TITLE: Equality of triangle inequality in complex numbers
QUESTION [7 upvotes]: $z$ and $w$ be nonzero complex numbers. How do I show that $|z+w|=|z|+|w|$ if and only if $z=sw$ for some real positive number $s$. 
I approached this by letting $z=a+ib$, and $w=c+id$, and kinda play around with it. I also tried to square both sides when proving forward direction, but I could not get it to work. Can anyone get me some ideas, maybe?
Thank you!

REPLY [5 votes]: Here's a cleaner proof of the second direction, which avoids coordinates.
Suppose $|z + w| = |z| + |w|$. Then
$$
\begin{align*}
|z|^2 + 2|zw| + |w|^2
&= (|z| + |w|)^2 \\
&= |z + w|^2 \\
&= (z + w)(\overline{z + w}) \\
&= z\overline{z} + z\overline{w} + w\overline{z} + w\overline{w} \\
&= |z|^2 + 2\mathrm{Re}(z\overline{w}) + |w|^2.
\end{align*}
$$
This implies $|zw| = \mathrm{Re}(z\overline{w})$. But because $|zw| = |z\overline{w}|$, so $z\overline{w}$ must be a nonnegative real number. Hence, $\arg(z) = \arg(w)$, and we conclude $z = sw$ for some $s > 0$.