I'm reviewing some elementary properties in complex analysis, and I came across with the following statement: $\log(z^m)=m\log(z)$, $\forall m\in\mathbb{Z}\setminus\{0\}$.
First, let's recall that by definition, we have $\log(z) = \left\{\ln|z| + i\theta \mid \theta\in \arg(z)\right\}$.
This is my attempt to prove one inclusion of this steatment:
Let $w \in\log(z)$. So we have $w = \ln|z^m|+i\theta$, for some $\theta\in\arg(z^m)$. Now we can use the argument property: $\arg(z^m)=\underbrace{\arg(z)+\dots+\arg(z)}_{m\ \text{times}}$.
So that implies the existence of $\theta_j\in\arg(z)$, $1\leq j \leq m$ such that $\theta = \displaystyle\sum_{j=1}^{m}\theta_j$
Therefore, using the property of real logarithm we obtain: \begin{equation*}w =\underbrace{\ln|z| + \cdots + \ln|z|}_{m\ \text{times}} + i(\theta_1+\cdots+\theta_m) = (\ln|z| + i\theta_1) + \cdots +(\ln|z| + i\theta_m)\end{equation*} Then, we have $w_k = \ln|z| + i\theta_k\in\log(z)$, $1\leq k\leq m$.
Hence, it was proved that $w=w_1 + \cdots + w_m \in m\log(z)$.
Is this correct? If so, my intuition tells me that the a similar argument proves the other inclusion. I don't know, maybe I miss something.