# Why: A holomorphic function with constant magnitude must be constant.

How can I prove the following assertion?

Let $f$ be a holomorphic function such that $|f|$ is a constant. Then $f$ is constant.

Edit: The more elementary the proof, the better. I'm working my way through a complex analysis workbook, and by this exercise, the only substantial thing covered has been the Cauchy-Riemann equations.

• What is the domain of $f$? If it's all of $\mathbb C$, then you can use Liouville's theorem. – lhf May 21 '12 at 18:57
• The domain is allowed to be any open set. – Mark May 23 '12 at 0:03
• I suspect that the domain $\Omega$ must be a connected open set for this result to bind. Otherwise, isn't it possible for $\vert f(z) \vert$ to to be constant, with $f'(z) = 0$, but $f(z)$ to take different values, each with the same modulus, on different components of $\Omega$? Note that some of the proofs given rely (perhaps tacitly) on integrating along a path . . . – Robert Lewis Sep 21 '15 at 1:26
• @RobertLewis I had a comment saying you were wrong but now I think you are right! Why not have two disconnected disks and the first disk maps to 1 and the second disk maps to i! – Mark Sep 21 '15 at 5:29
• Exactly my point! Thanks, and Cheers! – Robert Lewis Sep 21 '15 at 5:37

Let $f(z) = u(x,y) + iv(x,y)$, where $u(x,y),v(x,y) \in \mathbb{R}$. We then have that $$\lvert f \rvert^2 = u(x,y)^2 + v(x,y)^2 = \text{constant}$$ Hence, $$\frac{\partial (u^2 + v^2) }{\partial x} = 0 = \frac{\partial (u^2 + v^2) }{\partial y}$$ This implies $$u \frac{\partial u}{\partial x} + v \frac{\partial v}{\partial x} = 0 = u \frac{\partial u}{\partial y} + v \frac{\partial v}{\partial y}$$ Making use of the Cauchy-Riemann equations, we get $$u \frac{\partial v}{\partial y} + v \frac{\partial v}{\partial x} = 0 = -u \frac{\partial v}{\partial x} + v \frac{\partial v}{\partial y}$$ Multiply the left equation by $u$ and the right equation by $v$ and add them up to get $$(u^2 + v^2) \frac{\partial v}{\partial y} = 0$$ If $u^2+v^2 = 0$, then $f=0$ throughout the domain since $\lvert f \rvert = \text{constant}$. If $u^2+v^2 \neq 0$, then $\dfrac{\partial v}{\partial y}=0$ which also gives us $\dfrac{\partial v}{\partial x}=0$. Now by Cauchy-Riemann, we also get that $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial y}=0$. Hence, $f'(z) = 0$ which gives us that $f$ is constant.

Also if $f(z)$ happens to be entire, then the conclusion immediately follows from Liouville's theorem.

• This was the approach I was attempting. – Mark May 21 '12 at 19:21
• why $v_y = 0$ give us $v_x = 0$ ? this is not clear. – Richard Clare Jul 6 '17 at 14:24
• @RichardClare Geometrically rotate the domain by 90 degrees. $g(z) = g(x,y) = f( i z) = u(-y, x) + i v(-y, x)$ Clearly $g$ also satisfies the hypothesis since its magnitude is the same as $f$, so re-use the proof to show that the derivative of the imaginary component of $g$ as we travel along the $y$ axis is zero $D_2(\Im(g)) = 0$. But this is the same as $-D_1(\Im(f)) = v_x$ – Mark Jun 23 '18 at 4:31

If $$|f|$$ is constant, then the image of $$f$$ is a subset of a circle in $$\mathbb{C}$$. Apply the Open Mapping Theorem.

