# Define a branch cut of the argument depending on z

I asked the following question a few days ago: Analytic branch of the complex logarithm in the logarithmic spiral branch cut

Someone (not in the post) suggested me to take a branch of the argument like that:

$$f(z) = Arg_{|z| - 2\pi} (z) \tag{1}$$

in the domain $$D = \mathbb{C} \setminus \{te^{it} : t \ge 0\}$$, where $$Arg_{\alpha}$$ returns values of $$\arg(z)$$ in the range $$(\alpha, \alpha + 2\pi)$$.

So in this case, $$f(z) \in (|z| - 2\pi, |z|)$$.

In my opinion, it is equal to the following:

$$f(z) = Arg(e^{i(\pi - |z|)}\cdot z) + |z| - \pi \tag{2}$$

Since I want anything with angle of $$|z| -2\pi$$ to map to $$-\pi$$, compute its principal argument (which is the $$Arg(\cdot)$$) and then return it to $$|z| - 2\pi$$.

My questions are:

1. Is (1) even a branch of $$\arg(z)$$ in $$D$$? Is it continuous? I know that for a fixed $$\alpha$$, $$Arg_{\alpha}(z)$$ is indeed a continuous branch of $$\arg(z)$$, but in this case I can't figure it out.

2. If (1) is a branch of $$\arg(z)$$, does it define a holomorphic branch of $$\log(z)$$?

3. Is (2) correct? Does the equality hold?

4. Is there a general rule that defines a continuous branch of the argument when it depends on $$z$$?

Thanks for the help.

The concept of a branch of the argument function on a domain $$U \subset \mathbb C^* = \mathbb C \setminus \{0\}$$ should include continuity. If we drop this requirement, we would allow completely erratic argument functions. So the definition should be

A continuous function $$A :U \to \mathbb R$$ is called a branch of the argument function on $$U$$ if $$\lvert z \rvert e^{iA(z)} = z$$ for all $$z∈U$$.

Given such a branch $$A$$ we define $$\ln_A : U \to \mathbb C, \ln_A(z) = \ln^{\mathbb R}(\lvert z \rvert) + i A(z) .$$

Here $$\ln^{\mathbb R} : (0,\infty) \to \mathbb R$$ denotes the real logarithm. Clearly $$\ln_A$$ is a continuous function such that $$e^{\ln_A(z)} = e^{\ln^{\mathbb R}(\lvert z \rvert)}e^{i A(z)} = \lvert z \rvert e^{i A(z)} = z$$ for all $$z \in U$$. Note that if we take a non-continuous $$A$$, then also $$\ln_A$$ is not continuous so that it cannot be holomorphic.

We claim that $$\ln_A$$ is holomorphic for each branch $$A$$ of the argument function on $$U$$. That is, $$\ln_A$$ is a branch of the logarithm on $$U$$.

Consider $$z_0 \in U$$. Since the exponential map has a nowhere vanishing derivative, it maps an open neighborhood $$W$$ of $$\ln_A(z_0)$$ biholomorphically onto an open neighborhood $$V$$ of $$e^{\ln_A(z_0)} = z_0$$. Let $$E : W \to V$$ denote the biholomorphic restriction of the exponential map. Since $$\ln_A$$ is continuous, we find an open neighborhood $$V'$$ of $$z$$ in $$U$$ such that $$\ln_A(V') \subset W$$. Let $$V'' = V \cap V'$$. For $$z \in V''$$ we get $$E(\ln_A(z)) = e^{\ln_A(z)} = z = E(E^{-1}(z)$$, thus $$\ln_A(z) = E^{-1}(z)$$. This shows that $$\ln_A$$ is complex differentiable at $$z_0$$.

In fact we have a $$1$$-$$1$$-correspondence between branches of the argument function on $$U$$ and branches of the logarithm on $$U$$. To see this, let $$\Re, \Im : \mathbb C \to \mathbb R$$ denote the real and imaginary part functions $$\Re(x + iy) = x, \Im(x + iy) = y$$.

Given a branch $$\ell$$ of the logarithm on $$U$$, the function $$\Im \circ \ell$$ is a branch of the argument function on $$U$$: We have $$z = e^{\ell(z)} = e^{\Re(\ell(z))} e^{i \Im(\ell(z))}$$ which implies $$\lvert z \rvert = e^{\Re(\ell(z))}$$ and thus $$z = \lvert z \rvert e^{i \Im(\ell(z))}$$.

The associations $$A \mapsto \ln_A$$ has $$\ell \mapsto \Im \circ \ell$$ are inverse to each other:

1. $$\ln_{\Im \circ \ell} = \ell$$:
Since $$\lvert z \rvert = e^{\Re(\ell(z))}$$, we get $$\ln_{\Im \circ \ell}(z) = \ln^{\mathbb R}(\lvert z \rvert) + i (\Im \circ \ell)(z) = \ln^{\mathbb R}(e^{\Re(\ell(z))}) + i \Im(\ell(z)) = \Re(\ell(z)) + i \Im(\ell(z)) = \ell(z).$$

2. $$\Im \circ \ln_A = A$$:
We have $$\Im(\ln_A(z))) = \Im( \ln^{\mathbb R}(\lvert z \rvert) + i A(z)) = A(z)$$.

