Prove $|(a+bi)^n| = |(a+bi)|^n$

Question

I was resolving an exercise where I had to use $\lvert(a+bi)^4\rvert$ and put it as $\lvert a+bi\rvert^4$. Of course that $$r = \lvert a+bi\lvert = \sqrt{\left(a^2+b^2\right)}$$

I realized that I can say in general that $$\lvert\left(a+bi\right)^n\rvert = \left\lvert(a+bi)\right\rvert^n$$

However, I feel that this is a difficult approach. Can someone help me?

Answer 1

Write $a+bi=r(\cos\theta+i\sin\theta)$, then $$(a+bi)^n=r^n(\cos(n\theta)+i\sin(n\theta)). $$ Now $$|(a+bi)^n|=|r^n(\cos(n\theta)+i\sin(n\theta)|=r^n=|a+bi|^n. $$

Answer 2

$z= a+bi\\ |z| = \sqrt {z\bar z}\\ |z^n| = \sqrt {(z^n)(\overline {z^n})} = \sqrt {(z\bar z)^n} = |z|^n$

Answer 3

We can show by induction the more general result that if $z_j\in\mathbb{C}$, $|\prod_{j=1}^n z_j|=\prod_{j=1}^n|z_j|$. The main insight is that $\sqrt{a}{\sqrt b}=\sqrt{ab}$.

$j=1$ is clear. For $j=2$, note $(a+bi)(c+di)=(ac-bd)+(ad+bc)i$, so $|(a+bi)(c+di)|=\sqrt{(a^2+b^2)(c^2+d^2)}$. On the other hand, $|a+bi||c+di|=\sqrt{a^2+b^2}\sqrt{c^2+d^2}=\sqrt{(a^2+b^2)(c^2+d^2)}$.

Suppose the result holds for $j=m$. Then since $\mathbb{C}$ is closed under multiplication, $\prod_{j=1}^m z_j\in\mathbb{C}$, and the inductive step and base case $j=2$ give us $|\prod_{j=1}^{m+1}z_j|=|\prod_{j=1}^m z_j||z_{m+1}|=(\prod_{j=1}^m |z_j|)|z_{m+1}|=\prod_{j=1}^{m+1}|z_j|.$