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

\displaystyle \textbf{Question 1: }\text{The value of }i^{528}\text{ is}
\displaystyle \text{(a) }1\qquad\text{(b) }-1\qquad\text{(c) }i\qquad\text{(d) }-i
\displaystyle \text{Answer:}
\displaystyle \text{We know that }i^4=1.
\displaystyle i^{528}=i^{4\times132}=(i^4)^{132}=1^{132}=1.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 2: }\left[i^{19}+\left(\frac{1}{i}\right)^{25}\right]^2\text{ is equal to}
\displaystyle \text{(a) }4\qquad\text{(b) }-4\qquad\text{(c) }i\qquad\text{(d) }-i
\displaystyle \text{Answer:}
\displaystyle i^{19}=i^{16+3}=i^3=-i.
\displaystyle \text{Also, }\frac{1}{i}=-i.
\displaystyle \therefore \left(\frac{1}{i}\right)^{25}=(-i)^{25}=-i.
\displaystyle \therefore \left[i^{19}+\left(\frac{1}{i}\right)^{25}\right]^2=(-i-i)^2.
\displaystyle =(-2i)^2=4i^2=-4.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 3: }\frac{i^{592}+i^{590}+i^{588}+i^{586}+i^{584}}{i^{582}+i^{580}+i^{578}+i^{576}+i^{574}}\text{ is equal to}
\displaystyle \text{(a) }-1\qquad\text{(b) }1\qquad\text{(c) }-2i\qquad\text{(d) }i
\displaystyle \text{Answer:}
\displaystyle \text{Using }i^4=1,\text{ we reduce each power modulo }4.
\displaystyle i^{592}=1,\quad i^{590}=-1,\quad i^{588}=1,\quad i^{586}=-1,\quad i^{584}=1.
\displaystyle \therefore \text{Numerator}=1-1+1-1+1=1.
\displaystyle i^{582}=-1,\quad i^{580}=1,\quad i^{578}=-1,\quad i^{576}=1,\quad i^{574}=-1.
\displaystyle \therefore \text{Denominator}=-1+1-1+1-1=-1.
\displaystyle \therefore \frac{1}{-1}=-1.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 4: }i^n+i^{n+1}+i^{n+2}+i^{n+3}\text{ is equal to}
\displaystyle \text{(a) }0\qquad\text{(b) }1\qquad\text{(c) }-1\qquad\text{(d) }2
\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).
\displaystyle =i^n(0)=0.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The conjugate of }i^{-35}\text{ is}
\displaystyle \text{(a) }1\qquad\text{(b) }-1\qquad\text{(c) }i\qquad\text{(d) }-i
\displaystyle \text{Answer:}
\displaystyle i^{-35}=\frac{1}{i^{35}}.
\displaystyle i^{35}=i^{32+3}=i^3=-i.
\displaystyle \therefore i^{-35}=\frac{1}{-i}=i.
\displaystyle \text{The conjugate of }i\text{ is }-i.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }z_1=3+2i\text{ and }z_2=2-4i,\text{ then}
\displaystyle |z_1+z_2|^2+|z_1-z_2|^2\text{ is equal to}
\displaystyle \text{(a) }11\qquad\text{(b) }22\qquad\text{(c) }55\qquad\text{(d) }66
\displaystyle \text{Answer:}
\displaystyle z_1+z_2=(3+2i)+(2-4i)=5-2i.
\displaystyle \therefore |z_1+z_2|^2=5^2+(-2)^2=29.
\displaystyle z_1-z_2=(3+2i)-(2-4i)=1+6i.
\displaystyle \therefore |z_1-z_2|^2=1^2+6^2=37.
\displaystyle \therefore |z_1+z_2|^2+|z_1-z_2|^2=29+37=66.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The value of }\frac{i-i^3}{2}\text{ is}
\displaystyle \text{(a) }i\qquad\text{(b) }2i\qquad\text{(c) }-i\qquad\text{(d) }-2i
\displaystyle \text{Answer:}
\displaystyle \text{We know that }i^3=-i.
\displaystyle \therefore \frac{i-i^3}{2}=\frac{i-(-i)}{2}.
\displaystyle =\frac{2i}{2}=i.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{The real part of }\frac{(1+i)^2}{3-i}\text{ is}
\displaystyle \text{(a) }\frac{1}{3}\qquad\text{(b) }-\frac{1}{5}\qquad\text{(c) }-\frac{1}{3}\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle (1+i)^2=1+2i+i^2=2i.
\displaystyle \therefore \frac{(1+i)^2}{3-i}=\frac{2i}{3-i}.
\displaystyle \text{Multiplying the numerator and denominator by }3+i,\text{ we get}
\displaystyle \frac{2i}{3-i}\times\frac{3+i}{3+i}=\frac{6i+2i^2}{9-i^2}.
\displaystyle =\frac{-2+6i}{10}=-\frac{1}{5}+\frac{3}{5}i.
\displaystyle \therefore \text{The real part is }-\frac{1}{5}.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The conjugate of the complex number }\frac{2+5i}{4-3i}\text{ is}
\displaystyle \text{(a) }\frac{7-26i}{25}\qquad\text{(b) }\frac{7+26i}{25}\qquad\text{(c) }\frac{-7-26i}{25}\qquad\text{(d) }\frac{7+26i}{25}
\displaystyle \text{Answer:}
\displaystyle \frac{2+5i}{4-3i}=\frac{2+5i}{4-3i}\times\frac{4+3i}{4+3i}.
\displaystyle =\frac{(2+5i)(4+3i)}{4^2+3^2}.
\displaystyle =\frac{8+6i+20i+15i^2}{25}.
\displaystyle =\frac{-7+26i}{25}.
\displaystyle \text{Therefore, its conjugate is }\frac{-7-26i}{25}.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }z(2-i)=(3+i),\text{ then }z^{20}\text{ is equal to}
\displaystyle \text{(a) }2^{10}\qquad\text{(b) }-2^{10}\qquad\text{(c) }2^{20}\qquad\text{(d) }-2^{20}
\displaystyle \text{Answer:}
\displaystyle z=\frac{3+i}{2-i}.
\displaystyle =\frac{(3+i)(2+i)}{(2-i)(2+i)}=\frac{5+5i}{5}=1+i.
\displaystyle \therefore z^{20}=(1+i)^{20}.
\displaystyle \text{Now, }(1+i)^2=1+2i+i^2=2i.
\displaystyle \therefore (1+i)^{20}=\{(1+i)^2\}^{10}=(2i)^{10}.
\displaystyle =2^{10}i^{10}=2^{10}i^8i^2=-2^{10}.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If }4x+i(3x-y)=3+i(-6),\text{ then the values of }x\text{ and }y\text{ are}
\displaystyle \text{(a) }x=3,\ y=4\qquad\text{(b) }x=\frac{3}{4},\ y=\frac{33}{4}
\displaystyle \text{(c) }x=4,\ y=3\qquad\text{(d) }x=33,\ y=4
\displaystyle \text{Answer:}
\displaystyle \text{Equating the real parts, we get}
\displaystyle 4x=3\quad\Rightarrow\quad x=\frac{3}{4}.
\displaystyle \text{Equating the imaginary parts, we get}
\displaystyle 3x-y=-6.
\displaystyle 3\left(\frac{3}{4}\right)-y=-6.
\displaystyle \frac{9}{4}-y=-6\quad\Rightarrow\quad y=6+\frac{9}{4}=\frac{33}{4}.
\displaystyle \therefore x=\frac{3}{4},\quad y=\frac{33}{4}.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If }i^{103}=a+ib,\text{ then }a+b\text{ is equal to}
\displaystyle \text{(a) }1\qquad\text{(b) }-1\qquad\text{(c) }0\qquad\text{(d) }2
\displaystyle \text{Answer:}
\displaystyle i^{103}=i^{4\times25+3}=i^3=-i.
\displaystyle \text{Comparing }a+ib=-i,\text{ we get }a=0\text{ and }b=-1.
\displaystyle \therefore a+b=0+(-1)=-1.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }z_1=\sqrt{3}+i\sqrt{3}\text{ and }z_2=\sqrt{3}+i,\text{ then the quadrant in which }-\frac{z_1}{z_2}\text{ lies is}
\displaystyle \text{(a) I}\qquad\text{(b) II}\qquad\text{(c) III}\qquad\text{(d) IV}
\displaystyle \text{Answer:}
\displaystyle -\frac{z_1}{z_2}=-\frac{\sqrt{3}+i\sqrt{3}}{\sqrt{3}+i}.
\displaystyle =-\frac{(\sqrt{3}+i\sqrt{3})(\sqrt{3}-i)}{(\sqrt{3}+i)(\sqrt{3}-i)}.
\displaystyle =-\frac{(3+\sqrt{3})+i(3-\sqrt{3})}{4}.
\displaystyle =-\frac{3+\sqrt{3}}{4}-i\frac{3-\sqrt{3}}{4}.
\displaystyle \text{Both the real and imaginary parts are negative.}
\displaystyle \therefore -\frac{z_1}{z_2}\text{ lies in the third quadrant.}
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{The conjugate of }\frac{2-i}{(1-2i)^2}\text{ is}
\displaystyle \text{(a) }-\frac{2}{25}+\frac{11}{25}i\qquad\text{(b) }-\frac{2}{25}-\frac{11}{25}i
\displaystyle \text{(c) }\frac{2}{25}-\frac{11}{25}i\qquad\text{(d) }\frac{2}{25}-\frac{11}{25}i
\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)^2+4^2}.
\displaystyle =\frac{-6+8i+3i-4i^2}{25}.
\displaystyle =\frac{-2+11i}{25}=-\frac{2}{25}+\frac{11}{25}i.
\displaystyle \therefore \text{its conjugate is }-\frac{2}{25}-\frac{11}{25}i.

