Complex isosceles triangle... Let complex numbers $z_1,z_2,z_3$ be vertices of a right isosceles triangle with $z_3$ at the 90 degree angle. Show that: 
$(z_1-z_3)^2=2(z_1-z_3)(z_2-z_3)$
I have tried setting them equal and deriving a simpler problem from it, using Pythagorean's theorem, translating the vertices to the origin, but nothing is giving me this result. What am I missing?
 A: Let $P(z_1), Q(z_2)$ and $R(z_3)$ be the vertices of an isoceles triangle $PQR$ right angle of $Q$. Rotate $Q(z_2)$ about $P$ in anticlockwise direction, we have
$z_3 - z_1 = \frac{|z_3-z_1|}{|z_2-z_1|} (z_2-z_1)e^{\frac{i\pi}{4}}$
which is equivalent to $\frac{PR}{PQ}$$(z_2-z_1)e^{\frac{i\pi}{4}}$ = $\sqrt2$$(z_2-z_1)e^{\frac{i\pi}{4}}$  $\cdots (1)$
Similarly by rotating $Q(z_2)$ about $R(z_3)$ in clockwise direction, we have
$z_1-z_3=\frac{|z_1-z_3|}{|z_2-z_3|}(z_2-z_3)e^{\frac{-i\pi}{4}}$
which is equivalent to $\frac{PR}{QR}$$(z_2-z_3)e^{\frac{-i\pi}{4}}$=$\sqrt2$$(z_2-z_3)e^{\frac{-i\pi}{4}}$ $\cdots(2)$
Multipying $(1)$ and $(2)$ we get
$(z_1-z_3)^2=2(z_1-z_2)(z_2-z_3)$ as the desired result.
A: If the angle at $z_2$ is a right angle, the correct equation is
$$\begin{equation*}
(z_1-z_3)^2 = 2(z_1-z_2)(z_2-z_3)\tag{1}
\end{equation*}$$
as written by @Shobhit.
Without loss of generality, let
$z_1=a$, $z_2=0$, and $z_3 = ia$,
where $a>0$.
We have
$$(z_1-z_3)^2 = (a-i a)^2 = (1-i)^2a^2 = (1-2i-1)a^2 = -2i a^2$$
and
$$2(z_1-z_2)(z_2-z_3) = 2a(-ia) = -2ia^2.$$
Note: All other such triangles can be had by the transformation
$z_i\to \alpha z_i + \beta$,
where $\alpha,\beta\in\mathbb{C}$.
(This transformation rotates, stretches, and shifts the original triangle.)
Note that under this transformation
$$z_i-z_j \to \alpha(z_i-z_j),$$
and so
$$\begin{eqnarray*}
(z_1-z_3)^2 &\to& \alpha^2(z_1-z_3)^2 \\
2(z_1-z_2)(z_2-z_3) &\to& 2\alpha^2(z_1-z_2)(z_2-z_3).
\end{eqnarray*}$$
Thus, (1) holds for all isosceles right triangles. 
