Originally Posted by
mathaddict Let $\displaystyle z=a+bi$
$\displaystyle 2|a+bi|=|a+bi-3-3i|$
$\displaystyle 2|a+bi|=|a-3+(b-3)i|$
$\displaystyle 2(a+bi)^2=[a-3+(b-3)i]^2$ $\displaystyle \color{red}|a + bi| \neq (a + bi)^2$ and $\displaystyle \color{red}|(a - 3) + (b - 3)i| \neq [(a - 3) + (b - 3)i]^2$.
$\displaystyle 2(a^2+2abi-b^2)=[(a-3)^2+2(a-3)(b-3)i-(b-3)^2]$
$\displaystyle a^2-b^2+6a-6b+(2ab+6a+6b-18)i=0$
SO $\displaystyle a^2-b^2+6a-6b=0$ ---1
and $\displaystyle 2ab+6a+6b+18=0$ ----2
Now we have to solve the simultaneous equation
From 1
$\displaystyle (a-b)(a+b+6)=0$
$\displaystyle a=b$ , $\displaystyle a=-6-b$
Substitute into 2
$\displaystyle ab+3a+3b-9=0$
so when $\displaystyle a=b$ , $\displaystyle b^2+6b-9=0\Rightarrow b=-3\pm3\sqrt{2} $
and a will be the same .
when $\displaystyle a=-6-b$ ,$\displaystyle (-6-b)b+3(-6-b)+3b-9=0$
$\displaystyle b^2-6b-27=0$
b=9 or b=-3 , so you can find a
so you should get 4 different sets of z .
PLs check through my algebra work .