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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.

Answer:

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?

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Looks right to me... –  Aryabhata Feb 14 '12 at 20:03

1 Answer 1

up vote 6 down vote accepted

Yes.


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.

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