Let $D$ be a domain in $\mathbb{C}$ and $\phi:D\rightarrow D'$ a conformal map. Let $\alpha: [0,1]\rightarrow D$ be a closed curve in $D$. For a point $p\in D$, I think that the winding number $I(\alpha; p)$ of $\alpha$ with respect to $p$ should be equal to the winding number $I(\phi\circ\alpha;p)$ of $\phi\circ \alpha$ with respect to $\phi(a)$, but I don't know how to prove it. When I write out the definition, I don't see why they are equal, or how to use the condition that $\phi$ is conformal. Maybe they are not? $$I(\alpha;p)=\frac{1}{2\pi i}\int_\alpha \frac{1}{z-1} d z$$ $$I(\phi\circ\alpha;\phi(p))=\frac{1}{2\pi i}\int_{\phi\circ\alpha} \frac{1}{z-\phi(p)} dz=\frac{1}{2\pi i}\int_0^1\frac{\phi'(\alpha(t))\cdot \alpha'(t)}{\phi(\alpha(t))-\phi(p)}dz=\frac{1}{2\pi i}\int_\alpha \frac{\phi'(z)}{\phi(z)-\phi(p)}dz$$
I appreciate any help!
They don't have to be equal. Suppose that $D=D'=\mathbb C\setminus\{0\}$, that $\phi(z)=z^2$, that $\alpha(t)=e^{2\pi it}$, and that $p=0$. Then $I(\alpha;p)=1$ and $I(\phi\circ\alpha;p)=2$.