I'm reading Complex Variables and Applications by Brown and Churchill, and I get stuck in Section 65. Suppose
$$S(z) = \sum_{n=0}^\infty a_n (z-z_0)^n$$
is a function defined on the interior of the circle of convergence of the above power series. I know that $S(z)$ is analytic, and the following theorem holds.
Let $C$ be a closed curve in the circle of convergence in the power series $S(z)$, and $g(z)$ is continuous on $C$. Then the term-by-term intergration of $g(z) S(z)$ holds. i.e.
$$\int_C g(z) S(z) \,\mathrm{d}z = \sum_{n=0}^\infty a_n \int_C g(z)(z-z_0)^n \,\mathrm{d}z.$$
I can't understand the following claim.
If $\lvert g(z) \rvert = 1$ for each value of $z$ in the open disk bounded by the circle of power series $S(z)$, the fact that $(z-z_0)^n$ is entire when $n = 0,1,\dots$ ensures that
$$\int_C g(z)(z-z_0)^n \,\mathrm{d}z = \int_C (z-z_0)^n \,\mathrm{d}z = 0.$$
I tried thinking about
- Maximum modulus principle
- The modulus of $g$ is constant in the open disk, so there's a local minimum of the modulus of $g$, so $g$ is constant.
- But then I realized that $g$ is merely continuous. It is possible that $g$ isn't holomorphic.
- $$\left\lvert\int_C f\right\rvert\le\int_C\left\lvert f\right\rvert$$
- \begin{align}\left\lvert\int_C g(z)(z-z_0)^n\,\mathrm{d}z\right\rvert&\le\int_C\left\lvert g(z)\right\rvert\left\lvert(z-z_0)^n\right\rvert\,\mathrm{d}z\\ &=\int_C\left\lvert(z-z_0)^n\right\rvert\,\mathrm{d}z\end{align}
- But I can't get an integral whose value is zero.
I need this theorem to establish the fact that $S(z)$ is holomorphic in its circle of convergence (with Morera's theorem).