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I have a question for homework that asks

If $f(z) = u(z) + iv(z)$ is analytic on $\mathbb{D}$ (the open unit disk), and $u(z)$ extends to be continuous on $\overline{\mathbb{D}}$, then $$ f(z) = \int_0^{2 \pi} u(e^{i \theta}) \frac{e^{i \theta} + z}{e^{i \theta} - z} \frac{d \theta}{2 \pi} + iv(0) $$ for every $z \in \mathbb{D}$.

Since $u(z)$ is harmonic (because $f$ is analytic) on $\mathbb{D}$ and extends to be continuous on $\overline{\mathbb{D}}$, I know from a theorem that $$ u(z) = \int_0^{2 \pi} u(e^{i \theta}) \frac{e^{i \theta} + z}{e^{i \theta} - z} \frac{d \theta}{2 \pi} $$ for every $z \in \mathbb{D}$. Further, since $v$ is harmonic, I know that there is some $\varepsilon > 0$ such that $\{ |z - z_0| < \varepsilon \} \subseteq \mathbb{D}$ and $$ v(z) = \int_0^{2 \pi} v(z + \varepsilon e^{i \theta}) \frac{d \theta}{2 \pi} $$ for every $z \in \mathbb{D}$. Where I am having trouble is that pesky '$iv(0)$' in the expression above. Why are we evaluating $v$ at $0$?

tylerc0816
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1 Answers1

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You are missing a real part,

$$u(z) = \operatorname{Re} \int_0^{2\pi} u(e^{i\theta})\frac{e^{i\theta}+z}{e^{i\theta}-z}\,\frac{d\theta}{2\pi}.\tag{1}$$

Now, the integrand is holomorphic in $z$, thus

$$\tilde{f}(z) := \int_0^{2\pi} u(e^{i\theta})\frac{e^{i\theta}+z}{e^{i\theta}-z}\,\frac{d\theta}{2\pi}$$

defines a holomorphic function in the unit disk, and by $(1)$, we have $\operatorname{Re} \tilde{f} = \operatorname{Re} f = u$, so $f-\tilde{f}$ is a real-valued holomorphic function, hence constant.

Now,

$$\tilde{f}(0) = \int_0^{2\pi} u(e^{i\theta})\frac{e^{i\theta}+0}{e^{i\theta}-0}\,\frac{d\theta}{2\pi} = \int_0^{2\pi} u(e^{i\theta})\,\frac{d\theta}{2\pi} = u(0),$$

by the mean value property of harmonic functions, so

$$f(0) - \tilde{f}(0) = i \operatorname{Im} f(0).$$

Daniel Fischer
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  • I know this is an old post, so I may not get a reply, but I have a question. Maybe I'm missing it, but why are you evaluating $\tilde{f}(z)$ at $z = 0$? I could be for the mean value property, but I tried reading the statement on Wikipedia, and I found it confusing. Does the property tell you to evaluate it at $0$? If not, why does the evaluation at $0$ give you the result? – Nolan P Jun 22 '21 at 15:12