Geometry behind $\int_{0}^{2π}\frac{e^{ix}}{e^{ix}-z}~dx=2\pi(|z|<1)$ It's a nice exercise to prove, 
$$\int_{0}^{2π}\frac{e^{ix}}{e^{ix}-z}~dx=2\pi(|z|<1)$$ using Leibneiz's rule.But,what's the geometrical interpretation of this?Any idea?

I'm not sure about geometrical, but you can get the same result with complex analysis. Consider the path $\gamma$ defined by the counterclockwise rotation on $|z|=1$. Note here I am using $z$ as a variable and will call the constant $z$ from your statement by $z_0$. Then
$$\int_\gamma \frac{1}{z-z_0}dz$$
can be transformed by $z=e^{i\theta}$ to become
$$=i\int_{0}^{2\pi}\frac{e^{i\theta}}{e^{i\theta}-z_0}dz$$
However, we can also solve this integral using the Residue Theorem. Since $|z_0|<1$, there is exactly one pole inside $\gamma$. We can find the residue at this pole easily:
$$\text{Res}(z_0)=\lim_{z\to z_0} (z-z_0)\frac{1}{z-z_0}=1$$
Then
$$i\int_{0}^{2\pi}\frac{e^{i\theta}}{e^{i\theta}-z_0}dz=\int_\gamma \frac{1}{z-z_0}=2\pi i\text{Res}(z_0))=2\pi i$$
$$\int_{0}^{2\pi}\frac{e^{i\theta}}{e^{i\theta}-z_0}dz=2\pi$$
