$\text{If } |z_1| = |z_2|, \text{ show that } \frac{z_1 + z_2}{z_1-z_2} \text{is imaginary.} $ $\text{If } |z_1| = |z_2|, \text{ show that } \frac{z_1 + z_2}{z_1-z_2} \text{is imaginary.} $
The first thing I tried to do was to multiply both top and bottom by the conjugate of the denominator...
$$
\frac{z_1 + z_2}{z_1-z_2} \left( \frac{z_1 + z_2}{z_1+z_2} \right) \\
= \frac{z_1^2 + 2z_1z_2 + z_2^2}{z_1^2-z_2^2} 
$$
Then I $\text{Let }z_1,z_2 = x_1+iy_1,x_2+iy_2$ and expanded.. but then the equation was too big to work with. What I wanted to do was to simplify as much as I can, such as I did with $\frac{1-z}{1+z}$ which just equaled $\frac{-i\sin\theta}{1+\cos\theta}$ (after being written in Mod-Arg form, of course). So, what should I do from here on? Thanks in advance. 

Multiply instead by $\dfrac{\bar{z_1} + \bar{z_2}}{\bar{z_1} + \bar{z_2}}$ to get $\displaystyle \frac{|z_1|^2 + z_1\bar{z_2} + \bar{z_1}{z_2} + |z_2|^2}{|z_1|^2 + z_1\bar{z_2} -  \bar{z_1}{z_2} - |z_2|^2}$. The denominator is imaginary and the numerator is real.
