Geometric Solution for Equation with Complex Numbers Given we have two complex numbers $z_1$ and $z_2$ with $|z_1| = |z_2|$.
How can it be shown geometrically, that $\frac{z_1+z_2}{z_1-z_2}$ is purely imaginery?
 A: Consider the quadrilateral with vertices $0, z_1, z_1+z_2, z_2$. The condition $\lvert z_1\rvert = \lvert z_2 \rvert$ makes it a rhombus. Thus the two diagonals $z_1+z_2$ and $z_1 - z_2$ are orthogonal.
A: Here is an elementary way for checking that we don't have a real part for your complex fraction. Let we have $$z=\frac{z_1}{z_2}=\frac{x_1+iy_1}{x_2+iy_2}$$ then with a tedious hand calculation we get: $$z=\left(\frac{x_1x_2+y_1y_2}{x_2^2+y_2^2}\right)+i\left(\frac{y_1x_2-y_2x_1}{x_2^2+y_2^2}\right)$$ This is for $z_1$ and $z_2$. Now you set $Z_1=z_1+z_2$ and $Z_2=z_1-z_2$ in above result and by assuming $$x_1^2-x_2^2=-(y_1^2+y_2^2)$$ you'll get the final result. Sorry if this way is not a geometrically one.
A: Firstly $z_1$ and $z_2$ lie on a circle of radius $r=|z_1|=|z_2|$. Now wherever $A=z_1+z_2$ is, $B=z_1-z_2=z_1+z_2-(2z_2)$ is found by reflecting $z_1+z_2$ through the point $z_1$.
Now consider the circle with centre $z_1$ and radius $r$ (and so contains $A$ and $B$ - draw a picture). Now because $|z_1|=r$ we have that $\angle AOB$ is the angle in a semicircle and so is $\pi/2$. 
Therefore 
$$\arg A=\arg B\pm\pi/2\Rightarrow \arg A-\arg B=\pm\pi/2.$$
Now define the multiplication of complex numbers $z$ by $w$ geometrically: then $z\times w$ is $z$ is stretched by a factor $|w|$ and rotated through an angle $\arg w$ in the anti-clockwise sense. If you don't want to do that just show $$(r(\cos\theta+i\sin\theta))(s(\cos\alpha)+i\sin\alpha)=rs(\cos(\theta+\alpha)+i\sin(\theta+\alpha)).$$
Dividing by $w$ includes a clockwise rotation through an angle $\arg w$ - i.e. taking $\arg w$ away from $\arg z$. 
So we have 
$$\frac{z_1+z_2}{z_1-z_2}=\frac{A}{B}.$$
Now starting at $A$, dividing by $B$ includes a clockwise rotation through an angle $\arg B$ so 
$$\arg\frac{A}{B}=\arg A-\arg B,$$
and we have already shown that this is $\pm\pi/2$. Therefore 
$\displaystyle\frac{z_1+z_2}{z_1-z_2}$ is imaginary.
