If $z$ and $w$ are complex numbers such that $|(z+w)| = |(z-w)|$, prove that $\arg(z)-\arg(w) = \pm (\pi/2)$

Geometrically, consider the parallelogram with vertices $0,z,w,z+w$. Then the condition is that its two diagonals have the same length, so it must be a rectangle, that is $z$ and $w$ are orthogonal.

Algebraically, from $$(z+w)(\bar{z}+\bar{w})=(z-w)(\bar{z}-\bar{w})$$ one obtains $z\bar{w}+w\bar{z}=0$ and now just switch to polar notation.

Since $\arg(0)$ is not defined, let's assume that $z,w\ne0$. Then dividing by $\left|\,w\,\right|$ gives $$ \left|\,\frac zw-1\,\right|=\left|\,\frac zw+1\,\right| $$ This means that $\frac zw$ is equidistant from $-1$ and $1$; that is, $\frac zw$ is pure imaginary. Thus, $$ \arg\left(\frac zw\right)=\pm\frac\pi2 $$ which is the same as $$ \arg(z)-\arg(w)=\pm\frac\pi2 $$

$|z+w| = |z-w|$ means that $z$ is equidistant from $\pm w$, and so $z$ lies on the line bisecting the segment $[-w,w]$. This line is perpendicular to the segment, hence the result.