Question

If $|z-1-i|=1$ and $|w-1+i|=2$ where, $w,zϵC$ then maximum value of $\frac{1}{|4z-1{|}^2}+\frac{1}{|3w+1{|}^2}$ is

• A. 1
• B. 3
• C. 2
• D. 4

Answer 1

The correct option is C 2

We need to maximize $\frac{1}{|4z-1{|}^2}+\frac{1}{|3w+1{|}^2}$
$\Rightarrow \frac{1}{16{|z-\frac{1}{4}|}^2}+\frac{1}{9{|w+\frac{1}{3}|}^2}$
So we need to minimize ${|z-\frac{1}{4}|}^{2}and{|w+\frac{1}{3}|}^2$
Minimum ${|z-\frac{1}{4}|}^2={(\sqrt{{(1-\frac{1}{4})}^2+1^2}-1)}^2=\frac{1}{16}$
Minimum ${|w+\frac{1}{3}|}^2={(2-\sqrt{{(1+\frac{1}{3})}^2+1^2})}^2={(2-\frac{5}{3})}^2=\frac{1}{9}$
So answer is 2

$d_{min}=|z-\frac{1}{4}|$
$f_{min}=|w+\frac{1}{3}|$