The exercise:

Let $h$ be a function harmonic on $\{z\in\mathbb{C}: \rho_1 < |z| < \rho_2\}$. Using the fact that $h_x - ih_y$ is holomorphic, prove that there exist unique constants $(a_n)_{n\in\mathbb{Z}}$ and $b$ with $a_0, b\in\mathbb{R}$, such that \begin{align*} h(z) = Re\left(\sum_{-\infty}^\infty a_n z^n\right) + b\log|z|\qquad (\rho_1 < |z| < \rho_2) \end{align*}

My approach: we know that $g(z) = h_x - ih_y$ is holomorphic in the annulus, but we can't simply integrate it between $z_0$ and $z$ to find $f$ such that $\operatorname{Re} f = h$ because the domain is not simply connected. So, I proceed as follows: decompose $g(z)$ into Laurent series in the annulus, choose any $z_0$ from the annulus and integrate each member of the series: $$\int_{z_0}^z c_n z^n dz.$$ Since for every member except $-1$st, the integral is path-independent and an antiderivative exists, no problem here. As for the member $\frac{c_{-1}}{z}$, the integral from $z_0$ to $z$ depends on the path we choose. Here, I am having a problem with $c_{-1}$. If there was no complex constant, I would just say that the antiderivative is the complex logarithm and its imaginary part $i(\operatorname{Arg} z + 2k\pi)$ changes depending on the path, but the real part $\ln|z|$ is the same. Since we only need the real part, our expression is well-defined and we're done.

But since we have $c_{-1}(\ln|z| + i(\operatorname{Arg} z + 2k\pi)$, the real part would include $2k\pi$ and my argument doesn't work. Is there a reason for $c_{-1}$ to be real? Another point: in my case we extract the $\ln|z|$ and we have no element that becomes $a_0$, so in our case $a_0 = 0$. I guess, I could have $a_0 = \operatorname{Im}(c_{-1})\cdot \operatorname{Arg} z$, but I would have to define the $\operatorname{Log} z$ in an unambiguous manner, and I can't see clearly how...


You have

\begin{align} c_{-1} &= \frac{1}{2\pi i} \int_{\lvert z\rvert = \rho} g(z)\,dz\\ &= \frac{1}{2\pi i}\int_{\lvert z\rvert = \rho} (h_x(z) - i h_y(z))\, (dx + idy)\\ &= \frac{1}{2\pi i}\int_{\lvert z\rvert = \rho} (h_x(z)\,dx + h_y(z)\,dy) + \frac{1}{2\pi} \int_{\lvert z\rvert = \rho} (h_x(z)\,dy - h_y(z)\,dx) \end{align}

for every $\rho \in (\rho_1,\rho_2)$.

If you can see a reason why

$$\int_{\lvert z\rvert = \rho} (h_x(z)\,dx + h_y(z)\,dy) = 0,$$

you can see that $c_{-1}\in\mathbb{R}$.

  • $\begingroup$ Thank you Daniel! If you don't mind, I have two follow-up questions. First, I am having trouble understanding the last integral. If $|z| = \rho$, then $x$ goes from $-\rho$ to $\rho$ and back, so we can write for the first part of the integral: $\int_{\rho}^{-\rho} h_x(z) dx + \int_{-\rho}^{\rho} h_x(z) dx$ which gives $h(-\rho + iy) - h(\rho + iy) + h(\rho + iy) - h(-\rho + iy) = 0$... anyway, I'm quite sure I'm getting it wrong... $\endgroup$ – Igor Deruga May 7 '15 at 11:06
  • $\begingroup$ The second thing is about the $\operatorname{Log} z$ for the integral of the form $I = \int_{z_0}^z\frac{c_{-1}}{z} dz$. If we limit it to the paths in the domain $\{z = re^{i\phi}: \rho_1 < r < \rho2, \phi\in [0, 2\pi)\}$, we could say that our function has an antiderivative and we have $I = c_{-1}\operatorname{Log} z - c_{-1}\operatorname{Log} z_0$. Now, since it is well-defined, taking the real part to form the elements $a_0$ and $b$ is quite easy and in this case $a_0$ is not necessarily equal to zero. Do you think this approach could work too? $\endgroup$ – Igor Deruga May 7 '15 at 11:12
  • $\begingroup$ I think I can see the flaw in my approach: if I integrate the Laurent series, I get $f(z) - f(z_0)$. Since $f(z)$ part doesn't have the member $a_0$ within the sum (it becomes Log during the integration), then we have: $a_0 = f(z_0)$, but I am not sure that we can choose $z_0$ so that $f(z_0)\in\mathbb{R}$ as required... $\endgroup$ – Igor Deruga May 7 '15 at 11:23
  • $\begingroup$ $\int_{\lvert z\rvert = \rho}$ means we integrate over the circle with radius $\rho$ (counterclockwise), you parametrise $z = \rho e^{it},\; t\in [0,2\pi]$. Then $x(t) = \rho\cos t$, and $dx$ becomes $-\rho\sin t\,dt$, similar for $dy$. The integrand then becomes $\frac{d}{dt}\bigl(h(\rho e^{it})\bigr)$. If you're familiar with differential forms, you note that $h_x\,dx + h_y\,dy = dh$, and one integrates over a closed path. It doesn't quite work as you wrote it, since when moving from $-\rho$ to $\rho$ we move along a semicircle in the lower half-plane, $\endgroup$ – Daniel Fischer May 7 '15 at 11:34
  • $\begingroup$ and when going back, we move along a semicircle in the upper half-plane, so the paths don't cancel, and the integrals of only the part $h_x\,dx$ generally don't cancel. You can only apply the fundamental theorem of calculus to the partial derivative $h_x$ if you integrate over a line segment parallel to the real axis. Along other curves, you also need $h_y$ for the FTC. $\endgroup$ – Daniel Fischer May 7 '15 at 11:34

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