# For $u$ harmonic and $f = u+iv$ holomorphic, show that $f(z) = \frac{1}{\pi i} \oint_{|\zeta|=r} \frac{u(\zeta)}{\zeta - z} d\zeta - \overline{f(0)}$

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 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|

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.

• 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. Commented Jul 31, 2023 at 21:07
• 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 Commented Jul 31, 2023 at 23:57
• @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? Commented Aug 1, 2023 at 2:13
• math.stackexchange.com/questions/4088314/…
– Gary
Commented Aug 1, 2023 at 3:02
• $u(z)$ and $v(z)$ are shorthands for $u(x,y)$ and $v(x,y)$ ($z=x+\mathrm{i}y$).
– Gary
Commented Aug 3, 2023 at 23:15

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.
$$\frac{1}{2}(A-iB)=\frac{1}{2\pi i} \oint_{|\zeta|=r} \frac{u(\zeta)-iv(\zeta)}{\zeta - z} d\zeta$$
$$\overline{\frac{1}{2}(A-iB)}=-\frac{1}{2\pi i} \oint_{|\zeta|=r} \frac{u(\zeta)+iv(\zeta)}{\overline\zeta - \overline z} d\left(\overline \zeta \right)$$ 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 $$|z|, we have $$\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.