What's the complex derivatives of $\log|z|, z \in \mathbb{C} \setminus {0}$?

Question

I'm supposed to calculate the $\frac{\partial log|z|}{\partial z}$ and $\frac{\partial \log|z|}{\partial \overline{z}}$ for $z \in \mathbb{C}\setminus\{0\}$.

Is it really just straight forward chain rule replacing $|z|$ with$\sqrt{z\overline{z}}$? So e.g. $\frac{\partial}{\partial z}(\log|z|)=\frac{1}{2z}$. At least that would explain, why $z$ shouldn't become $0$.

The other part of my course's question is:

Let $f \in H(\mathbb{D})$ be rootless and $f' \in H(\mathbb{D})$, whereas $H(\mathbb{D})$ are the holomorphic functions on $\mathbb{D}$. Given $g: \mathbb{D} \to \mathbb{R},g(z)=\log|f(z)|$. Explain why $g$ is well-defined and show $\frac{\partial g}{\partial z}$ is holomorphic on $\mathbb{D}$.

And calculating $g'$ would show that it's a composition of holomorphic functions, right?

Answer 1

I will answer the first question partially, the second half of the first question is similar, by using the conjugate of the Wirtinger derivative. I also added a computation for the second question.

1. To verify your intuition, you can use the definition of the Wirtinger derivative as $2\frac{\partial}{\partial z} := \frac{\partial}{\partial x} - i \frac{\partial}{\partial y}$, which in this case, with $z = x + iy$, gives the following: $$\frac{1}{2} \left(\frac{\partial \log \sqrt{x^2+y^2}}{\partial x} - i \frac{\partial \log \sqrt{x^2+y^2}}{\partial y}\right) = \frac{1}{2} \frac{1}{|z|^2} \overline{z} = \frac{1}{2z}$$ This can be computed with the (real) chain rule.

The warning is, the complex chain rules look different in general.

2. Now for additional question about $g$, you have not specified what this $\mathbb{D}$ would be, and I suppose it should be the plane without origin as in the title. Then, one can use for example the (complex) chain rule with the answer to the first question to prove your additional question, using the rootlessness of f for well-definedness, and its holomorphic properties to compute and proof holomorphicity of $\frac{\partial{g}}{\partial{z}}$.

Explicitly, one way to proof that the function $\frac{\partial g}{\partial z}$ is holomorphic is by computing that $\frac{\partial}{\partial \overline{z}}$ of it vanishes. You can check (using, say, complex chain rule, the first computation, the product rule, and holomorphicity of $f$) that: $$\frac{\partial }{\partial \overline{z}}\frac{\partial g}{\partial z} = \frac{\partial }{\partial \overline{z}} \frac{f'(z)}{2 f(z)} = \frac{\partial }{\partial \overline{z}}(\frac{1}{f(z)}) f'(z) + \frac{1}{2f(z)}\frac{\partial f'(z)}{\partial \overline{z}} $$

Now notice that the second term vanishes as $f'(z)$ is holomorphic, and applying the product rule to the equation $$\frac{\partial 1}{\partial \overline{z}} = 0,$$ for $1 = \frac{f(z)}{f(z)}$ for any $z \in \mathbb{D}$, reveals that the first term does too, since $f(z)$ does not vanish.

Last remark: after proving the complex chain rule, one can prove that composition of holomorphic functions remains holomorphic.