If $\frac{(z+w)^2}{zw}$ is real, then $z, w$ and the origin of the complex plane are either colinear or vertext of triangle I need to prove that if $\frac{(z+w)^2}{zw}$ is real, then $z, w$ and the origin of the complex plane are either colinear or vertex of an isosceles triangle. 
I first tried some computations with the real part, like this:
$$Re\left(\frac{(z+w)^2}{zw}\right) = Re\left(\frac{z^ 2+2zw+w^2}{zw}\right) = \frac{z_1^2-z_2^2+2(z_1w_1-z_2w_2)+w_1^2-z_2^2}{z_1w_1-z_2w_2}$$
but them I remembered that I must set the imaginary part to $0$, because I'm not just intersted in the real part, I must also deal with the imaginary part neing $0$, so this is the equaiton:
$$\frac{z^2+2zw+w^2}{zw} = \frac{(z_1^2+2z_1z_2i-z_2^2)+2(z_1w_1+iz_1w_2+iz_2w_1-z_2w_2)+w_1^2+2w_1w_2-w_2^2}{z_1w_1+iz_1w_2+iz_2w_1-z_2w_2}=$$
$$\frac{z_1^2-z_2^2+2z_1w_1-2z_2w_2+w_1^2-w_2^2}{\cdots}+\cdots$$
Ok, I just realized that I'll also have to multiply by the conjugate below, and this will result in more squared terms. This is getting heavy, so I don't think this is the way to do it. Could somebody give me a help?
 A: If $\frac{(z+w)^2}{zw}$, $z\ne 0$, $w\ne 0$, is a real number, then its imaginary part is zero.  Therefore, we are seeking a relationship between $z=\rho e^{i\phi}$ and $w=re^{i\theta}$ such that
$$\begin{align}
0&=\text{Im}\left(\frac{(z+w)^2}{zw} \right)\\\\
&=\text{Im}\left(\frac{z}{w}+2+\frac{w}{z} \right)\\\\
&=\text{Im}\left(\frac{\rho}{r}e^{i(\phi-\theta)}+2+\frac{r}{\rho}e^{-i(\phi-\theta)} \right)\\\\
&=\left(\frac{\rho}{r}-\frac{r}{\rho}\right)\sin(\phi-\theta)\tag 1
\end{align}$$
This is true if either $\rho =r$ or $\phi = \theta\pm n\pi$.  
If $\rho =r$, then $0$, $z$, and $w$ form a triangle (unless $\phi =\theta \pm n\pi$ also).  
If $\phi =\theta \pm n\pi$, then $0$, $z$, and $w$ lie on the same line.
A: I think the computation is easier in polar notation. Let $z=ae^{i\theta}, w=be^{i\phi}$, where $a,b>0$. Then $$\frac{(z+w)^2}{zw}=\frac{a}{b}e^{i(\theta-\phi)}+\frac{b}{a}e^{i(\phi-\theta)}+2$$ and we want the imaginary part of this to be zero, that is $$\sin(\theta-\phi)(\frac{a^2}{b^2}-1)=0.$$ From this you get either $a=b$ (so that $|z|=|w|$, i.e. isosceles triangle) or $\theta=\phi, \theta=\phi+\pi$ (so that $0,z,w$ are on a line).
