1
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ Is $g$ real valued? $\endgroup$ – copper.hat Dec 22 '15 at 16:24
  • $\begingroup$ 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 $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 22 '15 at 16:27
  • $\begingroup$ 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. $\endgroup$ – Martin R Dec 22 '15 at 16:51
  • $\begingroup$ 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. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 22 '15 at 16:54
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.