Holomorphic function with constant absolute value 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:

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

For values of $z$ such that the matrix $A=\begin{pmatrix} u(z) &amp; v(z) \\ v(z) &amp; -u(z)\end{pmatrix}$ 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?
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?
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.