If $Im(z)>0$ and $Im(w)>0$, prove that $|z-\bar{w}|>|z-w|$ I'd like some feedback on my proof. Is it up to the standards of a graduate student? Are there any parts that remain unclear? Any suggestions on how to improve future proofs is also appreciated.
If $Im(z)>0$ and $Im(w)>0$, prove that $|z-\bar{w}|>|z-w|$
Let $z=x+iy$ and $w=a+ib\;$ both be complex numbers and $\;a,b,x,y \in \mathbb{R}$
$|z-\bar{w}|>|z-w| \iff |z-\bar{w}|^{2}>|z-w|^{2} \iff |z|^{2} -zw-\bar{z}\bar{w} +|w|^{2}>|z|^{2} -z\bar{w}-\bar{z}w +|w|^{2}$
$\iff -zw-\overline{zw}>z\bar{w}-\overline{z\bar{w}} \iff -2Re(zw)>-2Re(z\bar{w})$
Remember that $2\cdot Re(z) = z+\bar{z},\; \forall z \in \mathbb{C}$
As such, $-2Re(zw)>-2Re(z\bar{w}) \iff -ax+yb>-ax-yb \iff yb>-yb$
$ \iff Im(z)\cdot Im(w)>-Im(z)\cdot Im(w)$
which holds true since $Im(z)>0$ and $Im(w)>0$.
 A: Your proof is correct, comparing the squared distances is the right way to go.
You can shorten the argument to
$$
\begin{align}
 |z-\bar w|^2 - |z-w|^2 &= -zw - \bar z \bar w + z \bar w+\bar z w \\
 &= - (z- \bar z)(w-\bar w) \\
 &= - 2i \operatorname{Im}(z) \cdot 2i  \operatorname{Im}(w) \\
 &= 4 \operatorname{Im}(z) \cdot \operatorname{Im}(w) \\
 &> 0 \, .
\end{align}
$$
A geometric interpretation
The points $z, w, \bar w, \bar z$ form an isosceles trapezoid in the plane, with  $a=2\operatorname{Im}(z)$ and $b=2\operatorname{Im}(w)$ being the lengths of the parallel sides, $c=|z-w|$ the length of a leg, and $p=|z-\bar w|$ the length of the diagonal. So what we have derived is
$$
p = \sqrt{ab + c^2}
$$
and that is exactly the formula for the length of the diagonal. In particular, $p > c$.
A: Another geometric interpretation:  the real line is the perpendicuar bisector for the line-semgment from $w$ to $\overline{w}$.  Since $z$ is on the same side of the line as $w$, the distance $|z-w|$ is less than the distance $|z-\overline{w}|$.