\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{If }z=-5i^{-15}-6i^{-8}\text{ then }\bar{z}\text{ is equal to}
\displaystyle \text{(a) }-6-5i\qquad\text{(b) }-6+5i\qquad\text{(c) }6-5i\qquad\text{(d) }6+5i
\displaystyle \text{Answer:}
\displaystyle i^{-15}=\frac{1}{i^{15}}=\frac{1}{-i}=i.
\displaystyle i^{-8}=\frac{1}{i^8}=1.
\displaystyle \therefore z=-5i-6=-6-5i.
\displaystyle \therefore \bar{z}=-6+5i.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 16: }(1+i)^8+(1-i)^8\text{ is equal to}
\displaystyle \text{(a) }1\qquad\text{(b) }2\qquad\text{(c) }8\qquad\text{(d) }32
\displaystyle \text{Answer:}
\displaystyle (1+i)^2=1+2i+i^2=2i.
\displaystyle \therefore (1+i)^8=(2i)^4=16i^4=16.
\displaystyle (1-i)^2=1-2i+i^2=-2i.
\displaystyle \therefore (1-i)^8=(-2i)^4=16i^4=16.
\displaystyle \therefore (1+i)^8+(1-i)^8=16+16=32.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Which of the following options defines `imaginary number'?}
\displaystyle \text{(a) Square root of any number}\qquad\text{(b) Square root of positive number}
\displaystyle \text{(c) Square root of negative number}\qquad\text{(d) Cube root of number}
\displaystyle \text{Answer:}
\displaystyle \text{An imaginary number is obtained when we take the square root of a negative real number.}
\displaystyle \text{For example, }\sqrt{-1}=i.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{If }z=\frac{7-i}{3-4i},\text{ then }|z|^{14}\text{ is equal to}
\displaystyle \text{(a) }2^7\qquad\text{(b) }2^7i\qquad\text{(c) }-2^7\qquad\text{(d) }-2^7i
\displaystyle \text{Answer:}
\displaystyle |z|=\left|\frac{7-i}{3-4i}\right|=\frac{|7-i|}{|3-4i|}.
\displaystyle =\frac{\sqrt{7^2+(-1)^2}}{\sqrt{3^2+(-4)^2}}=\frac{\sqrt{50}}{5}.
\displaystyle =\sqrt{2}.
\displaystyle \therefore |z|^{14}=(\sqrt{2})^{14}=2^7.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 19: }\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{\pi}{4}\qquad\text{(d) }\frac{\pi}{6}
\displaystyle \text{Answer:}
\displaystyle \frac{1-i}{1+i}=\frac{(1-i)(1-i)}{(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{The complex number }-i\text{ lies on the negative imaginary axis.}
\displaystyle \therefore \arg(-i)=-\frac{\pi}{2}.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{If }Z_1=1+i,\ Z_2=2-i\text{ and }\overline{Z_1Z_2}=a+ib,\text{ then }a+b\text{ is equal to}
\displaystyle \text{(a) }2\qquad\text{(b) }1\qquad\text{(c) }3\qquad\text{(d) }4
\displaystyle \text{Answer:}
\displaystyle Z_1Z_2=(1+i)(2-i).
\displaystyle =2-i+2i-i^2=3+i.
\displaystyle \therefore \overline{Z_1Z_2}=3-i.
\displaystyle \text{Comparing with }a+ib,\text{ we get }a=3\text{ and }b=-1.
\displaystyle \therefore a+b=3-1=2.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

 

\displaystyle \text{CASE BASED/SOURCE BASED / PASSAGE BASED}

\displaystyle \textbf{Question 21: }\text{Conjugate of a complex number }z=x+iy\text{ is }x-iy\text{ and denoted by}
\displaystyle \bar{z}=x-iy.
\displaystyle \text{Based on above information answer the following questions:}
\displaystyle \text{(a) Find }x\text{ if }\sin x+i\cos2x\text{ and }\cos2x-i\sin2x\text{ are conjugate each other.}
\displaystyle \text{(b) Solve the equation }z^2+z.
\displaystyle \text{Answer:}
\displaystyle \text{(a) If the two complex numbers are conjugates, their real parts must be equal and}
\displaystyle \text{their imaginary parts must be equal in magnitude but opposite in sign.}
\displaystyle \therefore \sin x=\cos2x\quad\text{and}\quad\cos2x=\sin2x.
\displaystyle \text{Thus, }\sin x=\sin2x.
\displaystyle \sin2x-\sin x=0.
\displaystyle 2\cos\frac{3x}{2}\sin\frac{x}{2}=0.
\displaystyle \therefore \sin\frac{x}{2}=0\quad\text{or}\quad\cos\frac{3x}{2}=0.
\displaystyle \text{These values must also satisfy }\sin x=\cos2x\text{ and }\cos2x=\sin2x.
\displaystyle \text{No real value of }x\text{ satisfies both conditions simultaneously.}
\displaystyle \therefore \text{part (a), as printed, appears to contain an error.}
\displaystyle \text{(b) The equation is incomplete in the given question. The expression after }z^2+z
\displaystyle \text{is missing, so the equation cannot be solved uniquely.}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{An ant is moving around a few food pieces scattered on the floor along the curve}
\displaystyle \left|\frac{z-2}{z-3}\right|=2.
\displaystyle \text{(a) What is the shape of the path described by the ant}
\displaystyle \text{(b) Find the equation of the path described.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }z=x+iy.
\displaystyle \left|\frac{z-2}{z-3}\right|=2\implies|z-2|=2|z-3|.
\displaystyle \sqrt{(x-2)^2+y^2}=2\sqrt{(x-3)^2+y^2}.
\displaystyle \text{Squaring both sides, we get}
\displaystyle (x-2)^2+y^2=4\{(x-3)^2+y^2\}.
\displaystyle x^2-4x+4+y^2=4x^2-24x+36+4y^2.
\displaystyle 3x^2+3y^2-20x+32=0.
\displaystyle x^2+y^2-\frac{20}{3}x+\frac{32}{3}=0.
\displaystyle \left(x-\frac{10}{3}\right)^2+y^2=\frac{4}{9}.
\displaystyle \text{(a) Therefore, the path described by the ant is a circle.}
\displaystyle \text{(b) The equation of the path is }\left(x-\frac{10}{3}\right)^2+y^2=\frac{4}{9}.
\displaystyle \text{Its centre is }\left(\frac{10}{3},0\right)\text{ and radius is }\frac{2}{3}.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{The conjugate of a complex number }z\text{ is the complex number obtained by}
\displaystyle \text{changing the sign of its imaginary part. It is denoted by }\bar{z}.
\displaystyle \text{The modulus of a complex number }z=a+ib\text{ is defined as the non-negative real number}
\displaystyle \sqrt{a^2+b^2}.\text{ It is denoted by }|z|,\text{ i.e. }|z|=\sqrt{a^2+b^2}.

