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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.

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    $\begingroup$ That they are equal. Show that $|zw|=|z|\cdot|w|$ first and then do an induction to get what you want. $\endgroup$ – Mariano Suárez-Álvarez Feb 12 '16 at 17:25
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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 $|z^n|=|\rho^ne^{in\phi}|=\rho^n$.

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  • $\begingroup$ Ah, yes! This is, indeed, the obviousness I was looking for. Thanks! $\endgroup$ – thisisourconcerndude Feb 12 '16 at 17:28
  • $\begingroup$ You're welcome! My pleasure. - Mark $\endgroup$ – Mark Viola Feb 12 '16 at 17:28
  • $\begingroup$ Actually, now that I'm working through it, I'm a bit confused as to why we can ignore the $e^{in\phi}$ in the final line. Indeed, when $\phi$ is an integer multiple of $\pi / 2$ we have that $e^{in\phi} = 1$, and the above holds. But what about all other angles? $\endgroup$ – thisisourconcerndude Feb 12 '16 at 18:04
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    $\begingroup$ Note that $e^{in\phi}=\cos(n\phi)+i\sin(n\phi)$ and since $\sin^2(x)+\cos^2(x)=1$, we have $|e^{in\phi}|=\sqrt{\cos^2(n\phi)+\sin^2(n\phi)}=1$. $\endgroup$ – Mark Viola Feb 12 '16 at 18:20
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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 $$|z^2|=|x^2-y^2+2xyi|=\sqrt{x^4-2x^2y^2+y^4+4x^2y^2}=\sqrt{(x^2+y^2)^2}=x^2+y^2$$

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

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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

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

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