# Term-by-term integration of complex power series

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

1. 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.
2. $$\left\lvert\int_C f\right\rvert\le\int_C\left\lvert f\right\rvert$$
• \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).

• Is $g$ real valued? – copper.hat Dec 22 '15 at 16:24
• I suppose not. For your convenience, it is the $g$ on p.208 in Section 59 of the 7th ed. of the book math.s.chiba-u.ac.jp/~yasuda/ippansug/CV-bookfi.pdf – GNUSupporter 8964民主女神 地下教會 Dec 22 '15 at 16:27
• I assume that is a typographical error in the book. What the author probably meant is to apply the first formula to the special case $g(z) = 1$, on order to prove that the power series $S(z)$ is analytic in the interior of the circle of convergence. – Martin R Dec 22 '15 at 16:51
• Thank for your comment at first. I think about this for hours. If $g$ is continuous but not holomorphic, with $|g(z)| = 1$ in an open neighbourhood, I can neither prove that $g(z) \equiv 1 \,\forall z$ in the open neighbourhood, nor give a counterexample. – GNUSupporter 8964民主女神 地下教會 Dec 22 '15 at 16:54

I assume that is a typographical error in the book and should be $g(z) = 1$ instead of $|g(z)| = 1$.
What the author probably meant is to apply the first formula to the special case $g(z) = 1$, in order to prove that the power series $S(z)$ is analytic in the interior of the circle of convergence.