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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?

Aryabhata
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Jim_CS
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1 Answers1

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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.

Jonas Meyer
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  • Just a follow-up question. If a function's derivative $f'(z)$ is nule in a subset U of C, then is that enough to prove it is constant in U, like in real analysis? (even if it's not analytical) – Rye Jun 29 '20 at 09:39
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    @Rye: If it is a connected open set, yes. If the set is open, existence everywhere of the derivative implies it is analytical (very unlike in real analysis). If the set is connected but not open, I don't know. – Jonas Meyer Aug 04 '20 at 21:56