Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let G be a map in $\mathbb C$ ( $G \subseteq \mathbb C $) and $u: G\mapsto \Bbb C$ is harmonic function. Then show that $f: G\mapsto \Bbb C$

$$ f(x+iy)=\frac{du}{dx} - i\frac{du}{dy}$$

is holomorfic.

It was asking during my last exam. I answered like this : ƒ(x + i y) = u(x, y) + i v(x, y)

If u and v have continuous first partial derivatives and satisfy the CR equations, then it is holomorphic.
We have u(x,y)=$\frac{du}{dx}$ and v(x,y)=$- \frac{du}{dy}$ since $u$ is harmonic, u and v has continuous first partial derivatives. (1-is it true?) Because a harmonic function is a twice continuously differentiable function.

Now we have to proof it satisfies CR equations to show it is holomorphic.

firstly $\frac{d^2u}{dx^2}=-\frac{d^2u}{dy^2}$. Because $u$ is harmonic, so it satisfies laplace equation which is $\frac{d^2u}{dx^2}+\frac{d^2u}{dy^2}=0$. secondly $\frac{d^2u}{{dy}{dx}}=-(-\frac{d^2u}{{dx}{dy}})$. Because second order partial derivatives of u is continuous (2-is it true?)
so it satisfies two condition of CR equations hence it is holomorphic.

Was my answer right? if itis not can you tell me which phase was wrong.

share|improve this question
Your proof is good. I would also suggest editing the title of the question so it would be more informative. Something like "Show that the partial derivatives of a harmonic function fit into a hololomorphic function" or "Show that if $u$ is harmonic then $\frac{\partial u}{\partial z}$ is holomorphic". –  levap Dec 1 '12 at 4:31
okay done it. thanks.. –  lyme Dec 3 '12 at 23:07
You probably shouldn't use $u$ and $v$ to mean two different things. $u = \frac{\partial u}{\partial x}$ is not really what you mean. –  mrf Dec 3 '12 at 23:15
$u$ and u both are different. was that wrong? –  lyme Dec 4 '12 at 0:16
$u$ was a given function. Then you put $f = u + iv$, where ''$u$'' in fact is $\frac{\partial u}{\partial x}$, so you use $u$ to denote two different things: the original given function and the real part of $f$. (Simlarly for $v$.) –  mrf Dec 4 '12 at 7:51

Your Answer


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

Browse other questions tagged or ask your own question.