Construct $\varphi (z)$ such that $\int_{|z|=1} \frac{\varphi (z)}{z-w} dz =0$ I have this problem to complex analysis.
Construct $\varphi (z)$ a continuous function nonzero in $S^{1}$ such that $$\int_{|z|=1} \frac{\varphi (z)}{z-w} dz =0$$ for $|w|<1$.
I have the idea to take $\varphi (z) = (z-w)f(z)$ with $f(z)$ analytic function that is nonzero in $S^{1}$ and use Cauchy theorem for integral, but I am not sure that this is correct.
Note: English is not my first language so I am sorry if I did any mistake.

Take $\varphi(z)=\frac1z$. Then, if $w=0$,$$\int_{\lvert z\rvert=1}\frac{\varphi(z)}{z-w}\,\mathrm dz=\int_{\lvert z\rvert=1}\frac1{z^2}\,\mathrm dz=0.$$And, if $w\neq0$,\begin{align}\int_{\lvert z\rvert=1}\frac{\varphi(z)}{z-w}\,\mathrm dz&=\frac1w\int_{\lvert z\rvert=1}\frac1{z-w}-\frac1z\,\mathrm dz\\&=0,\end{align}because$$\int_{\lvert z\rvert=1}\frac1{z-w}\,\mathrm dz=\int_{\lvert z\rvert=1}\frac1z\,\mathrm dz=2\pi i.$$
