I'm reading a short paper here
Which attempts to prove/discuss all the properties of the complex logarithm, exponential, and power functions.
In the paper, the following statement is made on pg 9, for $k$ an integer
$z^{1/k}=|z|^{1/k}e^{iarg(z)/k}$
To prove this, I believe you would write:
$z^{1/k}=(|z|e^{i\cdot arg(z)})^{1/k}=\big(e^{Ln|z|}e^{i\cdot arg(z)}\big)^{1/k}=\big(e^{Ln|z|+i\cdot arg(z)}\big)^{1/k}=^*e^{Ln|z|/k+i\cdot arg(z)/k}=e^{Ln|z|/k}e^{i\cdot arg(z)/k}=|z|^{1/k}e^{iarg(z)/k}$
Why is the step labeled $=*$ true? The only properties we have for $e^z$ so far are that $e^{z+w}=e^{z}e^{w}$, $\frac{e^z}{e^w}=e^{z-w}$, $(e^{z})^n=e^{zn}$ for $n \in \mathbb{Z}$, and that $z^n=e^{n\cdot Ln(z)}$ for $n \in Z$. Or is there some other way to show the statement I referenced from pg 9 to be true?
I would assume that $z^{1/k}=|z|^{1/k}e^{i\arg z/k}$ is not as much a claim that is subject to proof, as it is a (tentative) definition of what $z^{1/k}$ means.
In order to see that this definition is reasonable, you might prove that $\bigl(|z|^{1/k}e^{i\arg z/k}\bigr)^k = z$. Along the way you would use the rule $(e^x)^k=e^{kx}$.
You should then also show that $z^{1/1}$ according to this definition agrees with what we already knows $z^1$ to be, namely $z$ itself.