In the book of Functions of One Complex Variable by Conway, at page 3, it is stated that
[...] On encountering an inequality one should always ask for necessary and sufficient conditions that equality obtains. From looking at a triangle and considering the geometrical significance of $$|z + w| \leq (|z| + |w|) \quad (z,w \in \mathbb{C}),$$ we are led to consider the condition $z = tw$ for some $t\in \mathbb{R}^{nn}$. [...]
Even though the author does not give an explicit proof for the fact that when equality holds in the triangle inequality for two complex numbers, then one of the complex numbers should be a multiple of the other; I have tried to prove this; even though I can show that what the author claims is true, I couldn't show that that is the only case where the equality hold.Here is what I have done;
Solution:
I have directly written $z = a + bi$, and $w = p+ qi$, and explicitly computed both parts of the equality, and in the end got
$$2abpq = (aq)^2 + (bp)^2,$$ i.e whenever the equality holds, this above must also hold.
It is clear that if $z = tw$ or $w = cz$ for some $t,c \in \mathbb{R}^{nn}$, then the above equality holds.
However, I would like to this is the only case where the inequality holds, so how can we do that ?
$|z+w|=|z|+|w|$ implies $|z|^{2}+|w|^{2} +2 Re (z\overset {-} w)=|z|^{2}+|w|^{2}+2|z||w|$, so $Re (z\overset {-} w)=|z||w|$. Now let $t \geq 0$ and consider $|z-tw|^{2}=|z|^{2}+t^{2}|w|^{2}-2t Re (z\overset {-} w)=|z|^{2}+t^{2}|w|^{2}-2t |z||w|$. Choose $t$ such that the right side is $0$ and you will get $z=tw$. To find $t$ you just have to solve the quadratic equation.