From $ \oint \frac{\phi(t)}{t - z} dt = 0$ prove that $ \oint \frac{t \phi(t)}{t - z} dt = 0$ Let $D$ - a simply connected bounded domain and $\phi(t)∈C(\partial D)$. And let $ \displaystyle \oint \frac{\phi(t)}{t - z} dt = 0$ $\forall z \notin \overline D$. Prove that $ \displaystyle \oint \frac{t \phi(t)}{t - z} dt = 0$ $\forall z \notin \overline D$. I can prove that $ \displaystyle \oint \frac{\phi(t)}{(t - z)^n} dt = 0$ and $ \displaystyle \oint \frac{t \phi(t)}{t - z} dt = \oint \frac{t^2 \phi(t)}{(t - z)^2} dt = ...=C$ $\forall z \notin \overline D$. But I can solve the initial prolem. Thanks for the help!

You didn't really say what you mean by $dt$ (some kind of arc length measure, I'm assuming). Ignoring this issue, we have that $$ \int \frac{t\phi(t)}{t-z}\, dt = \int \frac{(t-z+z)\phi(t)}{t-z}\, dt =\int\phi(t)\, dt , $$ so the (holomorphic) function $F(z)=\int\frac{t\phi(t)}{t-z}\, dt$ is constant, and $F(z)\to 0$ as $|z|\to\infty$. Thus $F=0$.