\displaystyle \text{(a) If }(x-iy)(3+5i)\text{ is the conjugate of }-6-24i,\text{ then find the value of }x+y.
\displaystyle \text{Answer:}
\displaystyle \text{The conjugate of }-6-24i\text{ is }-6+24i.
\displaystyle \therefore (x-iy)(3+5i)=-6+24i.
\displaystyle 3x+5y+i(5x-3y)=-6+24i.
\displaystyle \text{Equating the real and imaginary parts, we get}
\displaystyle 3x+5y=-6\qquad\text{and}\qquad5x-3y=24.
\displaystyle \text{Multiplying the first equation by }3\text{ and the second by }5,\text{ we get}
\displaystyle 9x+15y=-18\qquad\text{and}\qquad25x-15y=120.
\displaystyle \text{Adding, }34x=102\quad\Rightarrow\quad x=3.
\displaystyle 3(3)+5y=-6\quad\Rightarrow\quad5y=-15\quad\Rightarrow\quad y=-3.
\displaystyle \therefore x+y=3-3=0.

\displaystyle \text{(b) If }f(z)=\frac{7-z}{1-z^2},\text{ where }z=1+2i,\text{ then find }|f(z)|.
\displaystyle \text{Answer:}
\displaystyle |f(z)|=\left|\frac{7-z}{1-z^2}\right|=\frac{|7-z|}{|1-z^2|}.
\displaystyle \text{For }z=1+2i,\quad 7-z=7-(1+2i)=6-2i.
\displaystyle \therefore |7-z|=\sqrt{6^2+(-2)^2}=\sqrt{40}=2\sqrt{10}.
\displaystyle z^2=(1+2i)^2=1+4i+4i^2=-3+4i.
\displaystyle \therefore 1-z^2=1-(-3+4i)=4-4i.
\displaystyle |1-z^2|=\sqrt{4^2+(-4)^2}=\sqrt{32}=4\sqrt{2}.
\displaystyle \therefore |f(z)|=\frac{2\sqrt{10}}{4\sqrt{2}}=\frac{\sqrt{5}}{2}.
\displaystyle \\

\displaystyle \text{ASSERTION REASONING}

\displaystyle \textbf{Question 24: }\text{Assertion (A): If }i=\sqrt{-1}\text{ then }i^{4k}=1,\ i^{4k+1}=i,
\displaystyle i^{4k+2}=-1,\ i^{4k+3}=-i.
\displaystyle \text{Reason (R): }i^{4k}+i^{4k+1}+i^{4k+2}+i^{4k+3}=1
\displaystyle \text{(a) A is true, R is true; R is a correct explanation of A.}
\displaystyle \text{(b) A is true, R is true; R is not a correct explanation of A.}
\displaystyle \text{(c) A is true; R is false.}
\displaystyle \text{(d) A is false; R is true.}
\displaystyle \text{Answer:}
\displaystyle \text{We know that }i^4=1.
\displaystyle \therefore i^{4k}=(i^4)^k=1.
\displaystyle i^{4k+1}=i^{4k}\cdot i=i.
\displaystyle i^{4k+2}=i^{4k}\cdot i^2=-1.
\displaystyle i^{4k+3}=i^{4k}\cdot i^3=-i.
\displaystyle \therefore \text{Assertion (A) is true.}
\displaystyle \text{Now, }i^{4k}+i^{4k+1}+i^{4k+2}+i^{4k+3}
\displaystyle =1+i-1-i=0.
\displaystyle \therefore \text{Reason (R) is false.}
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{Assertion (A): For real roots of }ax^2+bx+c=0,\ D\geq0.
\displaystyle \text{Reason (R): The greatest value of }\lambda\text{ for which}
\displaystyle (2\lambda-1)x^2-4x+(2\lambda-1)=0\text{ has real roots is }1.
\displaystyle \text{(a) A is true, R is true; R is a correct explanation of A.}
\displaystyle \text{(b) A is true, R is true; R is not a correct explanation of A.}
\displaystyle \text{(c) A is true; R is false.}
\displaystyle \text{(d) A is false; R is true.}
\displaystyle \text{Answer:}
\displaystyle \text{For a quadratic equation }ax^2+bx+c=0\text{ to have real roots,}
\displaystyle D=b^2-4ac\geq0.
\displaystyle \therefore \text{Assertion (A) is true.}
\displaystyle \text{For }(2\lambda-1)x^2-4x+(2\lambda-1)=0,
\displaystyle D=(-4)^2-4(2\lambda-1)(2\lambda-1)\geq0.
\displaystyle 16-4(2\lambda-1)^2\geq0.
\displaystyle (2\lambda-1)^2\leq4.
\displaystyle -2\leq2\lambda-1\leq2.
\displaystyle -1\leq2\lambda\leq3.
\displaystyle -\frac{1}{2}\leq\lambda\leq\frac{3}{2}.
\displaystyle \therefore \text{the greatest value of }\lambda\text{ is }\frac{3}{2},\text{ not }1.
\displaystyle \therefore \text{Reason (R) is false.}
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{Assertion (A): If }|z_1+z_2|^2=|z_1|^2+|z_2|^2,\text{ then }\frac{z_1}{z_2}\text{ is purely imaginary.}
\displaystyle \text{Reason (R): If }z\text{ is purely imaginary, then }z+\bar{z}=0.
\displaystyle \text{(a) A is true, R is true; R is a correct explanation of A.}
\displaystyle \text{(b) A is true, R is true; R is not a correct explanation of A.}
\displaystyle \text{(c) A is true; R is false.}
\displaystyle \text{(d) A is false; R is true.}
\displaystyle \text{Answer:}
\displaystyle |z_1+z_2|^2=(z_1+z_2)(\bar{z}_1+\bar{z}_2).
\displaystyle =|z_1|^2+|z_2|^2+z_1\bar{z}_2+\bar{z}_1z_2.
\displaystyle \text{Given }|z_1+z_2|^2=|z_1|^2+|z_2|^2,
\displaystyle \therefore z_1\bar{z}_2+\bar{z}_1z_2=0.
\displaystyle \text{Dividing by }z_2\bar{z}_2,\text{ we get}
\displaystyle \frac{z_1}{z_2}+\frac{\bar{z}_1}{\bar{z}_2}=0.
\displaystyle \therefore \frac{z_1}{z_2}+\overline{\left(\frac{z_1}{z_2}\right)}=0.
\displaystyle \text{Since }z+\bar{z}=0\text{ implies that }z\text{ is purely imaginary,}
\displaystyle \therefore \frac{z_1}{z_2}\text{ is purely imaginary.}
\displaystyle \therefore \text{Assertion (A) is true.}
\displaystyle \text{Reason (R) is also true and correctly explains Assertion (A).}
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{Assertion (A): The equation }ix^2-3ix+2i=0\text{ has non-real roots.}
\displaystyle \text{Reason (R): If }a,b,c\text{ are real and }b^2-4ac\geq0,\text{ then the roots of the equation}
\displaystyle ax^2+bx+c=0\text{ are real and if }b^2-4ac<0,\text{ then the roots of the equation}
\displaystyle ax^2+bx+c=0\text{ are non-real.}
\displaystyle \text{(a) A is true, R is true; R is a correct explanation of A.}
\displaystyle \text{(b) A is true, R is true; R is not a correct explanation of A.}
\displaystyle \text{(c) A is true; R is false.}
\displaystyle \text{(d) A is false; R is true.}
\displaystyle \text{Answer:}
\displaystyle ix^2-3ix+2i=0.
\displaystyle \text{Dividing throughout by }i,\text{ we get}
\displaystyle x^2-3x+2=0.
\displaystyle (x-1)(x-2)=0.
\displaystyle \therefore x=1\text{ or }x=2.
\displaystyle \text{Thus, both roots are real. Hence, Assertion (A) is false.}
\displaystyle \text{For a quadratic equation with real coefficients, if }b^2-4ac\geq0,
\displaystyle \text{the roots are real, and if }b^2-4ac<0,\text{ the roots are non-real.}
\displaystyle \therefore \text{Reason (R) is true.}
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \text{VERY SHORT ANSWER}

