Animated Solution for Mathematics - Complex Numbers: For complex numbers $z$ and $w$, prove that $|z{|}^{2}w-|w{|}^{2}z=z-w$ if and only if $z=w$ or $z\bar{w}=1$.

Imagine standing on the complex plane, watching two numbers, $z$ and $w$, interact. They are not just points; they are vectors, rotations, and scalings all at once.

Today, we are going to prove a beautiful identity: $|z{|}^{2}w-|w{|}^{2}z=z-w$ holds if and only if $z=w$ or $z\bar{w}=1$. This is not just an algebraic exercise; it is a journey into the heart of complex symmetry.

Our first step is to recognize the power of the magnitude. We know that for any complex number $z$, the square of its magnitude is simply the number multiplied by its conjugate: $|z{|}^2=z\bar{z}$.

By substituting this into our original equation, we transform a geometric statement about magnitudes into a purely algebraic one:

Suddenly, the equation is no longer about distances; it is about the numbers themselves.

Now, let us bring order to the chaos. We move all terms to the left-hand side:

We can group the terms to reveal a hidden structure:

By factoring out $z$ from the first group and $w$ from the second, we get:

To clear the fog, let us introduce a new variable, $\alpha =z\bar{w}$. If we take the conjugate, we find $\bar{\alpha }=\bar{z}w$.

Substituting these into our factored equation gives us:

If we multiply the entire equation by $\bar{w}$, we get:

Substituting $\alpha =z\bar{w}$ and $|w{|}^2=w\bar{w}$, we arrive at:

Expanding this, we get:

Rearranging the terms, we get:

The left-hand side is a sum of magnitudes squared—it is strictly real. The term $(1+|w{|}^2)$ is also real and non-zero, which forces $\alpha$ to be a real number.

If $\alpha$ is real, then $\alpha =\bar{\alpha }$. We can now substitute this back into our earlier equation:

Factoring out $(\alpha -1)$, we obtain:

For this product to be zero, either $\alpha -1=0$ or $z-w=0$. This means $\alpha =1$ or $z=w$.

Since $\alpha =z\bar{w}$, the condition $\alpha =1$ is exactly $z\bar{w}=1$. We have arrived at our destination: the identity holds if and only if $z=w$ or $z\bar{w}=1$.