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I'm trying to show that

$$\log(z-z') = \log(-(z'-A)) - \sum_{n=1}^\infty \frac 1 n \left(\frac{z-A}{z'-A}\right)^n $$ where $A$ is a constant, and $\left|\dfrac{z-A}{z'-A}\right| < 1$.

If it matters, this question comes from a vector calculus problem where $A$ is the centroid of a region containing $z$, and $z'$ is from another region. Anyways, my attempt is:

We can split up the $\log$ term by adding and subtracting $A$: $$\log(z-z') = \log(z-A - (z' - A)) = \log(- (z' - A)) + \log(z-A)$$ so all that needs to be shown is that $$\log(z-A) \overset{\text{?}} = - \sum_{n=1}^\infty \frac 1 n \left(\frac{z-A}{z'-A}\right)^n \quad, $$ which is where I'm stuck. I can't find any expressions for power series of $\log$ which have a $1/n$ term like this.

Michael Hardy
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1 Answers1

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Recall that the Taylor series (See Here) for the complex function $f(z)=\log(1-z)$ is

$$\log(1-z)=-\sum_{n=1}^\infty \frac{z^n}{n}\tag 1$$

for $|z|<1$ and the branch cut does not intersect any point inside the unit disk.

Using $(1)$, we can write

$$\begin{align} \log(z-z')&=\log((z-A)-(z'-A))\\\\ &=\log(A-z')+\log\left(1-\frac{z-A}{z'-A}\right) \tag 2\\\\ &=\log(A-z')-\sum_{n=1}^\infty \frac{\left(\frac{z-A}{z'-A}\right)^n}{n} \end{align}$$

for $\left|\frac{z-A}{z'-A}\right|<1$ as was to be shown!

Note that $(2)$ is interpreted as a set equivalence (See Here). For any branch of $\log(z-z')$, there is some branch of $\log(A-z')$ and some branch of $\log\left(1-\frac{z-A}{z'-A}\right)$ for which $(2)$ holds. Conversely, it means for any branch of $\log(A-z')$ and any branch of $\log\left(1-\frac{z-A}{z'-A}\right)$, there is some branch of $\log(z-z')$ for which $(1)$ is true.

Mark Viola
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