Find the set of complex numbers $z$ which satisfy: $\left\lvert\frac{z-3}{z+3}\right\rvert=2$ Find the set of complex numbers $z$ which satisfy
$$\left\lvert\frac{z-3}{z+3}\right\rvert=2\text.$$
I need help on that one. Thank you.

Take square both sides then we get
$$|z-3|^2=4|z+3|^2.$$
Evaluate it (using the fact $|z|^2=z\overline{z}$ and $2\operatorname{Re}z=z+\overline{z}$) then
$$3|z|^2+30\operatorname{Re}z+27=0$$
Let $z=x+iy$. From above equation, we get
$$x^2+10x+y^2+9=0$$
and is equivalent to $(x+5)^2+y^2=4^2$

Details: From $|z-3|^2=4|z+3|^2$, we get
$$(z-3)\overline{(z-3)}=4(z+3)\overline{(z+3)}$$
and is equivalent to
$$(z-3)(\overline{z}-3)=4(z+3)(\overline{z}+3).$$
Expand the above equality then
$$z\overline{z}-3(z+\overline{z})+9=4(z\overline{z}+3(z+\overline{z})+9)$$
Note that $z\overline{z}=|z|^2$ and $z+\overline{z}=2\operatorname{Re}(z)$ (you can check it easily - take $z=x+iy$ and calculate it) so 
$$3|z|^2+30\operatorname{Re}z+27=0$$
