3
$\begingroup$

I'm reading the beginning of chapter 3 of Ahlfors where he starts to introduce analytic functions in regions, and I had a question on a part of Theorem 11:

Let $f=u+iv$ be an analytic function in a region $\Omega$. If the modulus of $f$ is constant, then $f$ must reduce to a constant.

His proof starts out as follows:

If $|f|=u^2+v^2$ is constant, then $ u\frac{\partial u}{\partial x} + v \frac{\partial v}{\partial x} = 0$, and $ u \frac{\partial u}{\partial y} + v \frac{\partial v}{\partial y} = -u \frac{\partial v}{\partial x} + v \frac{\partial u}{\partial x} = 0$.

My question is why does this imply that either $ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial x} = 0$ or that the determinant $u^2+v^2$ vanishes?

$\endgroup$

1 Answer 1

12
$\begingroup$

notice that you now have a solution $$\left(\begin{array}{cc} u & v\\ v & -u\end{array}\right)\left(\begin{array}{c} \frac{\partial u}{\partial x}\\ \frac{\partial v}{\partial x}\end{array}\right)=0$$ If the modulus (which is the determinant) is not zero then the matrix is invertible, so the solution to the system above must be zero.

$\endgroup$
1
  • $\begingroup$ Ah, cool. Thanks a lot! $\endgroup$
    – user1736
    Feb 5, 2011 at 18:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .