Expressing a function as exponential of another function in a given domain Can the functions $h_1(z) = 1+z^2$ and $h_2(z) = 1+\log(4+z)$ be expressed as the exponential of a function $f(z)$, where $f$ is holomorphic in the set $D = \{z : |z| < 2 \}$ ? 
More generally, when can a function be expressed as the exponential of another function?
 A: 
More generally, when can a function be expressed as the exponential of another function?

Since the exponential function has no zeros, a necessary condition is that the function to be expressed has no zeros.
By itself, that is not a sufficient condition, as for example $h(z) = z$ in $\mathbb{C}\setminus\{0\}$ shows.
If the domain of $h$ is simply connected, then $h$ being zero-free is sufficient. If a branch of the logarithm is defined on the image of $h$, then you can also write $h = e^f$ (with $f = \log \circ h$).
Generally, you can write $h = e^f$ on $\Omega$ if and only if $h$ has no zeros in $\Omega$ and the function $\frac{h'}{h}$ has a primitive in $\Omega$.
If $h = e^f$, then $\frac{h'}{h} = \frac{e^f\cdot f'}{e^f} = f'$, and conversely, if $\frac{h'}{h} = g'$ then $(h\cdot e^{-g})' = h'\cdot e^{-g} - g'\cdot h\cdot e^{-g} = (h' - g'h)e^{-g} = 0$, so $h\cdot e^{-g} \equiv c\neq 0$ and $h = e^{k + g}$ where $e^k = c$.
For your examples: $h_1(z) = 1+z^2$ has a zero (two, actually) in $D = \{z : \lvert z\rvert < 2\}$, hence cannot be written in the form $e^f$, and $h_2(z) = 1 + \log (4+z)$ has no zero in $D$, and can, since $D$ is simply connected, be written in the form $e^f$.
