What is the value of $\int_S \frac{\overline{f(z)}}{z-a}dz$ for an entire function $f(z)$, where $S$ is the unit circle.

Question

Suppose that $f(z)$ is an entire function and $S$ be the unit circle, then prove that $$\int_S \frac{\overline{f(z)}}{z-a} dz=2\pi i \cdot\left\{\begin{aligned}\overline{f(0)}~\quad~|a|<1 \\ \overline{f(0)}-\overline{f(\bar{a}^{-1})} \quad |a|>1 \end{aligned}\right.$$ How can I use the definition $\int_\gamma f(z)dz=\int_a^b f(\gamma(t))\gamma'(t)dt$ for a continuous function $f(z)$ on a piecewise smooth contour $\gamma:=\gamma(t),~a \leq \gamma \leq b$. Also how to use the analyticity of $f$ here?

Answer 1

By noting that $\overline{z} = \frac{1}{z}$ for $z \in S$, we have

$$ I := \int_{S} \frac{\overline{f(z)}}{z-a} \, \mathrm{d}z \stackrel{(z=w^{-1})}= \int_{-S} \frac{\overline{f(\overline{w})}}{\frac{1}{w}-a} \, \left(-\frac{\mathrm{d}w}{w^2} \right) = \int_{S} \frac{\overline{f(\overline{w})}}{w(1-aw)} \, \mathrm{d}w. $$

Since $g(w) := \overline{f(\overline{w})}$ is an entire function, we may apply the residue theorem.

• If $|a| < 1$, then the only pole inside of $S$ is $w=0$, and so, $$ I = 2\pi i \, \underset{w=0}{\mathrm{Res}} \, \frac{g(w)}{w(1-aw)} = 2\pi i g(0). $$

• If $|a| > 1$, then there are two poles, $w = 0$ and $w = \frac{1}{a}$, inside of $S$, and so, $$ I = 2\pi i \biggl( \underset{w=0}{\mathrm{Res}} \, \frac{g(w)}{w(1-aw)} + \underset{w=a^{-1}}{\mathrm{Res}} \, \frac{g(w)}{w(1-aw)} \biggr) = 2\pi i \bigl( g(0) - g(a^{-1}) \bigr). $$

Here, the following observation is useful:

If $P(z)$ and $Q(z)$ are analytic near $z = z_0$, $P(z_0) \neq 0$, and $Q(z)$ has a simple zero at $z = z_0$, then $$ \underset{z=z_0}{\mathrm{Res}} \, \frac{P(z)}{Q(z)} = \frac{P(z_0)}{Q'(z_0)}. $$