Proving an Inequality Involving the Modulus of the Difference of Moduli Prove the following inequality and give necessary and sufficient conditions for equality.
$\left| |z|-|w| \right| \leq |z-w|$ for complex numbers $z$ and $w$. 
I have the following:
By definition of the modulus of a complex number, then we get
$\sqrt{\sqrt{a^2+b^2} - \sqrt{c^2+d^2}} \leq \sqrt{(a+bi)-(c+di)} = \sqrt{z-w}$. After squaring both sides to clear a radical, I end up with
$|z|-|w| \leq z-w$. After squaring this inequality again, expanding  and collecting real and imaginary components, I have
$2\left[ b^2+d^2-\sqrt{a^2+b^2(c^2+d^2)}+ac-bd \right] \leq 2i(ab-ad-bc+cd)$. I know that I can factor the expression on the right as $2i[(a-c)(b-d)]$, but I still do not see how this is helping me achieve a solution to the problem. As for the sufficient and necessary conditions, I am still clueless still I have yet to solve the inequality. 
Any assistance received for this proof will be greatly appreciated, thanks in advance. By the way, this problem is coming straight from John B. Conway's Functions of One Complex Variable I.
 A: Hint:
$$
|z|=|(z-w)+w|\le|z-w|+|w|\implies |z|-|w|\le|z-w|.
$$
A: If you don't already know the triangle inequality, then you can argue like this. 
$$(|z| - |w|)^2 = |z|^2 - 2|z||w| + |w|^2 = |z|^2 - 2|z\bar{w}| + |w|^2 \le |z|^2 - 2\text{Re}(z\bar{w}) + |w|^2 = |z - w|^2$$
Therefore $||z| - |w|| \le |z - w|$. 
We also have that $||z| - |w|| = |z - w|$ if and only if $|z\bar{w}| = \text{Re}(z\bar{w})$, or $\text{Im}(z\bar{w}) = 0$. This is equivalent to the following condition: $z = 0$ or $w = 0$ or $z = \lambda w$ for some $\lambda \in \Bbb R\setminus\{0\}$. The reverse implication is established by verifying directly that $\text{Im}(z\bar{w}) = 0$ in each case. To see the forward implication, suppose $z\neq 0$ and $w\neq 0$. Then we can write $z = re^{i\theta}$ and $w = Re^{i\phi}$, where $r > 0$, $w > 0$, and $\theta, \phi \in (-\pi, \pi]$. If $\text{Im}(z\bar{w}) = 0$, then $\sin(\theta - \phi) = 0$, which implies $\theta - \phi = n\pi$ for some integer $n$. Since $\theta - \phi \in (-\pi, \pi]$, we have $n\in (-1,1]$. Thus $n = 0$ or $n = 1$. If $n = 0$, then $\theta = \phi$ and $z = \lambda w$ with $\lambda = r/R$. If $n = 1$, then $\theta = \phi + 1$ and $z = \lambda w$ with $\lambda = -r/R$.
