Proving $\operatorname{Log}(z)$ is differentiable on $\mathbb{C} \setminus \{x \in \mathbb{R}:x \leq 0\}$ This question is part of an assignment which I am solving.

Question :define $\log(z)=\log|z| + i\theta$ where $-\pi<\theta\leq\pi$ and $z=|z|e^{i\theta}$ ($z\neq0$).
Then prove that $\log(z)$ is differentiable on $\mathbb{C}\setminus \{x \in \mathbb{R}:x \leq 0\}$.

First I proved that $\log$ is continuous on $\mathbb{C}\setminus \{x \in \mathbb{R}:x \leq 0\}$.
Now, for differentiability I took
$$\lim_{h \to 0} \frac{(\log|z+h| - \log|z|)}{ h} =\frac{(\log(|1+h/z|)}{h} ,$$
but This is an indeterminate form.
So, kindly tell me what mistake I am making and tell me how to rightly approach it in a detailed way.

Also , part b is there which asks to prove that there is no power series $\sum_{n=0}^{\infty} a_{n} (z-c)^{n}$ convergent in $\mathbb{C}\setminus \{ x\in \mathbb{R}: x\leq 0 \}$ whose sum is $\log$ .

Kindly just give a hint for part b . I think b part will use a part and I could not solve a so b would be tough to do also.
 A: Part a: Let $U=\mathbb C\setminus (-\infty,0].$ You know $\log$ is continuous on $U.$ You also know $e^{\log z} =z$ on $U,$ which implies $\log$ is injective.
Let $z\in U.$ Choose $D(z,r)\subset U.$ If $0<|h|<r,$ then
$$1=\frac{(z+h)-z}{h} = \frac{e^{\log(z+h)}-e^{\log z}}{\log(z+h)-\log z}\cdot\frac{\log(z+h)-\log z}{h}.$$
This implies
$$\frac{\log(z+h)-\log z}{h}= \dfrac{1}{\dfrac{e^{\log(z+h)}-e^{\log z}}{\log(z+h)-\log z}}.$$
As $h\to 0,$ $\log(z+h)\to \log z.$ Thus the denominator on the right $\to \exp'(\log z) = \exp(\log z) = z.$ It follows that $\log'(z) = 1/z.$
Part b: Suppose there is such a power series $\sum_{n=0}^{\infty}a_n(z-c)^n.$ Note that $|z-c|$ can be made arbitrarily large while $z$ stays in $U.$ Thus the radius of convergence of the series is $\infty.$
For $z\in \mathbb C,$ let $f(z)=\sum_{n=0}^{\infty}a_n(z-c)^n.$ Then $f$ is an entire function that agrees with $\log z$ on $U.$ This forces $f$ to have discontinuities at each point of $(-\infty,0],$  just as $\log z$ does. That is a contradiction, showing no such power series exists.
