# Proving that holomorphic functions which map the real axis to the imaginary axis also map the imaginary axis to the real axis.

The actual question:

$$f(z)$$ is holomorphic in some neighbourhood of $$0$$, $$f(0) = 0$$, $$f'(0) =/= 0$$. Suppose $$f(z)$$ maps the real axis to the imaginary axis and suppose it maps the imaginary axis to a line $$l$$. Prove that $$l$$ is the real axis using the Cauchy Riemann equations.

Attempt:

It suffices to show that $$f$$ maps imaginary values to real values in a neighbourhood of $$0$$ since we know that $$l$$ is a line. To do this, we can show that $$v_y(z) = 0$$ for all imaginary values in a nbhd of $$0$$ since we know that $$v(0) = 0$$ and $$v_y(z) = 0$$ for imaginary values implies that $$v = 0$$ everywhere on the imaginary axis.

Write $$f = u + iv$$. Then $$f(x)$$ is imaginary for real $$x$$, so $$u(t) = 0$$ for all real $$t$$. This implies $$u_x(t) = 0$$ as $$u$$ is constant with respect to moving along the real axis (always $$0$$). By Cauchy Riemann, in a neighbourhood of $$0$$, we have $$u_x = v_y$$, so $$v_y(t) = 0$$ too. How do I now get $$v_y(z)$$?

• An alternative approach would be to show that holomorphic maps are conformal (i.e preserve angles of tangent vectors) Then integration should do the trick. – user587399 Aug 28 '19 at 20:21
• Are you supposed to use only the Cauchy-Riemann equations and no other theorems about holomorphic functions? – Eric Wofsey Aug 28 '19 at 20:22
• Also, I'm guessing you have a typo and the assumption should be $f'(0)\color{red}\neq0$. – Eric Wofsey Aug 28 '19 at 20:25
• I would be pessimistic for this result. If $f$ is holomorphic only on a vicinity of $0$ (say a disk D(0,R)) it can be extended into a non holomorphic function that it maps the real axis onto the imaginary axis without being holomorphic outside D. – Jean Marie Aug 28 '19 at 20:26
• @EricWofsey Surely---otherwise $z \mapsto z^2$, which satisfies $f'(0) = 0$ and maps both $\Bbb R$ and $i \Bbb R$ to $\Bbb R$, would be a counterexample. – Travis Willse Aug 28 '19 at 21:04

The Cauchy-Riemann equation gives that (locally invertible) holomorphic maps are conformal maps. If the real axis is mapped into the imaginary axis and the imaginary axis is mapped into a line $$\ell$$, the angle between the real and imaginary axis ($$90^\circ$$) has to be the same as the angle between the imaginary axis and $$\ell$$.

I am assuming $$f(0)=0$$ and $$f'(0)\color{red}{\neq} 0$$. Otherwise $$f(z)=iz^2$$ provides a counter-example.

Alternative way: let $$\sum_{n\geq 1}i^{-n}a_n z^n$$ be the Maclaurin series of $$f(z)$$, convergent over some neighbourhood $$D$$ of the origin. Since $$f$$ maps $$i\mathbb{R}\cap D$$ into $$\mathbb{R}$$, all the coefficients $$a_n$$ are real (and $$a_1\neq 0$$). There is some $$\omega\in S^1$$ (the unit vector giving the direction of $$\ell$$) such that $$\mathbb{R}\cap D$$ is mapped into $$\omega\mathbb{R}$$, so $$g(z) = \omega^{-1} f(z)$$ maps $$\mathbb{R}\cap D$$ into $$\mathbb{R}$$. By considering the Maclaurin series of $$g(z)$$ it follows that $$\omega^{-1}i^{-n} a_n$$ is real for any $$n\geq 1$$, so $$\omega=\pm i$$ and $$a_{2m}=0$$ for any $$m\geq 1$$.

I have exploited the following lemma: if $$h(z)=\sum_{n\geq 0}a_n z^n$$ is holomorphic in a neighbourhood $$U$$ of the origin and such that $$h$$ maps $$\mathbb{R}\cap U$$ to $$\mathbb{R}$$, then $$a_n\in\mathbb{R}$$ for any $$n\geq 0$$. Proof: $$a_n$$ is $$\frac{1}{n!}$$ times the $$n$$-th derivative of $$h$$ at the origin, which can be computed as the real limit $$\lim_{t\to 0^+}\frac{1}{t^n}\sum_{k=0}^{n}\binom{n}{k}(-1)^{n-k}\underbrace{h(kt)}_{\in\mathbb{R}}.$$ In simpler terms, the $$n$$-th complex derivative of $$h$$ at the origin has to agree with the $$n$$-th derivative of $$h$$, intended as a real function defined on $$\mathbb{R}\cap U$$.

• (+1) I should have thought of that… – José Carlos Santos Aug 28 '19 at 21:05
• Is there a way to do it without conformal maps? We haven't covered them yet (i.e we don't know that holomorphic functions are conformal maps). – John Aug 28 '19 at 23:22
• How do you get the existence of $w$? – John Aug 28 '19 at 23:46
• @Saad: from the hypothesis that the imaginary axis is mapped into a line. $\omega$ is just the unit vector giving the direction of $\ell$. – Jack D'Aurizio Aug 28 '19 at 23:57
• Thanks! Lastly, why does $a_{2m} = 0$ imply the result? And why do you need $f(0) = 0$ and $f'(0) =/= 0$? – John Aug 28 '19 at 23:59