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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.

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Din
    Dec 30, 2022 at 18:47
  • $\begingroup$ @Din See my update. $\endgroup$
    – Paul Frost
    Dec 31, 2022 at 13:07

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