\displaystyle \textbf{Question 28: }\text{Express the following complex number in the form }a+ib:
\displaystyle \frac{3-\sqrt{-16}}{1-\sqrt{-9}}
\displaystyle \text{Answer:}
\displaystyle \sqrt{-16}=4i\quad\text{and}\quad\sqrt{-9}=3i.
\displaystyle \therefore \frac{3-\sqrt{-16}}{1-\sqrt{-9}}=\frac{3-4i}{1-3i}.
\displaystyle \text{Multiplying the numerator and denominator by }1+3i,\text{ we get}
\displaystyle \frac{3-4i}{1-3i}\times\frac{1+3i}{1+3i}=\frac{(3-4i)(1+3i)}{1+9}.
\displaystyle =\frac{3+9i-4i-12i^2}{10}.
\displaystyle =\frac{15+5i}{10}=\frac{3}{2}+\frac{1}{2}i.
\displaystyle \therefore \text{The required form is }\frac{3}{2}+\frac{1}{2}i.
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{Find the modulus and conjugate of }\frac{(4+5i)^2}{(2+3i)^2}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }z=\frac{(4+5i)^2}{(2+3i)^2}=\left(\frac{4+5i}{2+3i}\right)^2.
\displaystyle \frac{4+5i}{2+3i}=\frac{(4+5i)(2-3i)}{(2+3i)(2-3i)}.
\displaystyle =\frac{8-12i+10i-15i^2}{4+9}=\frac{23-2i}{13}.
\displaystyle \therefore z=\left(\frac{23-2i}{13}\right)^2.
\displaystyle =\frac{529-92i+4i^2}{169}=\frac{525-92i}{169}.
\displaystyle \text{Now, }|z|=\left|\frac{(4+5i)^2}{(2+3i)^2}\right|.
\displaystyle =\frac{|4+5i|^2}{|2+3i|^2}=\frac{4^2+5^2}{2^2+3^2}.
\displaystyle =\frac{41}{13}.
\displaystyle \text{Also, }z=\frac{525}{169}-\frac{92}{169}i.
\displaystyle \therefore \bar{z}=\frac{525}{169}+\frac{92}{169}i.
\displaystyle \therefore \text{Modulus}=\frac{41}{13}\text{ and conjugate}=\frac{525+92i}{169}.
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{For any complex number prove that }|z^2|=|z|^2.
\displaystyle \text{Answer:}
\displaystyle \text{Let }z=a+ib.
\displaystyle \therefore z^2=(a+ib)^2=(a^2-b^2)+2abi.
\displaystyle |z^2|=\sqrt{(a^2-b^2)^2+(2ab)^2}.
\displaystyle =\sqrt{a^4-2a^2b^2+b^4+4a^2b^2}.
\displaystyle =\sqrt{a^4+2a^2b^2+b^4}.
\displaystyle =\sqrt{(a^2+b^2)^2}=a^2+b^2.
\displaystyle \text{Also, }|z|^2=\left(\sqrt{a^2+b^2}\right)^2=a^2+b^2.
\displaystyle \therefore |z^2|=|z|^2.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{Evaluate }i^{-999}.
\displaystyle \text{Answer:}
\displaystyle i^{-999}=\frac{1}{i^{999}}.
\displaystyle i^{999}=i^{4\times249+3}=i^3=-i.
\displaystyle \therefore i^{-999}=\frac{1}{-i}.
\displaystyle =\frac{i}{-i^2}=i.
\displaystyle \therefore i^{-999}=i.
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{Find the value of }\left(-\sqrt{-1}\right)^{4n+3},\text{ where }n\text{ is any natural number.}
\displaystyle \text{Answer:}
\displaystyle \text{Since }\sqrt{-1}=i,\text{ we have}
\displaystyle \left(-\sqrt{-1}\right)^{4n+3}=(-i)^{4n+3}.
\displaystyle =(-i)^{4n}(-i)^3.
\displaystyle =\{(-i)^4\}^n(-i)^3.
\displaystyle =1^n\cdot i=i.
\displaystyle \therefore \left(-\sqrt{-1}\right)^{4n+3}=i.
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{Evaluate }1+i^2+i^4+i^6+\cdots+i^{20}.
\displaystyle \text{Answer:}
\displaystyle i^2=-1,\quad i^4=1,\quad i^6=-1,\quad i^8=1,\ldots
\displaystyle \therefore 1+i^2+i^4+i^6+\cdots+i^{20}=1-1+1-1+\cdots+1.
\displaystyle \text{There are }11\text{ terms, with }6\text{ positive terms and }5\text{ negative terms.}
\displaystyle \therefore \text{The required value is }6-5=1.
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{Find the multiplicative inverse of }z=3-2i.
\displaystyle \text{Answer:}
\displaystyle \text{The multiplicative inverse of }z\text{ is }\frac{1}{z}.
\displaystyle \frac{1}{3-2i}=\frac{1}{3-2i}\times\frac{3+2i}{3+2i}.
\displaystyle =\frac{3+2i}{3^2+2^2}=\frac{3+2i}{13}.
\displaystyle \therefore \text{The multiplicative inverse is }\frac{3}{13}+\frac{2}{13}i.
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{Check if }z=\frac{3+2i}{2-3i}+\frac{3-2i}{2+3i}\text{ is purely real or purely imaginary.}
\displaystyle \text{Answer:}
\displaystyle \frac{3+2i}{2-3i}=\frac{(3+2i)(2+3i)}{(2-3i)(2+3i)}.
\displaystyle =\frac{6+9i+4i+6i^2}{13}=\frac{13i}{13}=i.
\displaystyle \frac{3-2i}{2+3i}=\frac{(3-2i)(2-3i)}{(2+3i)(2-3i)}.
\displaystyle =\frac{6-9i-4i+6i^2}{13}=\frac{-13i}{13}=-i.
\displaystyle \therefore z=i-i=0.
\displaystyle \text{Since the imaginary part of }z\text{ is }0,\text{ the number is purely real.}
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{If }z_1,z_2\text{ are }1-i\text{ and }-2+4i\text{ respectively, find }\mathrm{Im}\left(\frac{z_1z_2}{\bar{z}_1}\right).
\displaystyle \text{Answer:}
\displaystyle z_1=1-i,\quad z_2=-2+4i,\quad\bar{z}_1=1+i.
\displaystyle z_1z_2=(1-i)(-2+4i).
\displaystyle =-2+4i+2i-4i^2=2+6i.
\displaystyle \therefore \frac{z_1z_2}{\bar{z}_1}=\frac{2+6i}{1+i}.
\displaystyle =\frac{(2+6i)(1-i)}{(1+i)(1-i)}.
\displaystyle =\frac{2-2i+6i-6i^2}{2}=\frac{8+4i}{2}=4+2i.
\displaystyle \therefore \mathrm{Im}\left(\frac{z_1z_2}{\bar{z}_1}\right)=2.
\displaystyle \\

\displaystyle \textbf{Question 37: }\text{Find the real values of }x\text{ and }y\text{ if}
\displaystyle (1-i)x+(1+i)y=1-3i.
\displaystyle \text{Answer:}
\displaystyle (1-i)x+(1+i)y=(x+y)+i(y-x).
\displaystyle \therefore (x+y)+i(y-x)=1-3i.
\displaystyle \text{Equating the real and imaginary parts, we get}
\displaystyle x+y=1\qquad\text{and}\qquad y-x=-3.
\displaystyle \text{Adding the two equations, }2y=-2\quad\Rightarrow\quad y=-1.
\displaystyle \text{Substituting }y=-1\text{ in }x+y=1,\text{ we get}
\displaystyle x-1=1\quad\Rightarrow\quad x=2.
\displaystyle \therefore x=2,\quad y=-1.
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{If }\frac{1-i}{1+i}=a+ib,\text{ find }a\text{ and }b.
\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)(1-i)}.
\displaystyle =\frac{1-2i+i^2}{1-i^2}=\frac{-2i}{2}=-i.
\displaystyle \text{Comparing }a+ib=-i,\text{ we get}
\displaystyle a=0,\quad b=-1.
\displaystyle \\

