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. 

Let $x=a+bi, y=c+di,$ where $a,b,c,d \in \mathbb{R}.$ Then $$|x+y|=|(a+c)+(b+d)i|=\sqrt{(a+c)^2+(b+d)^2},$$ and $$|x|+|y|=\sqrt{a^2+b^2}+\sqrt{c^2+d^2}.$$ Thus, the inequality is equivalent to $$\sqrt{(a+c)^2+(b+d)^2} \leq \sqrt{a^2+b^2}+\sqrt{c^2+d^2},$$by squaring both sides, namely,$$ac+bd \leq \sqrt{(a^2+b^2)(c^2+d^2)}.$$
But in fact, by Cauchy's inequality, we have $$(a^2+b^2)(c^2+d^2) \geq (ac+bd)^2.$$ Hence,$$\sqrt{(a^2+b^2)(c^2+d^2)} \geq \sqrt{(ac+bd)^2}=|ac+bd|\geq ac+bd.$$
The equality holds if and only if $ad=bc$ and $ac+bd \geq 0.$
