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.

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

Can someone tell me if the following proposition is true ?

Theorem If $u=g + i h$ is a holomorphic function in $\Omega\subseteq \mathbb{C}$ and $\Omega$ is simply connected, then $v(z)=u(w)+ \int_\gamma \,g_x(z)-ih_y (z) \,dz$ is a primitive function of $u$ (where $w\in \Omega$ is fixed and $\gamma$ is some path from $w$ to $z$).

(I have come across (the implicit use of) this proposition by reading about something not really related to complex analysis and since I know very little about it, if wondered if it actually would be true taken out of context like this. I also wouldn't mind a proof, if it is true and someone would have the time.)

share|cite|improve this question
up vote 3 down vote accepted

I shall assume that $g$ and $h$ are realvalued. Since $u:=g+ih$ is holomorphic it follows from the the CR equations that $h_y=g_x$. Therefore for any curve $\gamma\subset\Omega$ connecting the point $z_0$ with a variable point $z$ one has $$\int_\gamma (g_x- i h_y)\ dz=(1-i)\int_\gamma g_x\ (dx+i dy) =(1-i)\int_\gamma(g_x\ dx + i h_y dy)=(1-i)\Bigl(g(z)-g(z_0)+i\bigl(h(z)-h(z_0)\bigr)\Bigr)=(1-i)\bigl(u(z)-u(z_0)\bigr)\ .$$ It follows that $$v(z):=u(z_0)+\int_{z_0}^z (g_x- i h_y)\ dz=i u(z_0)+(1-i) u(z)\ ,$$ which shows that your $v$ is more or less the given $u$ again, and not a primitive of $u$.

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.