\displaystyle \textbf{Question 39: }\text{If }a=\cos\theta+i\sin\theta,\text{ find the value of }\frac{1+a}{1-a}.
\displaystyle \text{Answer:}
\displaystyle \frac{1+a}{1-a}=\frac{1+\cos\theta+i\sin\theta}{1-\cos\theta-i\sin\theta}.
\displaystyle \text{Using }1+\cos\theta=2\cos^2\frac{\theta}{2},\quad1-\cos\theta=2\sin^2\frac{\theta}{2},
\displaystyle \text{and }\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2},\text{ we get}
\displaystyle \frac{1+a}{1-a}=\frac{2\cos^2\frac{\theta}{2}+2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}-2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}}.
\displaystyle =\frac{\cos\frac{\theta}{2}\left(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}\right)}{\sin\frac{\theta}{2}\left(\sin\frac{\theta}{2}-i\cos\frac{\theta}{2}\right)}.
\displaystyle \text{Since }\sin\frac{\theta}{2}-i\cos\frac{\theta}{2}=-i\left(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2}\right),
\displaystyle \frac{1+a}{1-a}=\frac{\cos\frac{\theta}{2}}{-i\sin\frac{\theta}{2}}.
\displaystyle =i\cot\frac{\theta}{2}.
\displaystyle \therefore \frac{1+a}{1-a}=i\cot\frac{\theta}{2}.
\displaystyle \\

\displaystyle \textbf{Question 41: }\text{If }z_1,z_2,z_3\text{ are complex numbers such that }|z_1|=|z_2|=|z_3|=
\displaystyle \left|1+\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}\right|=1,\text{ then find the value of }|z_1+z_2+z_3|.
\displaystyle \text{Answer:}
\displaystyle \text{Since }|z_1|=|z_2|=|z_3|=1,\text{ we have }\frac{1}{z_1}=\bar z_1,\ \frac{1}{z_2}=\bar z_2,\ \frac{1}{z_3}=\bar z_3.
\displaystyle \therefore \left|1+\bar z_1+\bar z_2+\bar z_3\right|=1.
\displaystyle \text{Taking conjugate inside the modulus, we get}
\displaystyle |1+z_1+z_2+z_3|=1.
\displaystyle \text{However, this condition alone does not determine a unique value of }|z_1+z_2+z_3|.
\displaystyle \text{Hence, as printed, the question appears to be incomplete or contains an error.}
\displaystyle \\

\displaystyle \textbf{Question 42: }\text{If }z_1=2-i\text{ and }z_2=-2+i,\text{ find}
\displaystyle \text{(i) }\mathrm{Re}\left(\frac{z_1z_2}{\bar z_1}\right)\qquad\text{(ii) }\mathrm{Im}\left(\frac{1}{z_1\bar z_1}\right)
\displaystyle \text{Answer:}
\displaystyle \text{Given, }z_1=2-i,\quad z_2=-2+i,\quad\bar z_1=2+i.
\displaystyle \text{(i) }z_1z_2=(2-i)(-2+i)=-3+4i.
\displaystyle \therefore \frac{z_1z_2}{\bar z_1}=\frac{-3+4i}{2+i}.
\displaystyle =\frac{(-3+4i)(2-i)}{(2+i)(2-i)}=\frac{-2+11i}{5}.
\displaystyle \therefore \mathrm{Re}\left(\frac{z_1z_2}{\bar z_1}\right)=-\frac{2}{5}.
\displaystyle \text{(ii) }z_1\bar z_1=(2-i)(2+i)=4+1=5.
\displaystyle \therefore \frac{1}{z_1\bar z_1}=\frac{1}{5}.
\displaystyle \text{Since }\frac{1}{5}\text{ is purely real, its imaginary part is }0.
\displaystyle \therefore \mathrm{Im}\left(\frac{1}{z_1\bar z_1}\right)=0.
\displaystyle \\

\displaystyle \text{SHORT ANSWER QUESTIONS}

\displaystyle \textbf{Question 43: }\text{If }(x+iy)^3=u+iv,\text{ then show that }\frac{u}{x}+\frac{v}{y}=4(x^2-y^2).
\displaystyle \text{Answer:}
\displaystyle \text{Given, }(x+iy)^3=u+iv.
\displaystyle (x+iy)^3=x^3+3x^2(iy)+3x(iy)^2+(iy)^3.
\displaystyle =x^3+3x^2yi-3xy^2-y^3i.
\displaystyle =(x^3-3xy^2)+i(3x^2y-y^3).
\displaystyle \text{Comparing real and imaginary parts, we get}
\displaystyle u=x^3-3xy^2\quad\text{and}\quad v=3x^2y-y^3.
\displaystyle \therefore \frac{u}{x}+\frac{v}{y}=\frac{x^3-3xy^2}{x}+\frac{3x^2y-y^3}{y}.
\displaystyle =x^2-3y^2+3x^2-y^2.
\displaystyle =4x^2-4y^2=4(x^2-y^2).
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 44: }\text{If }\left(\frac{1+i}{1-i}\right)^m=1\text{ then find the least positive integral value of }m.
\displaystyle \text{Answer:}
\displaystyle \frac{1+i}{1-i}=\frac{(1+i)(1+i)}{(1-i)(1+i)}.
\displaystyle =\frac{(1+i)^2}{1-i^2}=\frac{1+2i+i^2}{2}.
\displaystyle =\frac{2i}{2}=i.
\displaystyle \therefore \left(\frac{1+i}{1-i}\right)^m=i^m.
\displaystyle \text{For }i^m=1,\text{ the least positive integral value of }m\text{ is }4.
\displaystyle \therefore m=4.
\displaystyle \\

\displaystyle \textbf{Question 45: }\text{Simplify }(1+i)^4\left(1+\frac{1}{i}\right)^4.
\displaystyle \text{Answer:}
\displaystyle \frac{1}{i}=-i.
\displaystyle \therefore (1+i)^4\left(1+\frac{1}{i}\right)^4=(1+i)^4(1-i)^4.
\displaystyle =\{(1+i)(1-i)\}^4.
\displaystyle =(1-i^2)^4=2^4=16.
\displaystyle \therefore \text{The required value is }16.
\displaystyle \\

\displaystyle \textbf{Question 46: }\text{Find the value of }i^{45}+i^{46}+i^{47}+i^{48}.
\displaystyle \text{Answer:}
\displaystyle i^{45}=i^{4(11)+1}=i.
\displaystyle i^{46}=i^{4(11)+2}=-1.
\displaystyle i^{47}=i^{4(11)+3}=-i.
\displaystyle i^{48}=i^{4(12)}=1.
\displaystyle \therefore i^{45}+i^{46}+i^{47}+i^{48}=i-1-i+1=0.
\displaystyle \therefore \text{The required value is }0.
\displaystyle \\

\displaystyle \textbf{Question 47: }\text{If }z(2-i)=(3+i),\text{ then }z^{20}\text{ is equal to.}
\displaystyle \text{Answer:}
\displaystyle z=\frac{3+i}{2-i}.
\displaystyle =\frac{(3+i)(2+i)}{(2-i)(2+i)}.
\displaystyle =\frac{6+3i+2i+i^2}{4+1}=\frac{5+5i}{5}=1+i.
\displaystyle \therefore z^{20}=(1+i)^{20}.
\displaystyle \text{Now, }(1+i)^2=1+2i+i^2=2i.
\displaystyle \therefore (1+i)^{20}=\{(1+i)^2\}^{10}=(2i)^{10}.
\displaystyle =2^{10}i^{10}=2^{10}(-1)=-2^{10}.
\displaystyle \therefore z^{20}=-2^{10}=-1024.
\displaystyle \\

