Power series of the complex logarithm

Statement

Let $\operatorname{Log}$ be the principal branch of the logarithm. For each $z \in \mathbb{C}$ with $|z| < 1$, the series $\sum_{n \ge 1} z^n / n$ converges absolutely, and $$-\operatorname{Log}(1 - z) = \sum_{n=1}^{\infty} \frac{z^n}{n} = z + \frac{z^2}{2} + \frac{z^3}{3} + \cdots.$$

Remarks

Replacing $z$ by $-z$ gives $\operatorname{Log}(1 + z) = \sum_{n \ge 1} (-1)^{n+1} z^n / n$ for $|z| < 1$. The radius of convergence is $1$. For $|z| < 1$, the number $1 - z$ has positive real part, so it is not on the branch cut. Proof idea: both sides vanish at $0$, and both have the derivative $1/(1 - z) = \sum_{n \ge 0} z^n$ on the unit disc. At $z = -1$ the series still converges, and gives $\ln 2 = 1 - \frac{1}{2} + \frac{1}{3} - \cdots$.

Explanations

L1 human

Integrate the geometric series $1 + z + z^2 + \cdots = 1/(1 - z)$ term by term from $0$ to $z$. On the left you get $z + z^2/2 + z^3/3 + \cdots$, and on the right you get $-\operatorname{Log}(1 - z)$.