Proof of the power series of the complex logarithm

Builds on Principal branch Power series in the complex plane Disc of convergence Comparison test The sequence r^n tends to zero for |r| < 1 Sum of a finite geometric series Power series are holomorphic in the disc of convergence The principal logarithm is holomorphic off the branch cut Fundamental theorem of contour integration Principal value of the argument

Proof

We show that the series and $-\operatorname{Log}(1 - z)$ have the same derivative and agree at $0$ .

Let $D = \set{z : \abs{z} < 1}$ and $z \in D$ . Then $\abs{z^n / n} \le \abs{z}^n$ , and $\sum \abs{z}^n$ converges. By the comparison test, $\sum_{n \ge 1} z^n / n$ converges absolutely.
Consider the power series $\sum_{n \ge 1} z^n / n$ . For $0 \le s < 1$ , $s^n / n \le 1$ for all $n \ge 1$ . So its radius of convergence is at least $1$ , and $D$ lies in its disc of convergence.
Put $g(z) = \sum_{n=1}^{\infty} z^n / n$ for $z \in D$ . Since power series are holomorphic in the disc of convergence, $g$ is holomorphic on $D$ and $g'(z) = \sum_{n=1}^{\infty} z^{n-1}$ .
By the finite geometric sum, $\sum_{n=1}^{N} z^{n-1} = \frac{1 - z^N}{1 - z}$ . Since $\abs{z}^N \to 0$ by the limit of $r^N$ , we get $g'(z) = \frac{1}{1 - z}$ .
For $z \in D$ , $\operatorname{Re}(1 - z) \ge 1 - \abs{z} > 0$ . So $1 - z$ lies in the slit plane $S = \C \setminus (-\infty, 0]$ .
Put $h(z) = \operatorname{Log}(1 - z)$ for $z \in D$ , with $\operatorname{Log}$ the principal branch. Fix $z_0 \in D$ , $\zeta_0 = 1 - z_0$ , and $k \ne 0$ small. Then $\frac{h(z_0 + k) - h(z_0)}{k} = - \frac{\operatorname{Log}(\zeta_0 - k) - \operatorname{Log} \zeta_0}{-k}$ .
As $k \to 0$ , also $-k \to 0$ . Since $\operatorname{Log}$ is holomorphic on $S$ with derivative $1/\zeta$ , the quotient in step 6 tends to $-1/\zeta_0$ . So $h$ is holomorphic on $D$ with $h'(z) = -\frac{1}{1 - z}$ .
Put $F = g + h$ on $D$ . By steps 3, 4 and 7, $F$ is holomorphic on $D$ and $F'(z) = 0$ for all $z \in D$ .
Fix $z \in D$ and let $\gamma(t) = t z$ for $t \in [0, 1]$ , a path in $D$ . By the fundamental theorem of contour integration for $F$ and the continuous function $0$ , $F(z) - F(0) = \int_\gamma 0 \, dw = 0$ .
We have $g(0) = 0$ . Also $\operatorname{Arg} 1 = 0$ by the principal argument, since $1 = \cos 0 + i \sin 0$ . So $h(0) = \operatorname{Log} 1 = \ln 1 + 0 = 0$ , and $F(0) = 0$ .
By steps 9 and 10, $g(z) + \operatorname{Log}(1 - z) = 0$ . So $-\operatorname{Log}(1 - z) = \sum_{n=1}^{\infty} z^n / n$ for each $z \in D$ .

Idea

Differentiating the series term by term gives the geometric series $1/(1 - z)$ , which is also the derivative of $-\operatorname{Log}(1 - z)$ . Two functions on a disc with the same derivative differ by a constant, and both vanish at $0$ .

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Power series of the complex logarithm

Tag	`JM2RBQJ0MG`
URI	`m2://JM2RBQJ0MG`
Kind	proof
Writing	W0 stub
MSC	30B10
Proves	Power series of the complex logarithm
Uses	Principal branch, Power series in the complex plane, Disc of convergence, Comparison test, The sequence r^n tends to zero for \|r\| < 1, Sum of a finite geometric series, Power series are holomorphic in the disc of convergence, The principal logarithm is holomorphic off the branch cut, Fundamental theorem of contour integration, Principal value of the argument
Changed	4 November 2026

Proof of the power series of the complex logarithm

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