Some of the Cauchy type integral property Let $D$ - a simply connected bounded domain and $\phi(t) \in C(\partial D)$.
Prove that $ \displaystyle \oint \limits_{\partial D} \frac{\phi(t)}{t - z}dt = 0 $ $ \forall z \notin \overline D $ if and only if $ \displaystyle \oint \limits_{\partial D} t^n \phi(t) dt = 0$, $ n = 0, 1, ... $
I have no idea. From the complex analisys course I know that $ \displaystyle \oint \limits_{\partial D} \frac{\phi(t)}{t - z}dt$ is holomorphic and know the formula for it's derivatives. Also it is clear that the second condition is equivalent to $ \displaystyle \oint \limits_{\partial D} P(t) \phi(t) dt = 0$ for any polynom $ P(t) $. But I have no idea how to solve this problem.
Thanks for any help!
 A: For the implication $\Leftarrow$ : 
As you said, you have $\displaystyle \oint \limits_{\partial D} P(t) \phi(t) dt = 0$ for any polynom $P$. 
Consider the function $f: t \mapsto \frac{1}{t-z}$, since $z \notin \overline D$, it is continuous on $\overline D$, which is compact (closed and bounded). So with Stone-Weirstrass ($f$ is a function from $\overline D$ to $\mathbb{R}$), $\exists (p_n)$ sequence of polynomial functions, such as $p_n$ converges uniformly to $f$.
With the first remark we have $\forall n$, $\displaystyle \oint \limits_{\partial D} p_n(t) \phi(t) dt = 0$, and so $\displaystyle \oint \limits_{\partial D} \frac{\phi(t)}{t - z}dt = 0$ because the convergence is uniform, indeed : $$|| \phi (t)p_n(t)-\frac{\phi(t)}{t-z}||_{\infty} \le ||\phi||_{\infty}|| p_n(t)-\frac{1}{t-z}||_{\infty}$$
But $|| p_n(t)-\frac{1}{t-z}||_{\infty} \rightarrow 0$ and $||\phi||_{\infty} \le c$, because $\phi$ is continuous  on a compact, so $|| \phi (t)p_n(t)-\frac{\phi(t)}{t-z}||_{\infty} \rightarrow 0$ and so $\displaystyle \oint \limits_{\partial D} \frac{\phi(t)}{t - z}dt = 0$.
