For my course in complex analysis we have to prove that the trivial relation $e^{\ln{z}}=z$. We are given the series for $\ln z$:

$$f(w)=\sum_{n=0}^\infty (-1)^{n+1}\frac{w^n}{n}$$

$$\ln z = f(z-1)$$

I know that the series for $e^x$ is

$$e^{x}=\sum_{n=0}^\infty \frac{x^n}{n!}$$

I tried to solve

$$e^{\ln{z}}=\sum_{n=0}^\infty \frac{\left(\sum_{m=0}^\infty (-1)^{m+1}\frac{(z-1)^m}{m}\right)^n}{n!}$$

But just inserting the previous series into this does not yield a very convenient result. I think that if we expand the first power series around $z=0$ we would have already a problem (this makes sense since also $\ln 0$ does not exist. How and with what technique is this problem solved?

Edit (clarification):

We need to prove the relation given using the power series for $\ln{z}$ as definition.

  • 3
    $\begingroup$ The series for the logarithm must start at $1$, not at $0$. Power series manipulation seems very tedious. Would be differentiating $z\cdot e^{-\ln z}$ legitimate? $\endgroup$ – Daniel Fischer May 9 '14 at 10:21
  • $\begingroup$ we need to use the power series as definition for $\ln z$. $\endgroup$ – user1043065 May 9 '14 at 10:24
  • 2
    $\begingroup$ Yes, but that is no problem. You get $\exp'(z) = \exp(z)$ and $\ln'(z) = \frac{1}{z}$ from the power series, without assuming that the exponential and logarithm are inverses. $\endgroup$ – Daniel Fischer May 9 '14 at 10:29
  • $\begingroup$ well, it's very ugly because of z-1, so if you substitute z=b+1? your series is going to look better. $\endgroup$ – camel May 9 '14 at 10:37

First observe that $\sum_{n\geq1}(-1)^{n-1}\frac{(z-1)^n}{n}$ is absolutely convergent for $z\in B_1(1)\subseteq \mathbb{C}$, whence $$\log:B_1(1)\to\mathbb{C}, z\mapsto \sum_{n\geq1}(-1)^{n-1}\frac{(z-1)^n}{n}$$ is a well-defined analytic function. In particular we can differentiate term by term to our heart's content:

$$\partial_z\log (z)=\sum_{n\geq1}(-1)^{n-1}(z-1)^{n-1} =\sum_{n\geq0}(1-z)^n=\dfrac{1}{1-(1-z)}=\dfrac{1}{z}.$$

Put $\varphi:=\exp\circ \log:B_1(1)\to B_1(1)$. We claim that $\varphi=\operatorname{id}$. $\varphi$, being the composition of two analytic maps, is analytic, and by substituting term by term $\log(1)=0\implies \varphi(1)=1$. Via the chain rule

$$\partial_z \varphi(z)=\dfrac{\varphi(z)}{z}, \partial_z^2 \varphi(z)=0.$$

$\partial_z \varphi$ too is analytic (it is defined on $B_1(1)$!), and since its derivative vanishes it must be constant. Thus $\forall z\in B_1(1):\partial_z \varphi(z)=\partial_z\varphi(1)=\dfrac{\varphi(1)}{1}=1$. We are done by an appropriate form of the fundamental theorem of calculus.

Note: For reference purposes, this is an exercise from Lang's Complex Analysis, 3e (p. 67).


Let $\mathbb{N} = \{1,2,\cdots\}$. For each $\alpha = (\alpha_1, \cdots, \alpha_n)$, we write $|\alpha| = \alpha_1 + \cdots + \alpha_n$. Then

\begin{align*} e^{\log(1+z)} &= \sum_{n=0}^{\infty}\frac{1}{n!}\left(\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} z^k \right)^n \\ &= \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{\alpha\in\mathbb{N}^n} \frac{(-1)^{\alpha_1+\cdots+\alpha_n-n}}{\alpha_1 \cdots \alpha_n} z^{\alpha_1 + \cdots + \alpha_n} \\ &= \sum_{l=0}^{\infty} \Bigg( \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{\substack{\alpha \in \mathbb{N}^n \\ |\alpha| = l}} \frac{(-1)^{\alpha_1+\cdots+\alpha_n-n}}{\alpha_1 \cdots \alpha_n} \Bigg) z^l \end{align*}

Now we focus on the coefficient. For each given $l$,

$$ \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{\substack{\alpha \in \mathbb{N}^n \\ |\alpha| = l}} \frac{(-1)^{\alpha_1+\cdots+\alpha_n-n}}{\alpha_1 \cdots \alpha_n} = \frac{1}{l!} \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{\substack{\alpha \in \mathbb{N}^n \\ |\alpha| = l}} \binom{l}{\alpha} \prod_{i=1}^{n} (-1)^{\alpha_i-1}(\alpha_i-1)!. $$

Now notice that the $ \frac{1}{n!} \binom{l}{\alpha} \prod_{i=1}^{n} (\alpha_i-1)! $ is exactly the number of ways of splitting $\{1,\cdots,l\}$ into $n$ cycles of sizes $\alpha_1, \cdots, \alpha_n$. And identifying each of them as permutation on $\{1,\cdots,l\}$, the sign $(-1)^{\alpha_1-1}\cdots(-1)^{\alpha_n-1}$ is exactly the signature of that permutation. So if $S_l$ denotes the set of all permutations on $\{1,\cdots,l\}$, then

$$ \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{\substack{\alpha \in \mathbb{N}^n \\ |\alpha| = l}} \binom{l}{\alpha} \prod_{i=1}^{n} (-1)^{\alpha_i-1}(\alpha_i-1)! = \sum_{\sigma \in S_l} \operatorname{sign}(\sigma) = \begin{cases} 1, & l = 0, 1 \\ 0, & l \geq 2 \end{cases}. $$

Plugging this back proves

$$ e^{\log(1+z)} = 1+z $$

and therefore the claim follows.

Note: This argument is largely based on this posting.

  • $\begingroup$ Why is this equation $\sum_{n=0}^{\infty}\frac{1}{n!}\left(\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} z^k \right)^n = \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{\alpha\in\mathbb{N}^n} \frac{(-1)^{\alpha_1+\cdots+\alpha_n-n}}{\alpha_1 \cdots \alpha_n} z^{\alpha_1 + \cdots + \alpha_n}$ true ? $\endgroup$ – Matheus Manzatto Apr 22 '19 at 18:07

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