Proof of $|z^n| = |z|^n$, where $n \neq 0$

Question

With complex numbers, I can see intuitively that the modulus of $z^n$ must be equal to the modulus of $z$, all to the power of $n$, but I'm not sure how to prove it. Thanks

Answer 1

$$|a+ib||c+id|=\sqrt{a^2+b^2}\sqrt{c^2+d^2}=\sqrt{a^2c^2+a^2d^2+b^2c^2+b^2d^2}$$ and

$$|(a+ib)(c+id)|=|(ac-bd)+i(ad+bc)|=\sqrt{(ac-bd)^2+(ad+bc)^2}=\sqrt{a^2c^2+a^2d^2+b^2c^2+b^2d^2}$$

shows that $$|wz|=|w||z|.$$

Then you easily prove the claim by induction.

Answer 2

Let $z=r(\cos(\theta)+i\sin(\theta))$. By definition, we know that: $$|z|=|r(\cos(\theta)+i\sin(\theta))|=r$$ By De-Moivre's rule, we have: $$|z^n|=|r^n(\cos(n\theta)+i\sin(n\theta))|=r^n$$ Using the definition said before, we arrive at: $$|z^n|=r^n=|z|^n$$

Answer 3

Since $z\bar z\ge 0$ and $|z|=\sqrt{z\bar z}\quad$ then $$\quad|z^n|=\sqrt{z^n\overline {z^n}}=\sqrt{z^n\bar z^n}=\sqrt{(z\bar z)^n}=(\sqrt{z\bar z})^n=|z|^n$$

I agree though, you can object that we have to display a proof of $(\bar z)^n=\overline{z^n}$ ...

In which case we also have to go for induction as in Y.Daoust answer on

$\begin{align}\overline{uv}&=\overline{(a+ib)(x+iy)}=\overline{(ax-by)+i(ay+bx)}=(ax-by)-i(ay+bx)\\&=(ax-(-b)(-y))+i(a(-y)+(-b)x)=(a-ib)(x-iy)=\bar u\bar v\end{align}$