Show that $\exp \biggl(\sum_{n=1}^\infty (-1)^{n-1} \frac{z^n}{n} \biggr) = 1+z \ $ for $\ \lvert z \rvert < 1$

Question

I am working on the following exercise:

Show that $$\exp \biggl(\sum_{n=1}^\infty (-1)^{n-1} \frac{z^n}{n} \biggr) = 1+z$$ for all $z \in \mathbb{C}$ with $\lvert z \rvert < 1$.

Hint: $\frac{d}{dz} \frac{f(z)}{1+z} = ?$

I do not get how this hint is supposed to help. Following the hint I would have to use the quotient rule

$$\frac{u^\prime v - uv^\prime}{v^2}$$

, to evaluate $\frac{d}{dz} \frac{f(z)}{1+z}$, so I calculated the derivative

$$\frac{d}{dz} \exp \biggl(\sum_{n=1}^\infty (-1)^{n-1} \frac{z^n}{n} \biggr) = \biggl(\sum_{n=1}^\infty(-1)^{n-1}z^{n-1}\biggr) \exp \biggl(\sum_{n=1}^\infty (-1)^{n-1} \frac{z^n}{n} \biggr).$$

But I still can not see what this should be good for, when I use this on the quotient rule. Could you give me a hint?

Answer 1

We have, for $|z|<1$, $$ \frac{d}{{dz}}\frac{{f(z)}}{{1 + z}} = \frac{{f(z)\left( {\sum\limits_{n = 1}^\infty {( - 1)^{n - 1} z^{n - 1} } } \right)(1 + z) - f(z)}}{{(1 + z)^2 }} = f(z)\frac{{(1 + z)\sum\limits_{n = 1}^\infty {( - 1)^{n - 1} z^{n - 1} } - 1}}{{(1 + z)^2 }} \\ = f(z)\frac{{\sum\limits_{n = 1}^\infty {( - 1)^{n - 1} z^{n - 1} } - \sum\limits_{n = 1}^\infty {( - 1)^n z^n } - 1}}{{(1 + z)^2 }} = f(z)\frac{{1 - 1}}{{(1 + z)^2 }} = 0. $$ Thus $f(z)/(1+z)$ is a constant, which must be $f(0)/(1+0)=1$. Hence $f(z)=1+z$.

Answer 2

$$\exp[\sum_{n=1}^{\infty}(-1)^{n-1}\frac{z^n}{n}]= \exp[\ln(1+z)]=1+z, |z|<1.$$

Answer 3

Hint: What is the power series for $ln(z) $ ?