If $(z-2)/(z+2)$ is purely imaginary, find $z$ that satisfies this condition? 
If $(z-2)/(z+2)$ is purely imaginary, then what is the value of $z$ that satisfies this condition?

I am troubled solving this equation for a long time. I tried to put $z = a + i b$ and then I got $(a-2+ib)(a+2+ib)$ is imaginary. 
Not sure how to go now? Thank youvery much.

Putting $z=a+ib$, with $z\neq-2$, is a correct way to start. Then, you get \begin{align}\frac{z-2}{z+2}&=\frac{z-2}{(a+2)+ib}=\frac{(a-2)+ib}{(a+2)+ib}\cdot\frac{(a+2)-ib}{(a+2)-ib}\\[0.2cm]&=\frac{(a-2)(a+2)+b^2+ib(a+2-a+2)}{(a+2)^2+b^2}=\underbrace{\frac{(a-2)(a+2)+b^2}{(a+2)^2+b^2}}_{=\text{ real part }=0}+i\frac{4b}{(a+2)^2+b^2}\end{align} So, you get the equation $$(a-2)(a+2)+b^2=0\iff a^2+b^2=2^2 \iff |z|=2$$ which gives a set of possible solutions.
