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

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?

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 $$r \mapsto e^{i (b \log r + \alpha_0)}$$.

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 $$c_m \in \Bbb C$$.

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.

• The weakening of the hypotheses is quite nice! Mar 18 at 16:52
• @TravisWillse: Yes, the reflection principle can be a powerful tool. It works for “Condition B” as well :) Mar 18 at 16:57