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Math 382-0: Kursobjekt

Aaron Greicius

Section 1.11 Branches and harmonic functions

We will round out our discussion of differentiability with some miscellany, each item of which will be take up again in the course.

Subsection Branches

We threw the term branch around earlier when discussing notions of argument, logarithms, and complex powers. It is time at last to put this notion on firm, if somewhat formal ground.

Definition 1.11.1. Branch.

A complex set-valued function is a function \(\mathcal{F}\) whose domain is a subset \(D\subseteq \C\text{,}\) and whose images \(\mathcal{F}(z)\) are sets of complex numbers for all \(z\in \C\text{.}\)
A branch of a set-valued function \(\mathcal{F}\) with domain \(D\) is a function \(f\colon U\rightarrow \C\) satisfying the following properties:
  1. \(U\subseteq D\) is open and connected;
  2. \(f\) is continuous;
  3. \(f(z)\in \mathcal{F}(z)\) for all \(z\in U\text{.}\)

Definition 1.11.2. .

Given \(\alpha\in \R\text{,}\) we define \(R_\alpha\) to be the ray
\begin{equation*} R_\alpha=\{te^{i\alpha}\colon t\geq 0\}=\{z\in \C\colon \Arg z=\alpha+2\pi k \text{ for some } k\in \Z\}\cup\{0\}\text{,} \end{equation*}
and we define the \(\alpha\)-cut branch of \(\arg\) to be the function
\begin{align*} \Log_\alpha\colon \C-R_\alpha \amp \rightarrow \C \end{align*}
defined as follows: for all \(z\in \C-\R_{\alpha}\text{,}\) \(\Arg_{\alpha}(z)\) is the unique \(\theta\in (\alpha, \alpha+2\pi\) satisfying \(\alpha< \theta \leq \alpha+2\pi\text{.}\)
Similarly, we define the \(\alpha\)-cut branch of \(\log\) to be the function \(\Log_\alpha\colon \C-R_\alpha\colon \C\) defined as
\begin{equation*} \Log_\alpha(z)=\ln\abs{z}+i\Arg_\alpha(z)\text{;} \end{equation*}
and for a fixed \(w\in \C\text{,}\) we define the \(\alpha\)-cut branch of \(\mathcal{F}(z)=z^w\) to be the function
\begin{equation*} f(z)=e^{w(\ln\abs{z}+i\Arg_\alpha(z))}\text{.} \end{equation*}

Remark 1.11.3. \(\alpha\)-anchored branches.

It is easy to see that \(\Log_\alpha\) is indeed a branch of \(\arg\) for all \(\alpha\text{,}\) and as a result, that the \(\alpha\)-anchored branches of \(\log\) and the set-valued function \(\mathcal{F}(z)=z^w\) are also branches.
Note further that the principal branches of these three set-valued functions are just the \(-\pi\)-anchored branches in our new terminology.

Example 1.11.4. Relating \(\alpha\)-cut branches.

Given \(\alpha,\beta\in \R\) show that
\begin{equation*} \Arg_\beta(z)=\Arg_\alpha(e^{i(\alpha-\beta)}z)+\beta-\alpha\text{.} \end{equation*}
Hint.
First show that \(\arg(z)=\arg(e^{i(\alpha-\beta)}z)+\beta-\alpha\) for all \(z\text{,}\) where as always this must be interpreted as an equality of sets.
Solution.
This is left as a homework exercise.

Example 1.11.5. \(\Log_\alpha(z)\) is holomorphic.

Show that the branch \(f(z)=\Log_\alpha(z)\) is holomorphic on its domain \(\C-R_\alpha\) and satisfies \(f'(z)=\frac{1}{z}\)
Solution.
As in Example 1.10.15, we use the Cauchy-Riemann equations. Fix \(z_0=r_0e^{i\theta_0}\in \C-R_{\alpha}\text{.}\) Picking a small enough neighborhood in the \(r\theta\)-plane, we can assume that for \((r,\theta)\) close enough to \((r_0,\theta_0)\) we have \(z=re^{i\theta}\) if and only if \(r=\abs{z}\) and \(\Arg_\alpha z=\theta+2\pi k\) for some fixed \(k\in \Z\text{.}\) (This is because the map \((r,\theta)\mapsto \Arg_\alpha(e^{i\theta})\) is continuous.) We conclude that
\begin{equation*} \Log_\alpha(z)=\ln \abs{z}+i \Arg_\alpha z=\ln r+i(\theta+2\pi k) \end{equation*}
for some \(k\in \Z\text{,}\) and hence \(u(r,\theta)=r\) and \(v(r,\theta)=(\theta+2\pi k)\) for some \(k\text{.}\) We then proceed exactly as in Example 1.10.15 to show that the polar Cauchy-Riemann equations are satisfied, and that \(f'(z)=1/z\text{.}\)

Proof.

