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 $\operatorname{Re}(\alpha)$ is equal to	1	0
Q	If $x = p +i q$ is a complex number such that $x ^2=3+4 i$ and $x ^3=2+11 i$ 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 $\frac{|3 z +9-7 i|}{4}$ 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, $\bar{\alpha}=\frac{\frac{16}{ z }-\frac{16}{\bar{\omega}}}{16+\left(\frac{16}{ z }\right)\left(\frac{16}{\bar{\omega}}\right)}=-\alpha \Rightarrow \alpha+\bar{\alpha}=0 \Rightarrow \operatorname{Re}(\alpha)=0$.
(Q)$ x =\frac{ x ^3}{ x ^2}=\left(\frac{2+11 i}{3+4 i}\right) \times \frac{(3-4 i)}{(3-4 i)}=(2+i) $
$\therefore p =2, q =1$
$\Rightarrow ( p + q )=2+1=3$
(R) We have, $\overline{ z }=i z ^2 ; $ Put $z = x +i y$
$\Rightarrow x -i y =i\left( x ^2- y ^2+2 ixy \right)$
$\therefore $ On solving, we get
$z =0+i 0, i, \frac{\sqrt{3}}{2}-\frac{i}{2}, \frac{-\sqrt{3}}{2}-\frac{i}{2}$
$\Rightarrow 4$ solutions
(S) $|3 z +9-7 i| =|(3 z +6-3 i)+(3-4 i)| $
$ \leq|3 z +6-3 i|+|3-4 i|$
$ =3| z +2-i|+5 $
$ =20$