Does $\operatorname{Log}(z^n)=n\operatorname{Log}(z)+2ik\pi$?

Question

I study complex functions and I need to prove or disprove that for each $n$ that belongs to the natural numbers, there is a $k$ that belongs to the integers so for every $z$ that belongs to $\mathbb{C} \setminus (-\infty,0]$ I get $\operatorname{Log}(z^n)=n\operatorname{Log}(z)+2ik\pi$.
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Answer 1

$z$ can be expressed as $re^{i\theta}$ and arg $z$ as $\theta$ (I hope you know this form of a complex number. If you don't, a similar argument can be made for the trigonometric form instead). Note that $re^{i\theta}= re^{i\theta+ 2\pi ik}$ for all integers, k because $re^{i\theta+ 2\pi ik} = re^{i\theta} * e^{2\pi i k}$ and $e^{2\pi i k}$ is 1.

This is the part where you had an issue: $$i\,arg(z^n) = in\,arg\,z$$

Since $z = re^{i\theta+ 2\pi ik}=re^{i(\theta+ 2\pi k)}$ note $arg\,z = \theta + 2\pi k$,

$$i\,arg(z^n) = in(\theta + 2\pi k)$$ but $\theta$ is also arg z, Therefore;

$$i\,arg(z^n) = in(arg z + 2\pi k)$$ $$i\,arg(z^n) = in\, arg z + 2\pi i nk$$ nk is a product of two arbitrary integers and for some k, there will be an integer that can be expressed as such. Let that integer be m. Therefore, $$i\,arg(z^n) = in\, arg z + 2\pi i m$$ This is equivalent to what you wished to prove. (m is a dummy variable) The key insight is that $2\pi$ is a full rotation and lands you right back where you started. This means that say $\pi$ was the $\theta$, so would $3\pi$ because to get to $3\pi$, you would make a full rotation to land back right where $\pi$ was. I also want to point out that if you didn't show your attempt, I wouldn't know where exactly you had an issue.