In the text "Function Theory of One Complex Variable" Third Edition by Robert E. Greene and Steven G. Krantz i'm inquring if my proof to $\text{Proposition (1)}$ is valid ?

$\text{Proposition (1)}$

If $f$ is a holomorphic polynomial and if

$$\oint_{\partial D(0,1)}f(z) \overline z^{j} dz = 0, \, \, j = 0, 1, 2,.. $$

then prove that $f \equiv 0 $

To begin our quest to prove $(1)$, we write our choice of $f$ as,

$$p(z) = a_{n}(z-a_{1})…(z-a_{n}) = a_{n} \prod_{j}^{n}(z-a_{j}). \tag{1.2}$$

Putting everything together we have that,

$$\oint_{\partial D(0,1)} a_{n}\prod_{j}^{n}(z-a_{j}) \overline z^{j} dz = 0. \tag{1.3}$$

Using Feynman's Integration Trick and the Product rule finally,

\begin{align*} \partial_{\overline z} \bigg( \oint_{\partial D(0,1)} a_{n}\prod_{j}^{n}(z-a_{j}) \overline z^{j} dz \bigg) \tag{1.4} &= \\ \oint_{\partial D(0,1)} \bigg( \partial_{\overline z} a_{n} \prod_{j}^{n}(z-a_{j}) \overline z^{j} dz \bigg) &= 0 \tag{1.5} \end{align*}

  • $\begingroup$ What does (1.5) tell you? $\endgroup$ Aug 12, 2018 at 0:47
  • $\begingroup$ You could try proving this with Cauchy's Integrel Formula. It avoids this nasty integral. $\endgroup$ Aug 12, 2018 at 0:53
  • $\begingroup$ Yeah @SeanNemetz it seems so also I realized a problem with they way I defined $(1.2)$ with the $a_{n}$ it looks like $(1.4)$ to $(1.5)$ is bunk i'll have to find another to define $p(z)$ as a product to save the proof $\endgroup$
    – Zophikel
    Aug 12, 2018 at 3:21
  • $\begingroup$ Yeah I don't think you can use Feynman's trick with holomorphic stuff. $\bar{\partial}$ just isn't the same kind of operator. $\endgroup$ Aug 12, 2018 at 3:22

1 Answer 1


Firstly, $$I_{n,j}=\oint_{|z|=1} z^n \overline z^j dz=\oint_{|z|=1} (|z|^2)^{j}\cdot z^{n-j}dz=\oint z^{n-j}dz$$

Obviously, $I_{n,j}=2\pi i$ if $n-j=-1$ and equals $0$ otherwise.

Let $P(z)=\sum^k_{n=0}a_n z^n$ be a polynomial.

Then, $$\oint_{|z|=1}P(z)\overline z^jdz=\sum^k_{n=0}a_nI_{n,j}$$

Given the integral is zero, we need $a_{j-1}= 0$ for every $j\in\mathbb N$.

From here, we get $a_0=a_1=a_2=\cdots=0$.

The result about $I_{n,j}$ can be derived without Cauchy’s integral formula.

$$I_{n,j}=\oint_{|z|=1}z^n\overline z^jdz=\int^{2\pi}_0(e^{it})^n(e^{-it})^jie^{it}dt=\int^{2\pi}_0 i(e^{it})^{n-j+1}dt$$

which equals zero unless $n-j+1=0\implies n-j=-1$.

  • $\begingroup$ Are there any other ways to prove it ? $\endgroup$
    – Zophikel
    Aug 12, 2018 at 3:39
  • $\begingroup$ @Zophikel In what direction? $\endgroup$
    – Szeto
    Aug 12, 2018 at 3:44
  • $\begingroup$ @Zophikel Also you can recognize, on the unit circle, $\overline z=\frac1z$. $\endgroup$
    – Szeto
    Aug 12, 2018 at 3:55
  • $\begingroup$ In the direction of not using the Cauchy Integral Formula $\endgroup$
    – Zophikel
    Aug 12, 2018 at 3:58
  • $\begingroup$ @Zophikel Please see my edited answer. $\endgroup$
    – Szeto
    Aug 12, 2018 at 4:16

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