\displaystyle \textbf{Question 48: }\text{Evaluate }\sum_{n=1}^{13}\left(i^n+i^{n+1}\right),\text{ where }n\in N.
\displaystyle \text{Answer:}
\displaystyle \sum_{n=1}^{13}\left(i^n+i^{n+1}\right)=(i+i^2)+(i^2+i^3)+\cdots+(i^{13}+i^{14}).
\displaystyle \text{Since }i+i^2+i^3+i^4=i-1-i+1=0,
\displaystyle \sum_{n=1}^{12}i^n=0\quad\text{and}\quad\sum_{n=1}^{12}i^{n+1}=0.
\displaystyle \therefore \sum_{n=1}^{13}\left(i^n+i^{n+1}\right)=i^{13}+i^{14}.
\displaystyle i^{13}=i^{12}i=i,\qquad i^{14}=i^{12}i^2=-1.
\displaystyle \therefore \sum_{n=1}^{13}\left(i^n+i^{n+1}\right)=i-1=-1+i.
\displaystyle \therefore \text{The required value is }-1+i.
\displaystyle \\

\displaystyle \textbf{Question 49: }\text{Evaluate }1+i^2+i^4+i^6+\cdots+i^{2n}.
\displaystyle \text{Answer:}
\displaystyle 1+i^2+i^4+i^6+\cdots+i^{2n}=1+(-1)+(-1)^2+(-1)^3+\cdots+(-1)^n.
\displaystyle \text{This is a geometric progression with first term }1\text{ and common ratio }-1.
\displaystyle S=\frac{1-(-1)^{n+1}}{1-(-1)}=\frac{1-(-1)^{n+1}}{2}.
\displaystyle \therefore S=\frac{1+(-1)^n}{2}.
\displaystyle \therefore \text{the value is }1\text{ if }n\text{ is even, and }0\text{ if }n\text{ is odd.}
\displaystyle \\

\displaystyle \textbf{Question 50: }\text{Find the value of }(1+i)^6+(1-i)^3.
\displaystyle \text{Answer:}
\displaystyle (1+i)^2=2i.
\displaystyle \therefore (1+i)^6=(2i)^3=8i^3=-8i.
\displaystyle (1-i)^2=-2i.
\displaystyle \therefore (1-i)^3=(1-i)(-2i)=-2-2i.
\displaystyle \therefore (1+i)^6+(1-i)^3=-8i-2-2i.
\displaystyle =-2-10i.
\displaystyle \therefore \text{The required value is }-2-10i.
\displaystyle \\

\displaystyle \textbf{Question 51: }\text{Find the real values of }x\text{ and }y\text{ if }\frac{x-1}{3+i}+\frac{y-1}{3-i}=i.
\displaystyle \text{Answer:}
\displaystyle \frac{x-1}{3+i}+\frac{y-1}{3-i}=i.
\displaystyle \frac{(x-1)(3-i)+(y-1)(3+i)}{(3+i)(3-i)}=i.
\displaystyle \frac{3x+3y-6+i(y-x)}{10}=i.
\displaystyle \therefore 3x+3y-6+i(y-x)=10i.
\displaystyle \text{Equating the real and imaginary parts, we get}
\displaystyle 3x+3y-6=0\quad\text{and}\quad y-x=10.
\displaystyle \therefore x+y=2\quad\text{and}\quad y-x=10.
\displaystyle \text{Adding the two equations, }2y=12\quad\Rightarrow\quad y=6.
\displaystyle \text{Substituting }y=6\text{ in }x+y=2,\text{ we get}
\displaystyle x+6=2\quad\Rightarrow\quad x=-4.
\displaystyle \therefore x=-4,\quad y=6.
\displaystyle \\

\displaystyle \textbf{Question 52: }\text{Find all non-zero complex numbers }z\text{ satisfying }\bar{z}=iz^2.
\displaystyle \text{Answer:}
\displaystyle \text{Let }z=r(\cos\theta+i\sin\theta),\quad r>0.
\displaystyle \therefore \bar{z}=r(\cos\theta-i\sin\theta)=r\{\cos(-\theta)+i\sin(-\theta)\}.
\displaystyle iz^2=r^2\left\{\cos\left(2\theta+\frac{\pi}{2}\right)+i\sin\left(2\theta+\frac{\pi}{2}\right)\right\}.
\displaystyle \text{Equating the moduli, we get }r=r^2.
\displaystyle \text{Since }r>0,\text{ we get }r=1.
\displaystyle \text{Equating the arguments,}
\displaystyle -\theta=2\theta+\frac{\pi}{2}+2n\pi.
\displaystyle \therefore 3\theta=-\frac{\pi}{2}-2n\pi.
\displaystyle \therefore \theta=-\frac{\pi}{6}+\frac{2k\pi}{3},\quad k=0,1,2.
\displaystyle \text{For }k=0,\quad z=\cos\left(-\frac{\pi}{6}\right)+i\sin\left(-\frac{\pi}{6}\right)
\displaystyle =\frac{\sqrt{3}}{2}-\frac{i}{2}.
\displaystyle \text{For }k=1,\quad z=\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}=i.
\displaystyle \text{For }k=2,\quad z=\cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6}
\displaystyle =-\frac{\sqrt{3}}{2}-\frac{i}{2}.
\displaystyle \therefore z=\frac{\sqrt{3}-i}{2},\quad i,\quad-\frac{\sqrt{3}+i}{2}.
\displaystyle \\

\displaystyle \textbf{Question 53: }\text{If }|1-i|^x=2^x,\text{ find non zero integral solution.}
\displaystyle \text{Answer:}
\displaystyle |1-i|=\sqrt{1^2+(-1)^2}=\sqrt{2}.
\displaystyle \therefore (\sqrt{2})^x=2^x.
\displaystyle 2^{x/2}=2^x.
\displaystyle \therefore \frac{x}{2}=x.
\displaystyle \therefore x=0.
\displaystyle \text{Thus, the only integral solution is }x=0.
\displaystyle \text{Hence, there is no non-zero integral solution.}
\displaystyle \\

\displaystyle \textbf{Question 54: }\text{Solve the equation }|z+1|=z+2(1+i).
\displaystyle \text{Answer:}
\displaystyle \text{Let }z=x+iy.
\displaystyle \therefore |z+1|=\sqrt{(x+1)^2+y^2}.
\displaystyle \text{Also, }z+2(1+i)=x+2+i(y+2).
\displaystyle \text{Since }|z+1|\text{ is real, the imaginary part on the right must be zero.}
\displaystyle \therefore y+2=0\quad\Rightarrow\quad y=-2.
\displaystyle \therefore \sqrt{(x+1)^2+4}=x+2.
\displaystyle \text{Squaring both sides, we get}
\displaystyle (x+1)^2+4=(x+2)^2.
\displaystyle x^2+2x+5=x^2+4x+4.
\displaystyle \therefore 2x=1\quad\Rightarrow\quad x=\frac{1}{2}.
\displaystyle \therefore z=\frac{1}{2}-2i.
\displaystyle \\

\displaystyle \textbf{Question 55: }\text{Find real }\theta\text{ such that }\frac{3+2i\sin\theta}{1-2i\sin\theta}\text{ is purely real.}
\displaystyle \text{Answer:}
\displaystyle \frac{3+2i\sin\theta}{1-2i\sin\theta}\times\frac{1+2i\sin\theta}{1+2i\sin\theta}
\displaystyle =\frac{(3+2i\sin\theta)(1+2i\sin\theta)}{1+4\sin^2\theta}.
\displaystyle =\frac{3-4\sin^2\theta+8i\sin\theta}{1+4\sin^2\theta}.
\displaystyle \text{For the expression to be purely real, its imaginary part must be zero.}
\displaystyle \therefore 8\sin\theta=0.
\displaystyle \therefore \sin\theta=0.
\displaystyle \therefore \theta=n\pi,\quad n\in Z.
\displaystyle \\

