}{i-z}\right|=1
\displaystyle \therefore |z+i|=|z-i|
\displaystyle \text{Thus, }z\text{ is equidistant from }-i\text{ and }i.
\displaystyle \therefore z\text{ lies on the perpendicular bisector of the line joining }-i\text{ and }i.
\displaystyle \therefore z\text{ lies on the }x\text{-axis.}
\displaystyle \therefore \text{Correct option is (b).}
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\displaystyle \textbf{Question 41. }\text{If }z\text{ is a complex number, then}
\displaystyle \text{(a) }|\overline z|^2>|z|^2\qquad \text{(b) }|\overline z|^2=|z|^2\qquad \text{(c) }|\overline z|^2<|z|^2\qquad \text{(d) }|\overline z|^2\geq|z|^2
\displaystyle \text{Answer:}
\displaystyle \text{Let }z=x+iy
\displaystyle \therefore \overline z=x-iy
\displaystyle |z|^2=x^2+y^2,\qquad |\overline z|^2=x^2+y^2
\displaystyle \therefore |\overline z|^2=|z|^2
\displaystyle \therefore \text{Correct option is (b).}
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\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,}
\displaystyle |z_1z_2|=|z_1||z_2|
\displaystyle \therefore \text{Correct option is (a).}
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\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\text{-axis}\qquad \text{(b) circle with centre }(-1,0)\text{ and radius }1
\displaystyle \text{(c) }y\text{-axis}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle |z+1|=1
\displaystyle \therefore |(x+1)+iy|=1
\displaystyle \therefore (x+1)^2+y^2=1
\displaystyle \therefore z\text{ lies on a circle with centre }(-1,0)\text{ and radius }1.
\displaystyle \therefore \text{Correct option is (b).}
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