Modulus of z^-3?

Question

What is the result of $|z^{-3}|$ and how can one show it?

I know $z = e^{i\omega T}=cos(\omega T) + i\sin(\omega T)$, but I cant go further...

I would be glad if someone can explain further.

Answer 1

Recall Euler's formula:$$z=re^{i \theta}=r(\cos(\theta)+i\sin(\theta)),$$ and recall de Moivre's theorem: $$z^n=r^n(\cos(\theta)+i\sin(\theta))^n = r^n(\cos(n\theta)+i\sin(n\theta))$$

Here, choose $n=-3$:

$$z^{-3}=r^{-3}((\cos(\theta)+i\sin(\theta)))^{-3} = r^{-3}\cos(-3\theta)+ir^{-3}\sin(-3\theta)$$

We know that the modulo function is:

$$|z|=\sqrt{a^2+b^2}$$

So we simply plug in $a,b$:

$$ |z^{-3}|=\sqrt{(r^{-3}\cos(-3\theta))^2+(r^{-3}\sin(-3\theta))^2}$$ $$ |z^{-3}|=\sqrt{r^{-6}((\cos(-3\theta))^2+(\sin(-3\theta))^2})$$ $$ |z^{-3}|=\sqrt{r^{-6}\cdot1}$$ $$ |z^{-3}|=r^\frac{-6}{2}$$ $$ |z^{-3}|=r^{-3}$$

Answer 2

If $z$ is a complex number, we can write it as $z = re^{i\theta}$; then $|z| = |r|\cdot|e^{i\theta}|$. By the comment above, \begin{equation*} |z^n| = |z|^n = |r|^n |e^{i\theta}|^n. \end{equation*} Can you take it from there?