Question

List I		List II
P	Let $z,\omega ,\alpha$ be complex numbers such that $|z|=|\omega |=4$ and $\alpha =\frac{z-\bar{\omega }}{16+z\bar{\omega }}$, then $\mathrm{Re}(\alpha )$ is equal to	1	0
Q	If $x=p+iq$ is a complex number such that $x^2=3+4i$ and $x^3=2+11i$ where $i=\sqrt{-1}$, then $(p+q)$ is equal to	2	3
R	Number of complex numbers z satisfying the equation $\overline{z}=i{z}^2$, where $i=\sqrt{-1}$ is equal to	3	4
S	If $z\in C$ satisfies $|z+2-i|=5$ then the maximum value of $\frac{|3z+9-7i|}{4}$ is equal to	4	5

• A. P= 4 ,Q= 3 ,R= 2 ,S= 1
• B. P= 1 ,Q=2,R= 4 ,S= 3
• C. P= 1 ,Q=2,R= 3 ,S= 4
• D. P= 2 ,Q=1 ,R=3 ,S= 4

Answer 1

Answer: (C)

(P) We have, $\alpha =\frac{z-\bar{\omega }}{16+z\bar{\omega }}\Rightarrow \bar{\alpha }=\frac{\bar{z}-\omega }{16+\bar{z}\omega }$
But $z\overline{z}=\omega \bar{\omega }=16$
So, $\bar{\alpha }=\frac{\frac{16}{z}-\frac{16}{\bar{\omega }}}{16+\left(\frac{16}{z}\right)\left(\frac{16}{\bar{\omega }}\right)}=-\alpha \Rightarrow \alpha +\bar{\alpha }=0\Rightarrow \mathrm{Re}(\alpha )=0$.
(Q)$x=\frac{x^3}{x^2}=\left(\frac{2+11i}{3+4i}\right)\times \frac{(3-4i)}{(3-4i)}=(2+i)$
$\therefore p=2,q=1$
$\Rightarrow (p+q)=2+1=3$
(R) We have, $\overline{z}=i{z}^2;$ Put $z=x+iy$
$\Rightarrow x-iy=i\left(x^2-y^2+2ixy\right)$
$\therefore$ On solving, we get
$z=0+i0,i,\frac{\sqrt{3}}{2}-\frac{i}{2},\frac{-\sqrt{3}}{2}-\frac{i}{2}$
$\Rightarrow 4$ solutions
(S) $|3z+9-7i|=|(3z+6-3i)+(3-4i)|$
$\le |3z+6-3i|+|3-4i|$
$=3|z+2-i|+5$
$=20$