Maclaurin series of $\ln(1-z)$

Question

I'm not sure how to start on this problem other than the definition of $\ln(z)$, which is:

$$\ln(1-z) = \ln|1-z| + i(\arg z)$$

or I don't know what to make of $\ln(1-z)$.

Answer 1

For $z \in \mathbb C$ and $z \ne 1$ we have

$\log(1-z)=\log|1-z|+i \arg(1-z)$

If $|z|<1$, $f(z):=\log(1-z)$ is holomorphic and

$$f'(z)=-\frac{1}{1-z}.$$

Now write $f'$ as a geometric series, integrate and observe that $f(0)=0$

Answer 2

You can simply start from the real variable Taylor series:

$$\log(1+x) = \sum_{k = 1}^{+\infty} (-1)^{k+1} \frac{x^k}{k}$$

Then you substitute $x\to -x$ getting the series for

$$\log(1-x) = \sum_{k = 1}^{+\infty} (-1)^{k+1} \frac{(-x)^k}{k} = \sum_{k = 1}^{+\infty} (-1)^{k+1} (-1)^k \frac{x^k}{k}$$

Which is, manipulating easily the $(-1)$ terms:

$$\log(1-x) = \sum_{k = 1}^{+\infty} - \frac{x^k}{k}$$

Ore equivalently

$$\log(1-x) = -\sum_{k = 1}^{+\infty} \frac{x^k}{k}$$

Hence by going into the complex plane

$$x \to z$$

you can "simply" get (you can prove this without difficulties)

$$\boxed{\log(1-z) = -\sum_{k = 1}^{+\infty} \frac{z^k}{k}}$$