Let us now show that the function $$f : D \to \mathbb R$$ is well-defined.

The issue here is that the functions $$Arg_\alpha$$ are not defined on all of $$\mathbb C^*$$, but only on the sliced planes $$S_\alpha = \mathbb C \setminus \{te^{i\alpha} \mid t \in [0,\infty)\}$$. Thus we have to verify that $$z \in S_{\lvert z \rvert - 2\pi}$$ for all $$z \in D$$. Assume that $$z \notin S_{\lvert z \rvert - 2\pi}$$. Then $$z = te^{i(\lvert z \rvert -2\pi)} = te^{i\lvert z \rvert}$$ for some $$t \ge 0$$. This gives $$\lvert z \rvert = t$$, thus $$z = te^{it} \notin D$$, a contradiction.

Let us next show that $$(2)$$ is correct.

For $$z \in D$$ we have $$z = \lvert z \rvert e^{if(z)}$$, where $$f(z) \in (\lvert z \rvert - 2\pi, \lvert z \rvert)$$. Thus $$e^{i(\pi - \lvert z \rvert)} \cdot z = \lvert z \rvert e^{i(\pi - \lvert z \rvert +f(z))}$$, where $$\pi - \lvert z \rvert + f(z) \in (-\pi,\pi)$$. Thus $$e^{i(\pi - \lvert z \rvert)} \cdot z$$ is contained in the sliced plane on which the principal argument $$Arg$$ is defined; we have $$Arg(e^{i(\pi - \lvert z \rvert)} \cdot z) = \pi - \lvert z \rvert + f(z)$$ which is nothing else than $$(2)$$.

But it is obvious that the RHS of $$(2)$$ is a continuous function which means that $$f$$ is continuous.

We have now answered your questions 1. - 3. Question 4. is too vague to be answered.

Update:

Here is an approach avoiding to use the inverse function theorem for complex functions.

Let $$A : U \to \mathbb R$$ be a branch of the argument function. Then $$\cos(A(z)) + i \sin(A(z)) = e^{iA(z)} = \frac{z}{\lvert z \rvert}$$ Identifying $$\mathbb C$$ with $$\mathbb R^2$$ we get $$\cos(A(x,y)) = \frac{x}{\sqrt{x^2+y^2}}, \sin(A(x,y)) = \frac{y}{\sqrt{x^2+y^2}} .$$

$$A(x,y)$$ is contained in one or two one of the intervals of the form $$J^c_k = (k\pi, (k+1)\pi)$$ or $$J^s_k = (k\pi - \frac \pi 2, k\pi + \frac \pi 2)$$ with $$k \in \mathbb Z$$ (it is contained in exactly one of these intervals iff $$A(x,y)$$ is an integral multiple of $$\frac \pi 2$$). The intervals $$J^c_k$$ are mapped by $$\cos$$ bijectively onto $$(-1,1)$$, the intervals $$J^s_k$$ are mapped by $$\sin$$ bijectively onto $$(-1,1)$$.

Let us assume that $$A(x,y) \in J^c_k$$ (the case $$A(x,y) \in J^s_k$$ can be treated similarly). The inverse function $$(-1,1) \to J^c_k$$ is a branch of the arcus cosinus which we denote by $$\arccos_k$$. We know that $$\arccos_k$$ is continuously differentiable.

By continuity $$A$$ maps an open neigborhood $$V \subset U$$ of $$(x,y)$$ into $$J^c_k$$. Thus on $$V$$ we have $$A(x,y) = \arccos_k\left(\frac{x}{\sqrt{x^2+y^2}}\right). \tag{1}$$ This shows that $$A$$ is continuously differentiable on $$V$$.

Since this is true for all $$(x,y) \in U$$, we see that $$A$$ is continuously differentiable on $$U$$.

This shows that $$\ln_A(z) = \ln^{\mathbb R}(\lvert z \rvert) + i(A(z)$$ is real differentiable.

We can now compute the partial derivatives of $$u(x,y) = \ln^{\mathbb R}(\sqrt{x^2+y^2})$$ and $$v(x,y)= A(x,y)$$. Doing this for $$A$$ requires to treat of a lot of cases (we have to treat $$(1)$$ and its $$\sin$$-version, and in the derivatives we have to be careful with signs). Anyway, we should be able to verify the Cauchy–Riemann equations which assure complex differentiability. I have not done it, it is tedious work.

Another proof of complex differentiability without verifying the Cauchy–Riemann equations is based on A simple proof that a real differentiable local section of a holomorphic function is holomorphic.

• Thank you so much! You helped me understand this subject a lot!  Although, I have a question - is there a simpler way to prove that $\ln_A$ is holomorphic? Since I see you have used the inverse function theorem for complex functions, which we did not prove in my class when this topic of branches was taught.
– Din
Dec 30, 2022 at 18:47
• @Din See my update. Dec 31, 2022 at 13:07