= Holomorphic logarithm
{title2=$e^g=f,\quad g'=f'/f$}
A holomorphic logarithm of a nowhere-zero <holomorphic function> $f$ on a domain $U$ is a <holomorphic function> $g$ such that $e^{g(z)}=f(z)$ throughout $U$. Differentiating with the <chain rule> gives $g'=f'/f$. For $f(z)=z$, a <holomorphic logarithm> is a <branch of the complex logarithm>, and its <derivative> is $1/z$.
The <principal logarithm> is <holomorphic> on the slit <complex plane> $D=\mathbb C\setminus(-\infty,0]$, with
$$
\operatorname{Log}z=\log|z|+i\operatorname{Arg}z,\qquad -\pi<\operatorname{Arg}z<\pi,\qquad (\operatorname{Log}z)'=\frac1z.
$$
Here $\operatorname{Arg}$ is the chosen <complex argument>, and $\operatorname{Im}\operatorname{Log}z\in(-\pi,\pi)$. One may conventionally assign a principal value on the negative real axis, but that axis is excluded from this <holomorphic> domain. <Complex Analysis Lecture Notes by Dan Romik>[Romik, Complex Analysis Lecture Notes, section 13, items 109–110]{ref} For example, $1+i=\sqrt2e^{i\pi/4}$ gives $\operatorname{Log}(1+i)=\tfrac12\log2+i\pi/4$ and $(\operatorname{Log})'(1+i)=(1-i)/2$; all its other <complex logarithms> differ by $2\pi ik$, $k\in\mathbb Z$.
There is no <holomorphic logarithm> of $z$ on the punctured <complex plane> $\mathbb C\setminus\{0\}$. Such a function would be a <holomorphic primitive> of $1/z$, so its <contour integral> around every closed path would vanish. On the counterclockwise <unit circle>, however,
$$
\oint_{|z|=1}\frac{dz}{z}=\int_0^{2\pi}\frac{ie^{it}}{e^{it}}\,dt=2\pi i\ne0.
$$
This is the <period obstruction to a holomorphic antiderivative>, and also exactly the logarithm change in the <monodromy> diagram.
Every nowhere-zero <holomorphic function> $f$ on a <simply connected domain> $U$ does admit a <holomorphic logarithm>. Choose $a\in U$ and $b\in\mathbb C$ with $e^b=f(a)$, and define
$$
g(z)=b+\int_a^z\frac{f'(\zeta)}{f(\zeta)}\,d\zeta.
$$
The <Cauchy integral theorem> makes this <contour integral> independent of the path in $U$. Then $g'=f'/f$ and $(fe^{-g})'=e^{-g}(f'-fg')=0$; its value at $a$ is $1$, proving $e^g=f$. Any two choices on a <connected> domain differ by a constant $2\pi ik$, since their difference is <continuous> and takes values in the discrete set $2\pi i\mathbb Z$. <Complex Analysis Lecture Notes by Dan Romik>[Romik, section 12, Corollary 108, and section 13, Exercise 114]{ref}
For a computed example, take $f(z)=1+z$ on the <unit disc>, with $a=b=0$. Expanding $f'/f=1/(1+z)$ as a <geometric series> and integrating gives
$$
g(z)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}z^n}{n},\qquad g(1/2)=\log(3/2),\qquad e^{g(1/2)}=3/2.
$$
The construction also produces <holomorphic square roots>: $h=e^{g/2}$ satisfies $h^2=f$, so here $h(1/2)=\sqrt{3/2}$. More generally, $e^{\alpha g}$ gives a consistent <branch of a multivalued function> for <complex exponentiation> by $\alpha$. A <holomorphic logarithm> converts multiplicative expressions into additive ones while making the domain restrictions explicit.
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