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Show that if a holomorphic function has a constant absolute value, it must be a constant.

Suppose $f(z)=u(z)+iv(z)$ is holomorphic (where $z=x+iy$ is complex), and that $(u(z))^2+(v(z))^2=C$ for some constant $C$.

We have $\dfrac{\partial((u(z))^2+(v(z))^2)}{\partial x}= 0$. Applying the chain rule, we find that $$u(z)\cdot\dfrac{\partial u(z)}{\partial x} + v(z)\cdot\dfrac{\partial v(z)}{\partial x} = 0$$

Similarly, $$u(z)\cdot\dfrac{\partial u(z)}{\partial y} + v(z)\cdot\dfrac{\partial v(z)}{\partial y} = 0$$

Using the Cauchy-Riemann differential equations satisfied by any holomorphic function: $$u(z)\cdot\left(-\dfrac{\partial v(z)}{\partial x}\right) + v(z)\cdot\dfrac{\partial u(z)}{\partial x} = 0$$

We can write as a matrix form:

$$\begin{pmatrix} u(z) & v(z) \\ v(z) & -u(z)\end{pmatrix}\begin{pmatrix}\frac{\partial{u(z)}}{\partial{x}}\\\frac{\partial{v(z)}}{\partial{x}}\end{pmatrix} = 0$$

For values of $z$ such that the matrix $A=\begin{pmatrix} u(z) & v(z) \\ v(z) & -u(z)\end{pmatrix}$ is invertible, we get $\dfrac{\partial{u(z)}}{\partial{x}}=\dfrac{\partial{v(z)}}{\partial{x}}=0$. But what can we do for values of $z$ such that the matrix $A$ is not invertible?

PJ Miller
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    Look back at your hypothesis; $A$ has constant determinant; also $C$ is nonzero. – Marra Aug 24 '13 at 01:13
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    @GustavoMarra So, since the matrix $A$ has constant determinant $C\neq 0$, it is invertible for all values of $z$. Taking the inverse, we find $\dfrac{\partial u(z)}{\partial x} = \dfrac{\partial v(z)}{\partial x} = 0$. Similarly, $\dfrac{\partial u(z)}{\partial y} = \dfrac{\partial v(z)}{\partial y} = 0$. That means $u,v$ are constant, implying $f$ constant. – PJ Miller Aug 24 '13 at 01:43

3 Answers3

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If $C = 0$, then $u$ and $v$ are necessarily constant $0$, so we assume that $C \neq 0$. The determinant of $A$ is

$$u(z)(-u(z)) - v(z) v(z) = - (u(z)^2 + v(z)^2) = - C \neq 0$$

for any $z$. So you've got a an invertible matrix and a vector in its null space; so what must that vector be?

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The result follows directly from Liouville's theorem which says that an analytic function which is bounded in the complex plane is a constant. You can look up a proof for Liouville's theorem almost anywhere.

Betty Mock
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    This theorem holds even for non-entire functions. This proof does not work in that case. – Potato Aug 24 '13 at 02:37
  • Can you clarify this? We were told that the function is holomorphic and has constant absolute value. No smaller domain than the complex plane was indicated. So why is it not entire? – Betty Mock Aug 25 '13 at 15:32
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    I agree the problem is somewhat ambiguous. In fact, the problem is wrong as stated (consider two disjoint open domains, where $f(z)$ is $i$ on the first and $-i$ on the second). But in general, "holomorphic" means "holomorphic on some open connected domain $\Omega\subset \mathbb C$." – Potato Aug 26 '13 at 06:08
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This is a well-known theorem in Complex Analysis:

Liouville's Theorem: If a holomorphic function of one variable is bounded and entire, the function is necessarily constant.

Since your function has constant absolute value, it is bounded. By Liouville's Theorem it is therefore constant.

Daron
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