\displaystyle \textbf{Question 56: }\text{Show that }\left|\frac{z-2}{z-3}\right|=2\text{ represents a circle. Find centre and radius.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }z=x+iy.
\displaystyle \left|\frac{z-2}{z-3}\right|=2.
\displaystyle \therefore |z-2|=2|z-3|.
\displaystyle \sqrt{(x-2)^2+y^2}=2\sqrt{(x-3)^2+y^2}.
\displaystyle \text{Squaring both sides, we get}
\displaystyle (x-2)^2+y^2=4\{(x-3)^2+y^2\}.
\displaystyle x^2-4x+4+y^2=4x^2-24x+36+4y^2.
\displaystyle \therefore 3x^2+3y^2-20x+32=0.
\displaystyle x^2+y^2-\frac{20}{3}x+\frac{32}{3}=0.
\displaystyle \left(x-\frac{10}{3}\right)^2+y^2=\frac{100}{9}-\frac{32}{3}.
\displaystyle \therefore \left(x-\frac{10}{3}\right)^2+y^2=\frac{4}{9}.
\displaystyle \text{This is the equation of a circle.}
\displaystyle \therefore \text{Centre}=\left(\frac{10}{3},0\right)\text{ and radius}=\frac{2}{3}.
\displaystyle \\

\displaystyle \textbf{Question 57: }\text{Find smallest positive integer }n\text{ for which }\frac{1+i}{(1-i)^{n-2}}\text{ is a real number.}
\displaystyle \text{Answer:}
\displaystyle \frac{1+i}{(1-i)^{n-2}}.
\displaystyle \text{Since }1+i=\sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right),
\displaystyle 1-i=\sqrt{2}\left(\cos\left(-\frac{\pi}{4}\right)+i\sin\left(-\frac{\pi}{4}\right)\right).
\displaystyle \therefore \arg\left(\frac{1+i}{(1-i)^{n-2}}\right)=\frac{\pi}{4}-(n-2)\left(-\frac{\pi}{4}\right).
\displaystyle =\frac{(n-1)\pi}{4}.
\displaystyle \text{For the given complex number to be real, its argument must be an integral multiple of }\pi.
\displaystyle \therefore \frac{(n-1)\pi}{4}=k\pi,\quad k\in Z.
\displaystyle \therefore n-1=4k.
\displaystyle \therefore n=4k+1.
\displaystyle \text{For }k=0,\text{ we get }n=1.
\displaystyle \text{Indeed, for }n=1,\quad\frac{1+i}{(1-i)^{-1}}=(1+i)(1-i)=2,
\displaystyle \text{which is real.}
\displaystyle \therefore \text{the smallest positive integer is }n=1.
\displaystyle \\

\displaystyle \textbf{Question 58: }\text{Find the modulus and the amplitude of }\frac{1+2i}{1-3i}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }z=\frac{1+2i}{1-3i}.
\displaystyle =\frac{(1+2i)(1+3i)}{(1-3i)(1+3i)}.
\displaystyle =\frac{1+3i+2i+6i^2}{10}=\frac{-5+5i}{10}.
\displaystyle \therefore z=-\frac{1}{2}+\frac{1}{2}i.
\displaystyle \text{Therefore, }|z|=\sqrt{\left(-\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2}.
\displaystyle =\sqrt{\frac{1}{4}+\frac{1}{4}}=\frac{1}{\sqrt{2}}.
\displaystyle \text{Since the real part is negative and the imaginary part is positive, }z\text{ lies in the second quadrant.}
\displaystyle \tan\theta=\frac{\frac{1}{2}}{-\frac{1}{2}}=-1.
\displaystyle \therefore \theta=\frac{3\pi}{4}.
\displaystyle \therefore \text{Modulus}=\frac{1}{\sqrt{2}}\text{ and amplitude}=\frac{3\pi}{4}.
\displaystyle \\

\displaystyle \textbf{Question 59: }\text{If }(x+iy)(2-3i)=4+i,\text{ then find }x\text{ and }y.
\displaystyle \text{Answer:}
\displaystyle (x+iy)(2-3i)=4+i.
\displaystyle 2x-3xi+2iy-3i^2y=4+i.
\displaystyle (2x+3y)+i(2y-3x)=4+i.
\displaystyle \text{Equating the real and imaginary parts, we get}
\displaystyle 2x+3y=4\qquad\text{and}\qquad2y-3x=1.
\displaystyle \text{Multiplying the first equation by }2\text{ and the second by }3,\text{ we get}
\displaystyle 4x+6y=8\qquad\text{and}\qquad-9x+6y=3.
\displaystyle \text{Subtracting, }13x=5\quad\Rightarrow\quad x=\frac{5}{13}.
\displaystyle 2\left(\frac{5}{13}\right)+3y=4.
\displaystyle \therefore 3y=\frac{42}{13}\quad\Rightarrow\quad y=\frac{14}{13}.
\displaystyle \therefore x=\frac{5}{13},\quad y=\frac{14}{13}.
\displaystyle \\

\displaystyle \textbf{Question 60: }\text{Find }x\text{ and }y\text{ if }\frac{(1+i)^3}{1-i}+\frac{(1-i)^3}{1+i}=x+iy.
\displaystyle \text{Answer:}
\displaystyle (1+i)^2=2i\quad\Rightarrow\quad(1+i)^3=2i(1+i)=-2+2i.
\displaystyle (1-i)^2=-2i\quad\Rightarrow\quad(1-i)^3=-2i(1-i)=-2-2i.
\displaystyle \therefore \frac{(1+i)^3}{1-i}+\frac{(1-i)^3}{1+i}
\displaystyle =\frac{-2+2i}{1-i}+\frac{-2-2i}{1+i}.
\displaystyle \frac{-2+2i}{1-i}=\frac{(-2+2i)(1+i)}{(1-i)(1+i)}=-2.
\displaystyle \frac{-2-2i}{1+i}=\frac{(-2-2i)(1-i)}{(1+i)(1-i)}=-2.
\displaystyle \therefore x+iy=-2-2=-4.
\displaystyle \text{Comparing real and imaginary parts, we get}
\displaystyle x=-4,\qquad y=0.
\displaystyle \\

\displaystyle \text{LONG QUESTIONS}

\displaystyle \textbf{Question 61: }\text{If }x+iy=\sqrt{\frac{a+ib}{c+id}},\text{ then prove that }
\displaystyle (x^2+y^2)^2= \frac{a^2+b^2}{c^2+d^2}.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }x+iy=\sqrt{\frac{a+ib}{c+id}}.
\displaystyle \text{Squaring both sides, we get}
\displaystyle (x+iy)^2=\frac{a+ib}{c+id}.
\displaystyle \text{Taking modulus on both sides,}
\displaystyle |x+iy|^2=\left|\frac{a+ib}{c+id}\right|.
\displaystyle x^2+y^2=\frac{|a+ib|}{|c+id|}.
\displaystyle =\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}.
\displaystyle \text{Squaring both sides, we get}
\displaystyle (x^2+y^2)^2=\frac{a^2+b^2}{c^2+d^2}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 62: }\text{Find the modulus and argument of the complex number }\frac{1+7i}{(2-i)^2}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }z=\frac{1+7i}{(2-i)^2}.
\displaystyle (2-i)^2=4-4i+i^2=3-4i.
\displaystyle \therefore z=\frac{1+7i}{3-4i}.
\displaystyle =\frac{(1+7i)(3+4i)}{(3-4i)(3+4i)}.
\displaystyle =\frac{3+4i+21i+28i^2}{25}.
\displaystyle =\frac{-25+25i}{25}=-1+i.
\displaystyle \therefore |z|=\sqrt{(-1)^2+1^2}=\sqrt{2}.
\displaystyle \text{Since the real part is negative and the imaginary part is positive, }z\text{ lies in the second quadrant.}
\displaystyle \tan\theta=\frac{1}{-1}=-1.
\displaystyle \text{The reference angle is }\frac{\pi}{4}.
\displaystyle \therefore \arg z=\pi-\frac{\pi}{4}=\frac{3\pi}{4}.
\displaystyle \therefore \text{Modulus}=\sqrt{2}\text{ and argument}=\frac{3\pi}{4}.
\displaystyle \\

\displaystyle \textbf{Question 63: }\text{Find the value of }a\text{ such that the sum of the squares of the roots of the equation}
\displaystyle x^2-(a-2)x-(a+1)=0\text{ is least.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the roots of the equation be }\alpha\text{ and }\beta.
\displaystyle \alpha+\beta=a-2,\qquad \alpha\beta=-(a+1).
\displaystyle \text{Now, }\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta.
\displaystyle =(a-2)^2+2(a+1).
\displaystyle =a^2-4a+4+2a+2.
\displaystyle =a^2-2a+6.
\displaystyle =(a-1)^2+5.
\displaystyle \text{Since }(a-1)^2\geq0,\text{ the least value occurs when }a-1=0.
\displaystyle \therefore a=1.
\displaystyle \text{Hence, the required value of }a\text{ is }1.
\displaystyle \\