• In other words, any subset of the fixed circle cannot contain an open set, and therefore f is not an open map. – Mark May 21 '12 at 19:07
• @Mark: yes, exactly. – Pete L. Clark May 21 '12 at 20:13
• Hi Professor Clark, why is a subset of a circle a closed (and bounded) set? Wouldn't an arc be analogous to an open interval on the real line? I am trying to apply the open mapping theorem but am struggling with this line of reasoning. Thanks, @PeteL.Clark – User001 Nov 20 '15 at 0:46
• @Lebron: A subset of a circle is not necessarily closed. However, it has empty interior, which is what is being used. – Pete L. Clark Nov 20 '15 at 1:12
• @User001 You should be careful with the invocation of area, since that uses heavier tools (namely Lebesgue measure in R^2) and is not necessary here. Roughly speaking, you can see that it has an empty interior since the neighborhoods in R^2 are "two dimensional" while a circle is an arc which is "one dimensional." Thus, taking a neighborhood around any point of the circle, we can never have that neighborhood be a subset of the circle. – MCT Mar 6 '16 at 23:04

Pete's answer is quite sufficient, but I'll add another one which applies to more general maps (namely, complex-valued harmonic). Write $f=u+iv$, where $u$ and $v$ are real harmonic functions. As an exercise with the chain rule, verify the identity $$\Delta(u^2+v^2)=2|\nabla u|^2+2|\nabla v|^2$$ Conclude that if $u^2+v^2$ is constant, then so are $u$ and $v$.

• Triangle is laplacian while upside-down triangle is divergence, correct? – Mark May 21 '12 at 19:12
• @Mark Yes, $\Delta$ is the Laplacian. The meaning of $\nabla$ depends on context: in Vector Calculus $\nabla$ can be gradient, divergence, or curl, depending on what you apply it to. Here it is applied to a real-valued function, hence it is the gradient $\nabla u=\langle u_x,u_y\rangle$ . – user31373 May 21 '12 at 19:24

Another way: if $f(z_0) = u_0$ and $f'(z_0) = p \ne 0$, consider the directional derivative of $|f(z)|^2 = f(z) \overline{f(z)}$ at $z_0$ in the direction of some complex number $q$, i.e. \eqalign{\frac{d}{dt} \left(f(z_0 + tq) \overline{f(z_0 + tq)}\right) &= f(z_0 + tq) \overline{\frac{d}{dt} f(z_0 + t q)} + \overline{f(z_0 + tq)} \frac{d}{dt} f(z_0 + t q) \cr &= f(z_0 + t q) \overline{ q f'(z_0 + t q)} + \overline{f(z_0 + tq)} q f'(z_0 + t q)\cr} At $t=0$ this would be $u_0 \overline{q p} + \overline{u_0} q p = 2 \text{Re}(u_0 \overline{qp})$. If $u_0 \ne 0$ and $p \ne 0$, choose $q = u_0/p$ and this is $2 |u_0|^2 > 0$. But that's impossible if $|f(z_0)|^2$ is to be constant. So either $u_0 = 0$ (and then $|f(z)| = 0$ so $f(z) = 0$ everywhere) or $f'(z) = 0$ everywhere, and that implies $f$ is constant.

This follows from the Maximum Modulus Theorem. However, the proof that comes to mind of the Maximum Modulus Theorem using the Cauchy Integral Formula actually uses this fact, so that might be circular.

To prove this special case of the Maximum Modulus Theorem, Pete L. Clark has already mentioned that the Open Mapping Theorem applies.

An alternative is to use the Cauchy-Riemann equations. If $f(z)=f(x+iy)=u(x,y)+iv(x,y)$, by hypothesis there exists a constant $r\geq 0$ such that $u^2+v^2 = r^2$. It follows that $uu_x+vv_x=0$ and $uu_y+vv_y=0$. You also have $u_x=v_y$ and $u_y=-v_x$. A little algebraic manipulation yields $u(u_x^2+v_x^2)=v(u_x^2+v_x^2)=0$. Assuming $r\neq 0$, $u$ and $v$ are never simultaneously $0$, so it follows that $u_x^2+v_x^2\equiv 0$, which implies that $u_x=v_y=0$ and $v_x=-u_y=0$. Therefore $f$ is constant.

Another way to use the Cauchy-Riemann equations is to first map the circle $|z|=r$ into the real axis, either using a Möbius tranformation, or using logarithms. E.g., consider $g\circ f$, where $g(z)=i\dfrac{1+\frac{z}{r}}{1-\frac{z}{r}}$. (Such maps won't be defined on the entire circle, but it is enough to show that $f$ is locally constant.)