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Here's a question from a previous complex analysis qualifying exam that I'm honestly just stumped on:


Let $u$ be a harmonic function on the unit disc $D = \{z: |z|<1\}$, which is the real part of the holomorphic function $f$ on $D$. Show that for any $0<r<1$ we have

$$ f(z) = \frac{1}{\pi i} \oint_{|\zeta|=r} \frac{u(\zeta)}{\zeta - z} d\zeta - \overline{f(0)}, \quad \quad \text{for } |z|<r. $$


I know that $u$ harmonic means $u_{xx}+u_{yy} = 0$. The form of $f(z)$ given looks reminiscent of Cauchy's Integral Formula and we could say that since $f$ is holomorphic, we can write

$$f(z) = \frac{1}{2\pi i} \oint_{|\zeta|=r} \frac{f(\zeta)}{\zeta - z} d\zeta = \frac{1}{2\pi i} \oint_{|\zeta|=r} \left(\frac{u(\zeta)}{\zeta - z} + i \frac{v(\zeta)}{\zeta - z}\right) d\zeta$$

but where do I go from here?


UPDATE: Building on a suggestion from a comment, we also have $$\overline{f(z)} = \frac{1}{2\pi i} \oint_{|\zeta|=r} \frac{\overline{f(\zeta)}}{\zeta - z} d\zeta = \frac{1}{2\pi i} \oint_{|\zeta|=r} \left(\frac{u(\zeta)}{\zeta - z} - i \frac{v(\zeta)}{\zeta - z}\right) d\zeta$$

and thus $$\overline{f(0)} = \frac{1}{2\pi i} \oint_{|\zeta|=r} \frac{\overline{f(\zeta)}}{\zeta } d\zeta = \frac{1}{2\pi i} \oint_{|\zeta|=r} \left(\frac{u(\zeta)}{\zeta} - i \frac{v(\zeta)}{\zeta}\right) d\zeta.$$

I can add these two and with a bit of manipulation get $$f(z) + \overline{f(0)} = \frac{1}{\pi i} \oint_{|\zeta|=r} \frac{u(\zeta)}{\zeta - z} d\zeta - \frac{1}{2\pi i} \oint_{|\zeta|=r} \frac{z\overline{f(\zeta)}}{\zeta(\zeta - z)} d\zeta.$$

It remains to show that the last term goes to zero... and I am once again stuck.


UPDATE 2: I added a bounty to this question in hopes of getting a full, complete, and clear worked answer to this problem. I need to make sure I can do this kind of problem correctly before my own upcoming qual.

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  • $\begingroup$ Presumably you need to use the fact that $v$ is the (a) harmonic conjugate of $u$. Perhaps you should follow through a proof of the Cauchy Integral Formula. $\endgroup$ Jul 31, 2023 at 21:07
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    $\begingroup$ Try and see what happens when you subtract the two integrals (with $u$ and $v$) respectively; this would correspond to integrating $\overline {f(\zeta)}/(\zeta-z)$ and after some manipulation (eg power series of denominator and parametrization) you can see what happens and how that implies your result by adding $\endgroup$
    – Conrad
    Jul 31, 2023 at 23:57
  • $\begingroup$ @Conrad This gets me very close. I am left with a term that would need to go to zero: $-\frac{1}{2\pi i} \oint_{|\zeta| = r} \frac{zf(\zeta)}{\zeta(\zeta - z)} d\zeta$. What am I not seeing that sends this to zero? $\endgroup$
    – Serafina
    Aug 1, 2023 at 2:13
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    $\begingroup$ math.stackexchange.com/questions/4088314/… $\endgroup$
    – Gary
    Aug 1, 2023 at 3:02
  • $\begingroup$ $u(z)$ and $v(z)$ are shorthands for $u(x,y)$ and $v(x,y)$ ($z=x+\mathrm{i}y$). $\endgroup$
    – Gary
    Aug 3, 2023 at 23:15

1 Answer 1

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Let $A$ and $B$ be the following integrals: $$A=\frac{1}{\pi i} \oint_{|\zeta|=r} \frac{u(\zeta)}{\zeta - z} d\zeta$$ $$B=\frac{1}{\pi i} \oint_{|\zeta|=r} \frac{v(\zeta)}{\zeta - z} d\zeta$$

Proving this theorem amounts to showing that $\frac{1}{2}(A+iB)=f(z)$ and $\frac{1}{2}(A-iB)=\overline{f(0)}$. The first equation is clearly just the Cauchy integral formula. So, we will focus on the second one.

$$\frac{1}{2}(A-iB)=\frac{1}{2\pi i} \oint_{|\zeta|=r} \frac{u(\zeta)-iv(\zeta)}{\zeta - z} d\zeta$$

We take the complex conjugate of both sides.

$$\overline{\frac{1}{2}(A-iB)}=-\frac{1}{2\pi i} \oint_{|\zeta|=r} \frac{u(\zeta)+iv(\zeta)}{\overline\zeta - \overline z} d\left(\overline \zeta \right)$$ To be clear, if $d\zeta$ means integrating over a contour with parametrization $\zeta(t)$, then $d\left(\overline \zeta \right)$ means integrating over the contour with parametrization $\overline{\zeta(t)}$. Now, we substitute back in $f$ and use the identity $\overline \zeta = \frac{r^2}{\zeta} $ on a circle of radius $r$. $$\overline{\frac{1}{2}(A-iB)}=-\frac{1}{2\pi i} \oint_{|\zeta|=r} \frac{f(\zeta)}{\frac{r^2}{\zeta} - \overline z} d\left(\frac{r^2}{\zeta} \right)$$ We can now compute the differential as $d\left(\frac{r^2}{\zeta} \right) = -\frac{r^2}{\zeta^2}d\zeta$. $$\overline{\frac{1}{2}(A-iB)}=\frac{1}{2\pi i} \oint_{|\zeta|=r} \frac{f(\zeta)}{\zeta(1 - \frac{\zeta\overline z}{r^2})} d\zeta$$ This is now a closed contour integral of a holomorphic function. By noting that $|\zeta|=r$ and $|z|<r$, we have $\left|\frac{\zeta\overline z}{r^2}\right|<1 $ and we see that the only residue within the contour is at $z=0$. Therefore, the residue theorem gives the following. $$\overline{\frac{1}{2}(A-iB)}= \frac{f(0)}{1 - \frac{0 \cdot\overline z}{r^2}}=f(0)$$ Q.E.D.

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