Question

Match $\text{List I}$ with the $\text{List II}$ and select the correct answer using the code given below the lists :

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

Which of the following combinations is CORRECT?

• A. $\text{(C)}\to \text{(P)},\text{(D)}\to \text{(Q)}$
• B. $\text{(C)}\to \text{(Q)},\text{(D)}\to \text{(S)}$
• C. $\text{(C)}\to \text{(R)},\text{(D)}\to \text{(Q)}$
• D. $\text{(C)}\to \text{(R)},\text{(D)}\to \text{(S)}$

Answer 1

The correct option is D $\text{(C)}\to \text{(R)},\text{(D)}\to \text{(S)}$

$\left(C\right)$
We have, $\overline{z}=i{z}^2$
Put $z=x+iy$
$\Rightarrow x-iy=i(x^2-y^2+2ixy)$
On solving, we get
$x=-2xy\Rightarrow x=0$ or $y=-\frac{1}{2}$
and $-y=x^2-y^2$
For $x=0,$ we get $y=0,1$
and for $y=-\frac{1}{2},$ we get $x=\pm \frac{\sqrt{3}}{2}$
Hence, $z=0+i0,i,\frac{\sqrt{3}}{2}-\frac{i}{2},\frac{-\sqrt{3}}{2}-\frac{i}{2}$
$\Rightarrow 4$ solutions are there.

$\left(D\right)$
$|3z+9-7i|$
$=|(3z+6-3i)+(3-4i)|\le |3z+6-3i|+|3-4i|\le 3|z+2-i|+5\le 20$