Infinitely real-differentiable function with $f(0)=0$ but $\int_{\partial B_1(0)}\frac{f(z)}{z}dz\ne0$

Question

I'm searching for a infinitely real-differentiable function $f:\mathbb{C}\to\mathbb{C}$ with $f(0)=0$ but $$(*)\;\;\;\;\;\int_{\partial B_1(0)}\frac{f(z)}{z}dz\ne0$$ where $$B_r(z_0):=\left\{z\in\mathbb{C}:|z-z_0|<r\right\}$$ Cauchy's integral formula yields that $f$ can't be holomorphic.

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In addition: I want to show that $(*)$ holds, if $f$ is holomorphic and $\text{Re}(f(z))>0$ for all $z\in\partial B_1(0)$.

Answer 1

Put $f(z):=|z|^2=x^2+y^2$. Then $f(0)=0$, and $f(z)=1$ when $|z|=1$. It follows that $$\int_{\partial D}{f(z)\over z}\ dz=\int_0^{2\pi}{1\over e^{it}}\>i\>e^{it}\ dt=2\pi i\ne0\ .$$

Answer 2

Let $f:\mathbb{C}\to\mathbb{C}$ be holomorphic with $\Re (f(z))>0$ for all $z\in \partial B_1(0)$. By definition it holds $$I:=\int_{\partial B_1(0)}\frac{f(\zeta )}{\zeta}d\zeta =i\int_0^{2\pi}f(e^{it})dt$$ Since $e^{it}\in \partial B_1(0)$ for all $t\in [0,2\pi]$ we should be able to conclude $I\ne 0$.