What can be said about the relationship between the complex numbers $\lvert z\rvert^n$ and $\lvert z^n\rvert$?

Question

I've been playing around with this for a while without much progress. More precisely, I suppose, I'd like to know if one always less than or equal to the other? The fact that one never sees this in the usual properties of complex numbers leads me to believe it's either trivial or not generalizable. I can't seem to figure out which is the case, though.

We have that – since $\lvert z \rvert^2 = x^2 + y^2$ – if $n$ is even,

$$\lvert z \rvert^n = (x^2 + y^2)^{n/2}.$$

And, for $\lvert z^n \rvert$, we have

$$\lvert (x+iy)^n \rvert = \lvert (x^2+2ixy-y^2)^{n/2} \rvert$$

which doesn't seem very helpful.

Anyway, I feel like I'm overlooking something really obvious. Any help here would be appreciated.

Answer 1

It becomes apparent when using polar coordinates.

Let $z=\rho e^{i\phi}$.

Then, we have $|z|=\rho$ and thus $|z|^n=\rho^n$.

We also have $z^n=\rho^ne^{in\phi}$ and thus $|z^n|=|\rho^ne^{in\phi}|=\rho^n$.

Answer 2

You can try to prove $|z|^n=|z^n|$ by induction on $n$, following what you have tried to do.

For $n=1$, it is trivial. I sketch the case $n=2$. Let $z=x+iy$.

$$|z|^2=x^2+y^2$$ and $$|z^2|=|x^2-y^2+2xyi|=\sqrt{x^4-2x^2y^2+y^4+4x^2y^2}=\sqrt{(x^2+y^2)^2}=x^2+y^2$$

Answer 3

$$ (|z|^n)^2 = (|z|^2)^n = (z \bar z)^n= z^n (\bar z)^n = z^n \overline{(z^n)}=|z^n|^2 $$

Answer 4

If you're aware of the fact (which is easy to prove) that $|zw|=|z||w|$ for any complex $z,w$, then it's just a simple induction argument.

It's true for $n=2$ by the fact above. Now suppose it is true for $n$. Then

$$|z^{n+1}|=|z\cdot z^n|=|z||z^n|=|z||z|^n=|z|^{n+1}$$