Proving this polynomial is identically zero, as a result of this contour integration

Question

Problem: Let $p$ be a polynomial and suppose that $$ \int_{C_1(0)} p(z) \overline{z}^k dz = 0, \qquad k=0, 1, 2, \ldots $$ Prove that $p \equiv 0$. (Here $C_1(0)$ denotes the unit circle centered at origin).

My attempt: I parametrize the unit circle as $z(t) = e^{it}$. Then we have that $$ \int_0^{2 \pi} p(e^{it}) e^{-ikt} i e^{it} dt = i \int_0^{2 \pi} p(e^{it}) e^{i (1-k) t} dt = 0 $$ for all $k > 0$. I let $p(z) = a_n z^n + a_{n-1} z^{n-1} + \ldots + a_1 z + a_0$. I want to show that all the coefficients are zero. With this notation I have $$ i \int_{0}^{2 \pi} ( a_n e^{int} + a_{n-1} e^{(n-1)it} + \ldots + a_1 e^{it} + a_0) e^{i(1-k) t} dt =0. $$ I'm not sure how to show that it follows from this that all coefficients are zero. I think I need to use Cauchy's theorem somehow. Maybe I need to pick a specific $k$?

Answer 1

For $k=2$, we have \begin{align*} \int_{0}^{2\pi}p(e^{it})e^{-it}dt&=0\\ \sum_{l=0}^{n}\int_{0}^{2\pi}a_{l}e^{i(l-1)t}dt&=0\\ 2\pi a_{1}&=0\\ a_{1}&=0, \end{align*} can you proceed for other coefficients?

Answer 2

Here's an alternative method:

Hint The quantity $\overline{p(z)}$ is a linear combination of the monomials $\bar{z}^k$.

...which implies that $\int_{C_1(0)} p(z) \bar z \overline{p(z)} \,dz = 0$, but converting to a line integral using the usual parameterization gives $\int_0^{2 \pi} |p(e^{i \theta})|^2 d\theta = 0$.

Answer 3

The basic fact is that $$\frac{1}{2\pi}\int_0^{2\pi}e^{i(m-k)\theta}d\theta=\cases{0&if $m\ne k$\cr 1&if $m=k$}$$ So, your integral gives $a_{k-1}=0$.