# Thread: Complex Analysis argument question

1. ## Complex Analysis argument question

Could someone help me on this one.
Prove that arg(z^n)=nArg(z)+2kpi , where k is an integer and n is positive.

2. Originally Posted by frankdent1
Could someone help me on this one.
Prove that arg(z^n)=nArg(z)+2kpi , where k is an integer and n is positive.
If $\displaystyle z\not = 0$ then write $\displaystyle z = |z|e^{\theta i}$ where $\displaystyle \theta = \arg (z)$. This means $\displaystyle z^n = |z^n|e^{n\theta i}$ and that means $\displaystyle n\theta$ is the $\displaystyle \arg(z^n)$ (up to a factor of $\displaystyle 2\pi$).

3. Would you care to elaborate on the last part?

4. Originally Posted by frankdent1
Would you care to elaborate on the last part?
I am saying if $\displaystyle \theta$ is the argument of a non-zero complex number then we can write $\displaystyle z = |z|e^{i\theta}$ (and conversely). Now if we raise both sides to the $\displaystyle n$ we get $\displaystyle z^n = |z|^n \left( e^{i\theta} \right)^n$ but $\displaystyle |z|^n = |z^n|$ and $\displaystyle \left( e^{i\theta} \right)^n = e^{ni\theta}$ by de Moivre theorem. Thus, we can write $\displaystyle z^n$ as $\displaystyle |z^n|e^{ni\theta}$ and so $\displaystyle ni\theta$ is the argument of $\displaystyle z^n$, i.e. $\displaystyle n\theta = \arg (z^n)$ thus $\displaystyle n\arg (z) = \arg (z^n)$. I said "up to a factor of $\displaystyle 2\pi$" because as you know there can be many different argument differening by $\displaystyle 2\pi$.

5. Originally Posted by ThePerfectHacker
If $\displaystyle z\not = 0$ then write $\displaystyle z = |z|e^{\theta i}$ where $\displaystyle \theta = \arg (z)$. This means $\displaystyle z^n = |z^n|e^{n\theta i}$ and that means $\displaystyle n\theta$ is the $\displaystyle \arg(z^n)$ (up to a factor of $\displaystyle 2\pi$).
The red word factor is probably not the word I'd use since it implies multiplication of the multiples of 2 pi, whereas you are adding the multiples of 2 pi.

Originally Posted by frankdent1
Would you care to elaborate on the last part?
If you think of z^n as representing some complex number (1 + i, say) and then think of an Argand diagram showing 1 + i, you should be able to that adding (or subtracting) multiples of 2 pi 'keeps you' at 1 + i. Capisce?

### proof arg(z^n)=narg(z)

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