Which holomorphic functions have constant argument on rays from the origin? On circles centered at the origin?

Question

Multiples $$f(z) = c z, \qquad c \in \Bbb C \setminus \{0\},$$ of the identity function on $\Bbb C \setminus \{0\}$ trivially all satisfy the following special condition:

Condition A: All points on a given open ray $\{\arg z = \theta_0\}$ centered at the origin have the same argument under $f$, that is, whenever $\arg z = \arg w$ we also have $\arg f(z) = \arg f(w)$. Put another way, the function $\arg \circ f$ descends via $\arg$; in that case we can interpret the descent as a function $S^1 \to S^1$.

That $f(z) = c z$ satisfies Condition A is visible on an argument plot of the function, wherein each point $z$ is colored according to the argument $\arg f(z)$ (the case $c = 1$, i.e., the identity function, is shown).

Question A: What are all of the holomorphic functions that satisfy Condition A?

We can ask just as well for a dual condition:

Condition B: All points on a given circle $\{|z| = r_0\}$ centered at the origin have the same argument under $f$, that is, whenever $|z| = |w|$ we also have $\arg f(z) = \arg f(w)$. The function $\arg \circ f$ descend via the modulus function $|\cdot|$ to a function $\Bbb R_+ \to S^1$? The argument plot of such a function is rotationally invariant:

Question B: What are all of the holomorphic functions that satisfy Condition B?

Answer 1

Write the independent variable $z$ and dependent variable $w = f(z)$ in polar form, say, as $$z = r e^{i \theta}, \qquad f(z) = \rho e^{i \alpha} .$$ In particular, $\rho = |f(z)|$ and $\alpha = \arg f(z)$ are functions of $r, \theta$. Translating the Cauchy-Riemann equations into coordinates $(r, \theta), (\rho, \alpha)$ yields $$\phantom{(\textrm{CR})} \qquad \boxed{\frac{\rho_r}{\rho} = \frac{1}{r} \alpha_\theta, \qquad \frac{\rho_\theta}{\rho} = -r \alpha_r} . \qquad (\textrm{CR})$$

Question A: What are all of the holomorphic functions that satisfy Condition A?

Suppose that the holomorphic function $f(z) = \rho e^{i \alpha}$ satisfies Condition A. Differentiating along rays (i.e., with respect to $r$) gives that at an arbitrary point $z = r e^{i\theta}$, $$\alpha_r(z) = \left(\frac{\partial}{\partial r} \arg f\right)(z) = \lim_{s \to r} \frac{\arg f(s e^{i \theta}) - \arg f(r e^{i\theta})}{s - r} = 0 ,$$ so $\alpha$ is a function of $\theta$ alone. Since $f$ is holomorphic, it satisfies (CR), and in particular $\rho_\theta = -r \rho \cdot 0 = 0$, so $\rho$ is a function of $r$ alone. The other equation in (CR) gives that (inflicting a mild abuse of notation) $$\frac{r \rho_r(r)}{\rho(r)} = \alpha_\theta(\theta),$$ but the left- and right-hand sides depend only on $r, \theta$ respectively, so both sides are equal to some common constant $b$. Solving separately the o.d.e.s $\frac{r \rho_r(r)}{\rho(r)} = b$ and $\alpha_\theta(\theta) = b$ gives $$\rho(r) = \rho_0 r^b \qquad \textrm{and} \qquad \alpha(\theta) = b \theta + \alpha_0$$ for some real $\rho_0, \theta_0$. Thus, $$f(z) = f(r e^{i \theta}) = \rho e^{i \alpha} = (\rho_0 r^b) e^{i (b \theta + \alpha_0)} = c (r e^{i \theta})^b = c z^b ;$$ here $c := \rho_0 e^{i \alpha_0}$ can take on any value in $\Bbb C \setminus \{0\}$ and $b \in \Bbb R$. In summary, a holomorphic function satisfies Condition A iff it is locally given by some power function $$\color{#bf0000}{\boxed{f(z) = c z^b}}$$ with real exponent. Such a function extends to a function holomorphic on all of $\Bbb C \setminus \{0\}$, iff $z^b$ does, that is, iff $b \in \Bbb Z$. In either case, the descent $S^1 \to S^1$ is $\zeta \mapsto e^{i \alpha_0} \zeta^b$.

Question B: What are all of the holomorphic functions that satisfy Condition B?

An analogous argument shows that Condition B is equivalent to $\alpha_\theta = 0$, hence by (CR) $\rho_r = 0$, and a computation similar to the one for Condition A shows that a function $f(z)$ satisfies Condition B iff it is locally given by some power function $$\color{#bf0000}{\boxed{f(z) = c z^{i b}}}$$ with imaginary exponent, or more precisely, with $c \in \Bbb C \setminus \{0\}$, $b \in \Bbb R$. Except when $b = 0$, i.e., when $f$ is constant, no such function extends to be holomorphic on $\Bbb C \setminus \{0\}$. The argument plot after the definition of condition $b$ depicts the case $b = 1$. The descent $\Bbb R_+ \to S^1$ is $r \mapsto e^{i (b \log r + \alpha_0)}$.

Answer 2

For the first question (“Condition A”) it suffices to assume that (in a some neighbourhood of origin) $\arg f(z)$ is constant on two rays $\arg z= \alpha_1, \alpha_2$ such that $\alpha_2 - \alpha_1$ is not a rational multiple of $\pi$.

For simplicity assume that $$ \arg z = 0 \implies \arg f(z) = 0 \\ \arg z = \alpha \implies \arg f(z) = \beta \\ $$ with $\alpha \notin \pi \Bbb Q$. The general case can be reduced to this special case by considering $\tilde f(z) = f(e^{-i\alpha_1 z})$.

The first condition implies that $$ f(\bar z) = \overline{f(z)} $$ for all $z$, this is an application of the Schwarz reflection principle. Similarly, the second condition gives $$ f(e^{2i\alpha}\bar z) = e^{2i\beta}\overline{f(z)} $$ by applying the Schwarz reflection principle to $g(w) = e^{-i\beta}f(e^{i\alpha} z)$. It follows that $$ f(e^{2i\alpha} z) = e^{2i\beta} f(z) $$ for all $z$ in a neighbourhood of the origin. Now let $fz) = \sum_{n=0}^\infty c_n z^n$ be the Taylor series of $f$ at $z=0$. Then $$ \sum_{n=0}^\infty e^{2in\alpha}c_n z^n = \sum_{n=0}^\infty e^{2i\beta} c_n z^n $$ and therefore $c_n = 0$ or $e^{2in\alpha} = e^{2i\beta}$ must hold for every $n \ge 0$. But the latter can hold for at most one index $n$ because $\alpha$ is not a rational multiple of $\pi$. This shows that $f$ has necessarily the form $$ f(z) = c_m z^m $$ for some non-negative integer $m$ and some $c_m \in \Bbb C$.

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A similar reasoning works for the second question (“Condition B”). If $f$ is holomorphic in some annulus $r_1 < |z| < r_2$ and its argument is constant on two different circles (say $|z| = 1$ and $|z|=R \ne 1$) then reflection at those circles gives $$ f(R^2 z) = e^{2i\beta} f(z) $$ for some $\beta \in \Bbb R$. Substituting this in the Laurent series $fz) = \sum_{n=-\infty}^\infty c_n z^n$ then shows that $f$ is necessarily constant.