Vtyjkyuk

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 $textProposition (1)$ is valid ?

$textProposition (1)$

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,

$$p(z) = a_n(z-a_1)…(z-a_n) = a_n prod_j^n(z-a_j). tag1.2$$

Putting everything together we have that,

$$oint_partial D(0,1) a_nprod_j^n(z-a_j) overline z^j dz = 0. tag1.3$$

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

beginalign*
partial_overline z bigg( oint_partial D(0,1) a_nprod_j^n(z-a_j) overline z^j dz bigg) tag1.4 &= \
oint_partial D(0,1) bigg( partial_overline z a_n prod_j^n(z-a_j) overline z^j dz bigg) &= 0 tag1.5
endalign*

Firstly,
$$I_n,j=oint_ z^n overline z^j dz=oint_ (|z|^2)^jcdot z^n-jdz=oint z^n-jdz$$

Obviously, $I_n,j=2pi i$ if $n-j=-1$ and equals $0$ otherwise.

Let $P(z)=sum^k_n=0a_n z^n$ be a polynomial.

Then, $$oint_P(z)overline z^jdz=sum^k_n=0a_nI_n,j$$

Given the integral is zero, we need $a_j-1= 0$ for every $jinmathbb N$.

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.

$$I_n,j=oint_z^noverline z^jdz=int^2pi_0(e^it)^n(e^-it)^jie^itdt=int^2pi_0 i(e^it)^n-j+1dt$$

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