Show that $|z-z_1|^2 + |z-z_2|^2 = |z_1 - z_2|^2$ The problem is : if $z$ lies on a circle with diameter having endpoints $z_1$ and $z_2$ then show that $|z-z_1|^2 + |z-z_2|^2 = |z_1 - z_2|^2$ where $z, z_1, z_2 \in \mathbb{C}$.
The angle subtended by the diameter on any point on the circle is a right angle and thus $|z-z_1|$, $|z-z_2|$ and $|z_1-z_2|$ are the lengths of a right-angled triangle. The equation above then follows from the Pythagorean Theorem.
Now the equation for $z$ can also be written as $\left|z - \left(\dfrac{z_1+z_2}{2}\right)\right| = \dfrac{|z_1-z_2|}{2}$ since $\left(\dfrac{z_1+z_2}{2}\right)$ is the centre and $\dfrac{|z_1-z_2|}{2}$ is the radius of the circle. 
But since the locus of both the equations are the same, I figured that it should be possible to prove them equal. So here's what I did:
$$\begin{align}
\left|z - \left(\dfrac{z_1+z_2}{2}\right)\right| = \dfrac{|z_1-z_2|}{2} &\iff |2z - (z_1+z_2)| =|z_1-z_2| \\
&\iff |(z - z_1)+ (z-z_2)|^2=|z_1-z_2|^2 \end{align}$$
Using $|z_1+z_2|^2=|z_1|^2 + |z_2|^2 + \Re{(z_1\overline{z_2})}$,
$|z - z_1|^2+|z-z_2|^2  + \Re{((z-z_1)\overline{(z-z_2)})}=|z_1-z_2|^2$. Comparing it with what I have to show, it seems like I have to prove $\Re{((z-z_1)\overline{(z-z_2)})} =  0$, but I can't think of a way of doing that.
Thank you in advance!
Edit: To sum it up, my question is how to show that $$\left|z - \left(\dfrac{z_1+z_2}{2}\right)\right| = \dfrac{|z_1-z_2|}{2}\iff|z-z_1|^2 + |z-z_2|^2 = |z_1 - z_2|^2$$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,$where $z, z_1, z_2 \in \mathbb{C}$.
 A: If $z=x+iy,z_k=x_k+iy_k$ for $k=1,2$
the real part of $(z-z_1)(\overline{z-z_2})$ is $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)$
Using Article $145$ The elements of coordinate geometry by Loney 
$$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$$
A: After reading the comment, the question seems to be

If $z_1$ and $z_2$ are at different ends of the diameter of a circle, and $z$ is also on that circle, prove
  $$
\mathrm{Re}((z-z_1)(\overline{z-z_2}))=0
$$

Method 1: One way to see this is to note that $\mathrm{Re}(z\bar{w})=z\cdot w$ (dot product as vectors). Since $z-z_1$ and $z-z_2$ are perpendicular, $(z-z_1)\cdot(z-z_2)=0$.
Method 2: Another way of looking at this, is to let $c$ be the center of the circle and let $w=z-c$, $w_1=z_1-c$, and $w_2=z_2-c$. Then $w_2=-w_1$ and $|w|=|w_1|=|w_2|=r$
$$
\begin{align}
(z-z_1)(\overline{z-z_2})
&=(w-w_1)(\overline{w-w_2})\\
&=w\bar{w}+w_1\bar{w}_2-w\bar{w}_2-\bar{w}w_1\\
&=w\bar{w}-w_1\bar{w}_1+(w\bar{w}_1-\bar{w}w_1)\\
&=r^2-r^2+2i\,\mathrm{Im}(w\bar{w}_1)\\
&=2i\,\mathrm{Im}(w\bar{w}_1)
\end{align}
$$
which is pure imaginary. Therefore,
$$
\mathrm{Re}((z-z_1)(\overline{z-z_2}))=0
$$
A: WLOG suppose we have the circle $|z|=r\;$ and that $\,z_1=-r\;,\;\;z_2=r\;,\;z=re^{it}\;,\;\;0<t<\pi\;$ , then
$$|z-z_1|^2+|z-z_2|^2=|r(\cos t+1)+ir\sin t|^2+|r(\cos t-1)+i\sin t|^2=$$
$$=r^2(\cos^2t+2\cos t+1)+r^2\sin^2t+r^2(\cos^2t-2\cos t+1)+r^2\sin^2t=$$
$$=4r^2=|z_1-z_2|^2$$
