$\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. 
 A: How can you use the condition $| z_1 | = | z_2 |$? This invites to use the representation of complex numbers in polar coordinates. 
So if one writes $z_1 = r e^{i \theta_1}$ and $z_2 = r e^{i \theta_2}$, the question amounts to show that $\frac{e^{i \theta_1} + e^{i \theta_2}}{e^{i \theta_1} - e^{i \theta_2}}$ is imaginary. Diviving the numerator and denominator by $e^{i(\theta_1 - \theta_2)/2}$ leads to the result. 
A: 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.
A: $$\frac{z_1+z_2}{z_1-z_2}+\frac{\overline z_1+\overline z_2}{\overline{z_1-z_2}}$$
$$=\frac{2\overline z_1z_1-2\overline z_2z_2}{|z_1-z_2|^2}=0$$
as $z\overline z=|z|^2=|\overline z|^2 $
and as $x+iy+\overline{x+iy}=2x$
A: 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 $(a+c)(a-c)+(b+d)(b-d)=a^2+b^2-(c^2+d^2)=|z_1|^2-|z_2|^2$
