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.
Best Answer
Hint: $$ |z|=|(z-w)+w|\le|z-w|+|w|\implies |z|-|w|\le|z-w|. $$