$\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. 

Setting $z_1=a+ib,z_2=c+id$
$$\frac{z_1+z_2}{z_1-z_2}=\frac{a+c+i(b+d)}{a-c+i(b-d)}$$
$$=\frac{\{a+c+i(b+d)\}\{a-c-i(b-d)\}}{(a-c)^2+(b-d)^2}$$
Clearly, the real part of the numerator is $(a+c)(a-c)+(b+d)(b-d)=a^2+b^2-(c^2+d^2)=|z_1|^2-|z_2|^2$
