A tricky complex numbers if and only if proof

Question

For complex numbers $z$ and $w$ prove that $$|z|^2w -|w|^2z = z-w\quad \iff\quad z=w\quad\text{or}\quad z\bar{w}=1.$$

How would you go about solving this problem?

Answer 1

Hint: I think it should be sufficient that you write $z=a+ib$ and $w=c+id$ and work out the algebra on your equation.

Answer 2

Rearranging gives

$(|z|^2 + 1) w = (|w|^2 + 1) z$ .

If one of $w, z$ is zero, so is the other, and $w = z$.

If on the other hand neither is zero, we can rearrange so that each variable appears only on one side:

$\frac{w}{|w|^2 + 1} = \frac{z}{|z|^2 + 1}$.

If we take the modulus of both sides we get $\Phi(|w|) = \Phi(|z|)$, where $\Phi(t) := \frac{t}{t^2 + 1}$. In particular, $|w| = |z|$ or $|w||z| = 1$.

In the former case, substituting $|w| = |z|$ in either of the display equations gives $w = z$ after canceling.

In the latter case, substituting $|z| = \frac{1}{|w|}$ in the previous display equation, rearranging and canceling gives $z \bar{w} = 1$ as desired.

Answer 3

If one of $z$ or $w$ is zero the statement is true. If not, think about $$\frac{|z|^2+1}{|w|^2+1}=z\bar{w}\in\mathbb R.$$