Question:
If for the complex numbers $z$ satisfying $|z-2-2 i| \leq 1$, the maximum value of $|3 i z+6|$ is attained at $\mathrm{a}+i \mathrm{~b}$, then $\mathrm{a}+\mathrm{b}$ is equal to
Solution:
$|z-2-2 i| \leq 1$
$|x+i y-2-2 i| \leq 1$
$|(x-2)+i(y-2)| \leq 1$
$(x-2)^{2}+(y-2)^{2} \leq 1$
$|3 i z+6|_{\max }$ at $a+i b$
|3il $\left|z+\frac{6}{3 i}\right|$
$3|z-2 i|_{\max }$
From Figure maximum distance at $3+2 \mathrm{i}$
$a+i b=3+2 i=a+b=3+2=5$ Ans.
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