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.

$$\begin{align}
\left\lvert\frac{z-3}{z+3}\right\rvert=2\\
\left\lvert1-\frac{6}{z+3}\right\rvert=2\\
\left\lvert\frac16-\frac{1}{z+3}\right\rvert=\frac13\\
\left\lvert\frac16-Q\right\rvert=\frac13
\end{align}$$
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 $\left\lvert z+5\right\rvert=4$.
