Definition of $\log z $ Assume I want to define an analytic branch of $\log z $ in the region $D(i,1)$ which is the disk centered at $i $ with radius $1$.
Since the multivalued $\log z $ given by
$$ \log z
= \log(re^{i\theta})
= \log|r| + i\operatorname{Arg}(re^{i\theta})
= \log|r| + i(\theta+2\pi k) $$
I'll have to specify which $k $ I'm taking.
So, it seems natural to me to exclude the negative imaginary axis, i.e., for and $ z $ we'll take $ \operatorname{Arg}(z)\in\left(-\frac{1}{2}\pi,\frac{3}{2}\pi\right) $.
My questions are:

*

*Is this means that I defined $\log z $ on the whole plane excpet for the negative part of the imaginary axis? (is this region even simply connected?)


*How can I show that this definition actually defines an analytic function? it is obvious that
$$ e^{\log\left(z\right)}
= e^{\log|r|+i\operatorname{Arg}(z)}
= e^{\log|r|} \cdot e^{i\operatorname{Arg}(z)}
=|r|e^{i\operatorname{Arg}(z)}
=z $$
But the same equations I wrote could be valid for any definition of the $\log $, how exactly the choice of the range for $\theta $ being expressed here? and how can I show that this function is analytic?
Thanks.
 A: *

*It indeed means that you defined $\log(\cdot)$ on the complex plane except for the non-positive imaginary axis, i.e., the region
$$ \mathcal{D} = \mathbb{C} \setminus (-i[0, \infty)). $$
To show that $\mathcal{D}$ is simply connected, you may verify that $\mathcal{D}$ is a star-shaped domain, in the sense that there exists a point $z_0 \in \mathcal{D}$ such that every $z \in \mathcal{D}$ can be joined to $z_0$ through a line segment in $\mathcal{D}$.


*One way is to verify that your definition satisfies the Cauchy–Riemann equations. Note that
$$ \frac{\partial}{\partial x} \log |r| = \frac{x}{x^2+y^2}, \qquad \frac{\partial}{\partial y} \log |r| = \frac{y}{x^2+y^2} $$
and that any branch of $\arg(\cdot)$ satisfies
$$ \frac{\partial}{\partial x}\arg(z) = -\frac{y}{x^2+y^2}, \qquad \frac{\partial}{\partial y}\arg(z) = \frac{x}{x^2+y^2}. $$
Alternatively, you may realize your $\log(\cdot)$ as an inverse of an appropriate restriction of $\exp(\cdot)$. Together with the fact that $\exp(\cdot)$ has nowhere-vanishing derivative, the inverse function theorem for analytic functions will guarantee the analyticity of the inverse.
A: *

*Yes, the way you defined it, you don't have to restrict yourself to $D(\mathrm i,1)$. It's valid for the whole slit plane, with the slit being the non-positive imaginary axis (not just the negative imaginary axis, since $0$ also needs to be excluded). This is indeed simply connected. The upper half plane is homeomorphic to a slit plane (with the non-negative real axis missing) via the homeomorphism $z\mapsto z^2$, and the two slit planes are clearly homeomorphic. Also, the half plane is homeomorphic to the entire plane (which is simply connected), so the slit plane is, too. And since homeomorphisms preserve simply connectedness, the slit plane is simply connected.


*Your function
$$\log:\mathbb C\backslash[0,-\mathrm i\infty)\to\{z\in\mathbb C~\vert~\operatorname{Im}z\in(-\frac12\pi,\frac32\pi)\}$$
is the inverse of
$$\exp:\{z\in\mathbb C~\vert~\operatorname{Im}z\in(-\frac12\pi,\frac32\pi)\}\to \mathbb C\backslash[0,-\mathrm i\infty),$$
which is analytic, bijective, and has nowhere vanishing derivative. As such, it's inverse is also analytic (also with nowhere vanishing derivative, btw.).