We only provide a proof for the statements concerning \(\log\text{.}\) The claims for \(\arg\) were shown on a homework assignment. Similarly, the special case \(\mathcal{F}(z)=z^w\) with \(w=1/n\) for power functions is left as a homework exercise. The case for general \(w\) is slightly more difficult, but not beyond our means. Nonetheless we omit the proof here.
Suppose \(f\) and \(g\) are two branches of \(\log\) on the open connected set \(U\text{.}\) It follows that for all \(z\in U\) there is an integer \(k\in \Z\) such that \(f(z)=g(z)+2\pi i k\text{.}\) This means that the function \(h\colon U\rightarrow \C\) defined as \(h(z)=f(z)-g(z)\) is a continuous function and satisfies
\begin{equation*} h(U)\subseteq \{2\pi i k \colon k\in \Z\}\text{.} \end{equation*}
Since \(\{2\pi i k\colon k\ in \Z\}\) is clearly a discrete subset of \(\C\text{,}\) and since \(h\) is continuous and \(U\) is connected, it follows that \(h\) is a constant function: i.e., there is a \(k\in \Z\) such that \(h(z)=2\pi i k\) for all \(z\in U\text{.}\) We conclude that there is a \(k\in \Z\) such that \(f(z)=g(z)+2\pi i k\) for all \(z\in U\text{,}\) as desired.
For the second statement, assume \(f\) is a branch of \(\log\) on \(U\text{,}\) and let \(z_0\in U\text{.}\) We wish to show \(f'(z_0)=1/z_0\text{.}\) Write \(z_0=r^{i\alpha}\text{,}\) and let \(\Log_{-\alpha}\) be the corresponding branch of \(\log\) defined on \(V=\C-R_{-{\alpha}}\text{.}\) Since \(U\) and \(V\) are open, so is \(U\cap V\text{.}\) Since \(z_0\in U\cap V\text{,}\) we can find an open ball \(B_{r}(z_0)\subseteq U\cap V\text{.}\) Our two functions \(f\) and \(\Log_{-\alpha}\) are both branches of \(\log\) on \(B_r(z_0)\text{.}\) Since \(B_r(z_0)\) is connected, we conclude that there is a \(k\in \Z\) such that \(f(z)=\Log_{-\alpha}(z)+2\pi i k\) for all \(z\in B_{r}(z_0)\text{.}\) Taking derivatives, we then have
\begin{align*} f'(z) \amp =\frac{d}{dz}\Log_{-\alpha}(z)+\frac{d}{dz}(2\pi i k)\\ \amp = \frac{1}{z}+0 \amp \knowl{./knowl/xref/eg_alpha_cuts.html}{\text{Example 1.11.5}} \\ \amp = \frac{1}{z} \end{align*}
for all \(z\in B_{r}(z_0)\text{.}\) In particular, we have \(f'(z_0)=1/z_0\text{,}\) as desired.

Example 1.11.7. Branches of cube-root function.

Let \(U=\{z\in \C\colon \Im z> 0 \text{ or } \Re z < 0\}\text{.}\)
  1. Show that there are exactly three branches of \(\mathcal{F}(z)=z^{1/3}\) on \(U\) and give explicit formulas for each.
  2. Verify that (1.46) holds for one of your three branches.
Solution.
Since \(U\subseteq \C-R_{\alpha}\text{,}\) where \(\alpha=7\pi/4\text{,}\) the function
\begin{align*} f(z)\amp =e^{\frac{1}{3}\Log_{\alpha}z}\\ \amp = e^{1/3(\ln\abs{z}+i\Arg_{\alpha} z)}\\ \amp =r^{1/3}e^{i(\Arg_{\alpha} z)/3} \amp (r=\abs{z}) \end{align*}
is a branch of \(z^{1/3}\text{.}\) According to Theorem 1.11.6, the other branches are of the form
\begin{equation*} g(z)=e^{2\pi i k/3}f(z)\text{.} \end{equation*}
Letting \(k\) vary in this manner, we see that the multiplying factor \(e^{2\pi i k/3}\) ranges over the three cube-roots of unity: i.e., the distinct branches of \(z^{1/3}\) on \(U\) are
\begin{equation*} f(z), \zeta f(z), \zeta^2 f(z)\text{,} \end{equation*}
where \(\zeta=e^{2\pi i/3}=\cos(2\pi/3)+i\sin(2\pi/3)\text{.}\) Again using, Theorem 1.11.6, we have
\begin{align*} f'(z)\amp =\frac{1}{3}\, \frac{f(z)}{z} \\ \amp =\frac{\abs{z}^{1/3}e^{i(\Arg_{\alpha} z)/3}}{3\abs{z}e^{i \Arg_{\alpha}(z)}}\\ \amp = \frac{1}{3}\abs{z}^{-2/3}e^{-(2\Arg_{\alpha} z)i/3}\\ \amp = \frac{1}{3} e^{-\frac{2}{3}\Log_{\alpha} z}\\ \amp = \frac{1}{3} h(z)\text{,} \end{align*}
where \(h(z)=e^{-\frac{2}{3}\Log_{\alpha} z}\) is the \(7\pi/4\)-cut branch of \(z^{-2/3}\text{.}\) In other words, we have a derivative formula very much in the spirit of \(\frac{d}{dx}(x^{1/3})=\frac{1}{3}x^{-2/3}\text{.}\)

Subsection Harmonic functions

Definition 1.11.8. Harmonic functions.

Let \(U\subseteq \R^2\) be an open set. A function \(u\colon U\rightarrow \R\) is harmonic if it satisfies the Laplace’s equation
\begin{equation*} u_{xx}(x,y)+u_{yy}(x,y)=0 \end{equation*}
for all \((x,y)\in U\text{.}\)
A pair of harmonic conjugates on \(U\) is a pair \((u,v)\) of harmonic functions on \(U\) satisfying
\begin{align*} v_x \amp =-u_y \amp v_y=u_x\text{.} \end{align*}

Proof.

This is left as a homework exercise.

Example 1.11.10. Harmonic conjugates.

Show that \(u\colon \R^2\rightarrow \R\) defined as \(u(x,y)=\cos x\cosh y\) is harmonic and find a harmonic conjugate of \(u\text{.}\)
Solution.
We have \(\cos z=u(x,y)+v(x,y)i\text{,}\) where \(u(x,y)=\cos x\cosh y\) and \(v(x,y)=-\sin x\sinh y\text{.}\) Since \(\cos\) is holomorphic on \(\C\) (and the partials of \(u\) and \(v\) are sufficiently nice), we conclude that (a) \(u\) and \(v\) are both harmonic, and (b) \(v\) is a harmonic conjugate of \(u\text{.}\)
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