When will equality holds in reverse triangle inequality? Prove the reverse triangle inequality :$|z\pm w|\ge||z|-|w||$ for all $z, w \in \mathbb C$, with equality holds if and only if either $z$ or $w$ is a real multiple of another.
I have proved the inequality but not able to solve for equality.
It will be great if someone can help me in this.
What I have tried is this:
let $z=cw$ for some c $\in \mathbb R$
Consider $||z|-|w||=||cw|-|w||=||c||w|-|w||=(||c|-1|)|w|$
On the other hand: $|z-w|=|cw-w|=|(c-1)w|=(|c-1|)|w|$
from here, clearly both these are not equal if $c\lt 0$.
FORWARD PART:
Given, $|z-w|=||z|-|w||$ on squaring it and solving it, I get $Re(zw^-)=|z||w|$
And for $|z+w|=||z|-|w||$ , I get $Re(zw^-)= -|z||w|$
$w^- \;denotes\; w \;conjugate$
How to proceed further ?
 A: 
equality holds if and only if either $z$ or $w$ is a real multiple of another

You just proved that $c$ needs to be $\ge 0$ when you choose the $-$ sign in $|z \pm w|$. Similarly $c \le 0$ will work for the $+$ alternative.
So for an arbitrary $c \in \mathbb R$ at least one of the inequalities will become an equality. Granted, the wording of the problem could have been better.

[EDIT] To answer the question about proving the forward implication. For $w = 0$ the equality holds trivially, and $w = 0 z$. Otherwise let $c = \frac{z}{w} \in \mathbb C$ so that $z = c w$.

It remains to prove that if $|c w \pm w| = ||c w| - |w||$ then $c \in \mathbb R$. Squaring both sides and canceling out the common terms gives in the end:
$$
\pm (c + \bar c) = 2 |c|
$$
But $c + \bar c = 2 Re(c)$, so $|Re(c)|=|c|$, which means $c \in \mathbb R$, thus $z = c w$ is a real multiple of $w$.
A: For equality to hold
$$
Re(z\bar{w})=|z||w|\implies Re\big(|z|e^{\alpha}.|w|e^{-\beta}\big)=Re\big(|z||w|e^{i(\alpha-\beta)}\big)=|z||w|\\
|z||w|\cos(\alpha-\beta)=|z||w|\implies\cos(\alpha-\beta)=1\\
\implies\alpha-\beta=0\implies\arg z=\arg w\implies z=nw
$$
