Proof: complex analysis

Question

I need to prove these Identities: $$|z-a||z|=|z\bar z-az|$$

And

Given: $S=|z|$ I have to show $S^m=|z^m|=|z|^m$

As an aside:

What does this sum converge to?

$Cos(\theta)+cos(2\theta)+...+cos(n\theta)$

Work done so far:

I tried to break the modulus into into its definition, but it ends up in a complicated mess: $$|z-a||z|=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}|z|=$$ $$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\sqrt{x^2+y^2}=\sqrt{[(x_1-x_2)^2+(y_1-y_2)^2](x^2+y^2)}$$

From here the circular factorization starts...

was wondering if there is an easier way to prove these identities, without the need to rely on $z=x+bi$ and $|z|=\sqrt{x^2+y^2}$

Edit: For the first identity $a\in \mathbb C$ is fixed and $|z|=1$, but $z$ is not fixed.

Thanks in advance!

Answer 1

The first identity is true if $a$ is real. For $a=i$ it is not true: take $z=i$ to get a contradiction. For the first two identities first verify that $|z_1 z_2|=|z_1||z_2|$ for all complex numbers $z_1,z_2$ and that $|\overline {z}|=|z|$. We get $|z-a||z|=|\overline {z-a}||z|=|\overline {z}-a||z|=|z(\overline {z}-a)|=|z\overline {z}-az|$. For the last identity note that $\cos(k\theta) =Re (e^{i\theta})$ and use the formula for a geometric sum.

Answer 2

The first identity can be simplified by $|z|$, giving

$$|z-a|=|\overline z-a|.$$

If $a$ is real, $a=\overline a$; otherwise the identity is wrong.