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.
 A: 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 
$$3|z|^2+30\operatorname{Re}z+27=0$$
A: Putting  $\;z=x+iy\;,\;\;x,y\in\Bbb R\;$ :
$$\left|\frac{z-3}{z+3}\right|=2\iff |(x-3)+iy|^2=4|(x+3)+iy|^2\iff$$
$$x^2-6x+9+y^2=4x^2+24x+36+4y^2\iff$$
$$x^2+10x+y^2=-9\iff (x-5)^2+y^2=16$$
Thus, the set of complex numbers fulfilling the above condition is a circle of radius four.
A: $$\begin{align}
\left\lvert\frac{z-3}{z+3}\right\rvert=2\\
\left\lvert1-\frac{6}{z+3}\right\rvert=2\\
\left\lvert\frac16-\frac{1}{z+3}\right\rvert=\frac13\\
\left\lvert\frac16-Q\right\rvert=\frac13
\end{align}$$
where $Q=\frac{1}{z+3}$. This last equation shows us $Q$ is along a circle of radius $\frac13$ centered at $\frac16$. The reciprocal of a circle in the complex plane is another circle. The circle $Q$ is on has a diameter passing through $-1/6$ and $1/2$ on the real line. So after we take the reciprocal, we have a circle with a diameter on the real line passing through $-6$ and $2$, and so its center is at $-2$ and radius is $4$. 
So $z+3$ is on this circle. And then finally, $z$ is on a circle of radius $4$ centered at $-5$. That is, the solution set is the same as for $\left\lvert z+5\right\rvert=4$.