\displaystyle \textbf{Question 64: }\text{For what values of }x\text{ and }y\text{ are the numbers }3+ix^2y\text{ and}
\displaystyle x^2+y+4i\text{ conjugate of each other.}
\displaystyle \text{Answer:}
\displaystyle \text{Since the two complex numbers are conjugates, their real parts are equal and}
\displaystyle \text{their imaginary parts are equal in magnitude but opposite in sign.}
\displaystyle \therefore 3=x^2+y\qquad\text{and}\qquad x^2y=-4.
\displaystyle \text{From }3=x^2+y,\text{ we get }y=3-x^2.
\displaystyle \text{Substituting in }x^2y=-4,\text{ we get}
\displaystyle x^2(3-x^2)=-4.
\displaystyle x^4-3x^2-4=0.
\displaystyle (x^2-4)(x^2+1)=0.
\displaystyle \text{Since }x\text{ is real, }x^2=4.
\displaystyle \therefore x=\pm2.
\displaystyle y=3-x^2=3-4=-1.
\displaystyle \therefore x=\pm2,\quad y=-1.
\displaystyle \\

\displaystyle \textbf{Question 65: }\text{If }z=x+iy\text{ and }w=\frac{|1-iz|}{|z-i|}\text{ and }|w|=1\text{ then show that }z\text{ is purely}
\displaystyle \text{real.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }z=x+iy.
\displaystyle 1-iz=1-i(x+iy)=1+y-ix.
\displaystyle \therefore |1-iz|^2=(1+y)^2+x^2.
\displaystyle \text{Also, }z-i=x+i(y-1).
\displaystyle \therefore |z-i|^2=x^2+(y-1)^2.
\displaystyle \text{Since }|w|=1,\text{ we have }\frac{|1-iz|}{|z-i|}=1.
\displaystyle \therefore |1-iz|=|z-i|.
\displaystyle \text{Squaring both sides, we get}
\displaystyle (1+y)^2+x^2=x^2+(y-1)^2.
\displaystyle (1+y)^2=(y-1)^2.
\displaystyle 1+2y+y^2=y^2-2y+1.
\displaystyle \therefore 4y=0\quad\Rightarrow\quad y=0.
\displaystyle \therefore z=x+i(0)=x.
\displaystyle \therefore z\text{ is purely real.}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 66: }\text{Prove that maximum value of }|z|+|z-1|\text{ is }1.
\displaystyle \text{Answer:}
\displaystyle \text{As printed, the question contains an error. The expression }|z|+|z-1|
\displaystyle \text{has no maximum value, since it can increase without bound as }|z|\text{ increases.}
\displaystyle \text{However, its minimum value is }1.
\displaystyle \text{Using the triangle inequality,}
\displaystyle |z|+|z-1|\geq|z-(z-1)|.
\displaystyle =|1|=1.
\displaystyle \therefore |z|+|z-1|\geq1.
\displaystyle \text{Equality holds when }z\text{ lies on the real line between }0\text{ and }1.
\displaystyle \therefore \text{the minimum value of }|z|+|z-1|\text{ is }1.
\displaystyle \\

\displaystyle \textbf{Question 67: }\text{If }|z+1|=\sqrt{2}|z-1|\text{ prove that }z\text{ describes a circle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }z=x+iy.
\displaystyle |z+1|=\sqrt{2}|z-1|.
\displaystyle \sqrt{(x+1)^2+y^2}=\sqrt{2}\sqrt{(x-1)^2+y^2}.
\displaystyle \text{Squaring both sides, we get}
\displaystyle (x+1)^2+y^2=2\{(x-1)^2+y^2\}.
\displaystyle x^2+2x+1+y^2=2x^2-4x+2+2y^2.
\displaystyle \therefore x^2+y^2-6x+1=0.
\displaystyle x^2-6x+9+y^2+1-9=0.
\displaystyle \therefore (x-3)^2+y^2=8.
\displaystyle \text{This is the equation of a circle.}
\displaystyle \therefore \text{Centre}=(3,0)\text{ and radius}=2\sqrt{2}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 68: }\text{If }z_1,z_2\text{ are complex numbers such that }\frac{2z_1}{3z_2}\text{ is purely imaginary}
\displaystyle \text{number then find }\left|\frac{z_1-z_2}{z_1+z_2}\right|.
\displaystyle \text{Answer:}
\displaystyle \text{Since }\frac{2z_1}{3z_2}\text{ is purely imaginary and }\frac{2}{3}\text{ is real,}
\displaystyle \frac{z_1}{z_2}\text{ is also purely imaginary.}
\displaystyle \text{Let }\frac{z_1}{z_2}=iy,\quad y\in R.
\displaystyle \therefore z_1=iyz_2.
\displaystyle \left|\frac{z_1-z_2}{z_1+z_2}\right|=\left|\frac{iyz_2-z_2}{iyz_2+z_2}\right|.
\displaystyle =\left|\frac{iy-1}{iy+1}\right|.
\displaystyle =\frac{|iy-1|}{|iy+1|}.
\displaystyle =\frac{\sqrt{(-1)^2+y^2}}{\sqrt{1^2+y^2}}.
\displaystyle =\frac{\sqrt{1+y^2}}{\sqrt{1+y^2}}=1.
\displaystyle \therefore \left|\frac{z_1-z_2}{z_1+z_2}\right|=1.
\displaystyle \\

\displaystyle \textbf{Question 69: }\text{If }|z|=1\text{ then prove that }\frac{z-1}{z+1}\text{ is purely imaginary. What can you conclude}
\displaystyle \text{if }z=1.
\displaystyle \text{Answer:}
\displaystyle \text{Since }|z|=1,\text{ we have }z\bar{z}=1.
\displaystyle \therefore \bar{z}=\frac{1}{z}.
\displaystyle \text{Let }w=\frac{z-1}{z+1}.
\displaystyle \therefore \bar{w}=\frac{\bar{z}-1}{\bar{z}+1}.
\displaystyle =\frac{\frac{1}{z}-1}{\frac{1}{z}+1}=\frac{1-z}{1+z}.
\displaystyle =-\frac{z-1}{z+1}=-w.
\displaystyle \therefore w+\bar{w}=0.
\displaystyle \text{Hence, }\frac{z-1}{z+1}\text{ is purely imaginary.}
\displaystyle \text{If }z=1,\text{ then }\frac{z-1}{z+1}=\frac{1-1}{1+1}=0.
\displaystyle \therefore \text{for }z=1,\text{ the value of the expression is }0.
\displaystyle \\

\displaystyle \textbf{Question 70: }\text{If the real part of }\frac{z+2}{z-1}=4,\text{ then show that the locus of the point}
\displaystyle \text{representing }z\text{ in the complex plane is a circle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }z=x+iy.
\displaystyle \therefore \frac{z+2}{z-1}=\frac{x+2+iy}{x-1+iy}.
\displaystyle \text{Multiplying the numerator and denominator by }x-1-iy,\text{ we get}
\displaystyle \frac{z+2}{z-1}=\frac{(x+2+iy)(x-1-iy)}{(x-1)^2+y^2}.
\displaystyle \text{The real part is }\frac{(x+2)(x-1)+y^2}{(x-1)^2+y^2}.
\displaystyle \text{Given that the real part is }4,\text{ we have}
\displaystyle \frac{x^2+x-2+y^2}{(x-1)^2+y^2}=4.
\displaystyle x^2+x-2+y^2=4(x^2-2x+1+y^2).
\displaystyle 3x^2-9x+3y^2+6=0.
\displaystyle x^2-3x+y^2+2=0.
\displaystyle \left(x-\frac{3}{2}\right)^2+y^2=\frac{1}{4}.
\displaystyle \text{This is the equation of a circle.}
\displaystyle \therefore \text{Centre}=\left(\frac{3}{2},0\right)\text{ and radius}=\frac{1}{2}.
\displaystyle \text{Hence proved.}
\displaystyle \\

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