There is no 'nice' complex logarithm Two problems are meant to establish that no 'nice' logarithm function exists for complex numbers.
The first is
Let $U$ be an open set in $\mathfrak{C}\setminus\{0\}$.
Suppose $h:U \to \mathfrak{C}$ is a continuous function, such that $$e^{h(z)}=z \ \ \ \text{for every} \ z\in U$$
Show that $h$ is holomorphic by computing the defining formula for the derivative. $$lim_{w \to z} \frac{h(w)-h(z)}{w-z}$$
While the second is to prove no such function can exist on $\mathfrak{C}\setminus\{0\}$.
My attempt so far has been to write $h(z)=f(z)+ig(z)$  which then gives $f(z)=\ln|z|$ and $g(z)=\arg(z)$ 
which contradicts the assumption of continuity, proving the second part. This does not prove the first part, which is that if such an $h$ did exist then $h$ is holomorphic
 A: You can differentiate both sides of the relation to obtain that
$$e^{h(z)}\cdot h'(z) = 1 \Longrightarrow h'(z) = \frac{1}{e^{h(z)}} = \frac{1}{z}$$
which I suppose is the formula the question asks for. Notice that $z=0 \notin U$.
A: Hint for 1: Fix $z$; use local injectivity of $\exp$ and continuity of $h$ to deduce that $h$ is locally injective, i.e., that $h(w) \neq h(z)$ if $w \neq z$ are sufficiently close. Now write
$$
\frac{h(w) - h(z)}{w - z}
  = \frac{h(w) - h(z)}{e^{h(w)} - e^{h(z)}}
  = \frac{h(w) - h(z)}{e^{h(z)}(e^{h(w) - h(z)} - 1)},
$$
and use holomorphicity of $\exp$ (including your knowledge of the derivative of $\exp$) and continuity of $h$ to evaluate the limit as $w \to z$.
Hint for 2: Integrate $h'(z)\, dz$ around the unit circle. If there exists a holomorphic $h$ away from the origin, what is the integral equal to? What is the integral actually equal to?

For posterity, if $h(z) = u(z) + iv(z)$ with $u$ and $v$ real-valued, you can't deduce $u(z) = \ln|z|$ and $v(z) = \arg(z)$, only
$$
\operatorname{Re} z = e^{u(z)} \cos v(z),\qquad
\operatorname{Im} z = e^{u(z)} \sin v(z).
$$
(The formula for $u$ is all right, but there are infinitely many "working choices" of $v$, each a branch of $\arg$.)
