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.
 A: Another way would be to see that if $x$ is purely imaginary, then $x + \bar{x} = 0$
$$\frac{z-2}{z+2} + \frac{\bar{z}-2}{\bar{z}+2} = 0$$
Assume $z \neq -2$, then on solving you have:
$$z\bar z -2 \bar z + 2z -4 + z\bar z -2 z + 2\bar z -4 =0 \\
z\bar z = 4$$
Or that $|z| = 2$. This is a circle with radius $2$ and centre origin. Obviously note that $z = -2$ is not a solution.
A: Let $0 \ne b \in \mathbb R$
\begin{align}
   \frac{z-2}{z+2} &= bi \\
   z-2 &= bzi+2bi \\
   (1-bi)z&= 2(1+bi) \\
   z &= 2\frac{1+bi}{1-bi} \\
   z &= 2\dfrac{(1-b^2)+2bi}{1+b^2}
\end{align}
A: $$\frac{a-2+ib}{a+2+ib} = \frac{(a-2+ib)(a+2-ib)}{(a+2-ib)(a+2-ib)}$$
Now, simplify the expression, and remember how $(x+iy)(x-iy)=x^2+y^2$ is a real number. Therefore, you will get something like
$$\frac{\text{something} + (\text{something else})\cdot i}{\text{some real number}}$$
which is the same as
$$\frac{\text{something}}{\text{some real number}} + \frac{\text{something else}}{\text{some real number}}\cdot i$$
A: 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.
