<s>[INST] On the property $\lvert x + y \rvert \leq \lvert \lvert x \rvert + \lvert y \rvert \rvert$ for complex numbers Suppose we have complex numbers, $x$ and $y$. My question is: we know, from the triangle inequality, that $\lvert x + y \rvert \leq \lvert \lvert x \rvert + \lvert y \rvert \rvert$. It makes sense, geometrically, when the strict inequality would hold. Whenever, however, would we have equality? Would it be necessary that $\lvert x \rvert = \rvert y \rvert$, or is there some other necessary condition?
Thanks. 
 [/INST] \begin{align*}
&|x + y| = |x| + |y| \\
\implies \, &|x + y|^2 = (|x| + |y|)^2 \\
\implies \, &|x + y|^2 = |x|^2 + |y|^2 + 2|x||y| \\
\implies \, &(x + y)\overline{(x + y)} = x\overline{x} + y\overline{y} + 2|x||y| \\
\implies \, &x\overline{x} + y\overline{y} + x\overline{y} + y\overline{x} = x\overline{x} + y\overline{y} + 2|x||y| \\
\implies \, &x\overline{y} + \overline{x\overline{y}} = 2|x||y| \\
\implies \, &2\Re(x\overline{y}) = 2|x\overline{y}| \\
\end{align*}
So, we have a complex number $z$ such that $\Re(z) = |z|$. If $z = a + ib$, then
$$a = \sqrt{a^2 + b^2} \implies a^2 = a^2 + b^2 \implies b^2 = 0 \implies b = 0,$$
hence $z$ is real. Moreover, $z$ must be positive. Therefore, $x\overline{y} \in [0, \infty)$. The argument of $x\overline{y}$ is $0$, hence the difference between the arguments of $x$ and $y$ is $0$. That is, $x$ and $y$ must lie on the same ray from the origin, and therefore must be a positive real multiple of each other.
</s>