1

I got a question asking to prove that $|z^n|$ and $|z|^n$ are the same thing. But what is the difference between keeping the exponent in or out of the modulus?

J W
  • 2,184
7up234
  • 7
  • 1
    You can prove the statement by induction: Clearly, $|z^1| = |z|^1$. Then by induction, $$|z^n| = |z^{n-1}\cdot z| = |z^{n-1}||z| = |z|^{n-1}|z| = |z|^n.$$ To prove this, and probably to understand what you are wondering about, you just need to understand why $|ab| = |a||b|$. –  Oct 17 '22 at 18:12
  • 3
    $|(re^{i\theta})^n|=|(r^ne^{in\theta})|=r^n=|re^{i\theta}|^n$ – Bob Dobbs Oct 17 '22 at 18:13
  • @2211: I guess you first need to now that $|a b | = |a| |b|$. Not sure what the OP can assume though. – Fabian Oct 17 '22 at 18:13
  • @2211 Thanks! Do you think you could explain it using Polar Coordinates too? – 7up234 Oct 17 '22 at 18:51
  • $|z|^2=z\bar z$ is also a path to explore. But then you are similarly confronted to ${\bar z}^n=\bar{z^n}$. – zwim Oct 17 '22 at 19:05

2 Answers2

1

Putting the exponent inside means you are first computing the power of the complex number in complex multiplication, THEN taking the modulus of it to extract the real number size.

Putting it outside means you are first taking the real number modulus of the number, then taking that size to the power.

They are the same because in general, the modulus is multiplicative, aka $$|ab|=|a||b|$$

Alan
  • 16,582
-3

It probably suffices to just look at an example.

take $z=1+i$. Then:

  • $|z|^2=(\sqrt{2})^2 = 2$.
  • $|z^2| = |2i| = 2$.

So basically you'll need to go from the definition and prove $|z_1|\cdot|z_2| = |z_1z_2|$.

YYYYZZZZ
  • 17
  • No, it is not enough to look at a particular example. $(3,4,5)$ is a Pythagorean triple, but this does not prove that the Pythagorean theorem is true for all right triangles. – cpiegore Oct 19 '22 at 15:44
  • Of course the example does not give you a proof. On the other hand, OP did not ask for a proof, but asked for clarification on the notation. – YYYYZZZZ Oct 26 '22 at 20:19