Determining whether $f(z)=\ln r + i\theta$ (with domain $\{z:r\gt , 0\lt \theta \lt 2\pi\}$) is analytic

Question

Define $$f(z)=\ln r + i\theta$$ on the domain $\{z:r\gt , 0\lt \theta \lt 2\pi\}$.

This domain is just a punctured disk of radius $\ln r$, correct?

How does one determine whether this is analytic, I can't see how I would take the CREs

$$u(r,\theta) = \ln r$$ $$v(r,\theta) = \theta.$$

Should I convert this back to $x+iy$ form and proceed? How can I do such a thing with what appears to be a punctured open neighborhood?

How do I show that the function is analytic and find its derivatives?

Answer 1

The domain is the disk of radius $r$ with the closed positive real half-axis removed; in particular this is not a punctured disk.

A good way to check analyticity of $f$ is to write the C.-R. equations in polar coordinates. (The alternative is to write $f$ in rectangular coordinates, but it is slightly tricky to handle taking derivatives of the imaginary part of $f$: At least for $\theta \in (0, \frac{\pi}{2})$, the imaginary part of $f$ coincides with $\arctan\left(\frac{y}{x}\right)$, but this latter function is not defined on the $y$-axis, and it does not agree with the imaginary part of $f$ in the other three quadrants.)

Here, $$f(z) = \ln z,$$ where $\ln$ is the branch of the (natural) logarithm function whose argument takes values in $(0, 2 \pi)$, or in rectangular coordinates, $$f(x + iy) = \frac{1}{2} \ln(x^2 + y^2) + i \arg(x, y),$$ where $\arg$ here denotes the function that returns the anticlockwise angle in $(0, 2\pi)$ from the positive $x$-axis to the ray from the origin through $(x, y)$.

Answer 2

The domain given is $U=\mathbb {C}\setminus [0,\infty).$ I'm not sure why people are talking about a punctured disc.

Assuming the basic properties of $e^z,$ we can do the following. First, $f(re^{it}) = \ln r +it$ is continuous and 1-1 in $U.$ Second, $e^{f(z)} = z$ for all $z\in U.$ So fix a point $a\in U.$ Then for small $h$ we have

$$1 = \frac{e^{f(a+h)} - e^{f(a)}}{h} = \frac{e^{f(a+h)} - e^{f(a)}}{f(a+h)-f(a)}\cdot \frac{f(a+h)-f(a)}{h}.$$

Thus

$$\left (\frac{e^{f(a+h)} - e^{f(a)}}{f(a+h)-f(a)}\right )^{-1} = \frac{f(a+h)-f(a)}{h}.$$

As $h\to 0,$ $f(a+h) \to f(a),$ hence the left side tends one over the derivative of $e^z$ at $f(a).$ Because $(e^z)' = e^z,$ this is just $1/e^{f(a)} = 1/a.$ Therefore $f'(a) = 1/a$ as expected. This shows $f$ is analytic on $U.$