If for the complex numbers
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.