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I'm having trouble with this one: Let $u$ be a real-valued harmonic function on $D(0,1)$, and let $\gamma$ be a closed curve in that disk. Then $\int_\gamma u=0.$

I'm supposed to prove or disprove this statement. I'm inclined to believe it's true. What I have so far is: Since $u$ is real-valued and harmonic, there exists a harmonic function $v$ such that $f=u+iv$ is holomorphic on $D(0,1)$. Now since $\gamma$ is a closed curve in $D(0,1)$ it follows that $\int_\gamma f=\int_\gamma(u+iv)=0$.

I don't know what to do from here, I don't think I can conclude that $\int_\gamma u=0$ just from this last part.

mi986
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1 Answers1

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After working on it I believe this is a counterexample. Let $f(z)=z=x+iy=Re(f)+iIm(f)$. Then $f(z)$ is entire. Furthermore for $u=x$ we have $u_{xx}+u_{yy}=0$. Thus $u$ is harmonic. In particular it is harmonic on $D(0,1)$.

Take $\gamma=e^{i\theta}$ for $0\leq\theta\leq 2\pi.$ Then

\begin{align*}\int_{\gamma}u=\int_\gamma Re(f) &=\int_0^{2\pi}Re(e^{i\theta})\cdot ie^{i\theta}d\theta\\ &=\int_{0}^{2\pi}\cos\theta(i\cos\theta-\sin\theta)\\ &=\int_{0}^{2\pi}(i\cos^2\theta-\cos\theta\sin\theta)\\ &=\frac{1}{2}\int_0^{2\pi}\left(i+i\cos2\theta-\sin2\theta\right)\\ &=\frac{1}{2}\left(i\theta+\frac{i}{2}\sin2\theta+\frac{1}{2}\cos2\theta\right)\bigg|_{0}^{2\pi}\\ &=\frac{1}{2}\left(2\pi i\right)\\ &=\pi i\neq0. \end{align*}

mi986
  • 610