Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $u: \mathbb{R}^2 \to \mathbb{R}$ be a differentiable function. Prove that if the complex function

$f(x + iy) = u(x,y) + iu(x,y)$

is analytic in $\mathbb{C}$ then it is a constant function.


If $f$ is a analytic it satisfies the Cauchy Riemann equations. So $u_x = u_y$ and $u_x=-u_y$

This can only happen when $u_x$ and $u_y$ are equal $0$.

As the partial derivatives of $f$ are $0$, $f$ must be a constant function.

Is that correct?

share|cite|improve this question
Looks right to me... – Aryabhata Feb 14 '12 at 20:03
up vote 6 down vote accepted


Another way to think of it: $f(x+iy)=(1+i)u(x,y)$ has range contained in the line $\{t(1+i):t\in\mathbb R\}$, so $f$ is constant by the open mapping theorem, or by Liouville's theorem applied to $\frac{1}{f-1}$.

Or $g(x+iy)=(1+i)^3f(x+iy)=-4u(x,y)$ (or simply $\frac{1}{1+i}f = u$) is real valued, which makes applying the Cauchy-Riemann equations to $g$ a little more immediately show that $g$ is constant.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.