# How do we find a path $\gamma$ with winding number $1$ and $2$ relative to points $1$ and $2$, respectively?

Let $\gamma :[a, b]\to\Omega\subseteq\mathbb{C}$ denote a parametric piecewise continuously differentiable path in $\Omega$ and $$\text{ind}_{\gamma}(z):=\frac{1}{2\pi i}\int_{\gamma}\frac{1}{\zeta-z}d\zeta$$ denote its winding number relative to $z\in\mathbb{C}$. I want to find $\gamma$ with $$\text{ind}_{\gamma}(1)=1\;\;\;\text{and}\;\;\;\text{ind}_{\gamma}(2)=2$$ How would we do that?

We have to find a nice closed curve $\gamma\subset{\mathbb C}$ which winds once around $1$ and two times around $2$. This is a problem involving geometric intuition, and not integrals.
A possible solution: Draw two circles touching at ${5\over2}$, one centered at ${3\over2}$, the other centered at $2$, and concatenate them. This amounts to the following parametrization: $$\gamma:\quad t\mapsto\cases{{3\over2}+e^{it} \quad&(0\leq t\leq 2\pi) \cr 2+{1\over2}e^{it} &(2\pi\leq t\leq 4\pi) .\cr}$$ When you compute ${\rm ind}_\gamma$ at $1$ and $2$ using your complex analysis toolbox you shall see that the required values appear.
• I'm still unsure. Let $z_0\in\mathbb{C}$, $r>0$ and $B_r(z_0):=\left\{z\in\mathbb{C} : |z-z_0|<r\right\}$. Then we've got $$\int_{\partial B_r(z_0)}\frac{1}{\zeta -z}=2\pi$$ This maintains our intuition, but since this result is independent of the center of the circle ... Commented May 26, 2014 at 14:13