Let $\gamma=\gamma_1+\gamma_2+\gamma_3$, where $\gamma_1(t)=e^{it}$ for $0\leq t\leq 2\pi$, $\gamma_2(t)=-1+2e^{-2it}$ for $0\leq t\leq 2\pi$, and

Question

Let $\gamma=\gamma_1+\gamma_2+\gamma_3$, where $\gamma_1(t)=e^{it}$ for $0\leq t\leq 2\pi$, $\gamma_2(t)=-1+2e^{-2it}$ for $0\leq t\leq 2\pi$, and $\gamma_3(t)=1-i+e^{it}$ for $\pi/2\leq t\leq 9\pi/2.$ Determine all the values assumed by $n(\gamma, z)$ as $z$ varies over $\mathbb{C}\setminus|\gamma|.$

I do not understand this problem very well but I have tried to do the following:

$n(\gamma, z)=\frac{1}{2\pi i}\int_{\gamma}\frac{dw}{w-z}=\frac{1}{2\pi i}[\int_{\gamma_1}\frac{dw}{w-z}+\int_{\gamma_2}\frac{dw}{w-z}+\int_{\gamma_3}\frac{dw}{w-z}]=\frac{1}{2\pi i}[2\pi i-2\pi i+2\pi i]=1$

For the integral formula of Cauchy, taking $f(w)=1$, which is analytic. This is OK? Thank you very much.

Answer 1

The answer depends upon where $z$ is. If it is outside the area bounded by $\gamma_1$, $\gamma_2$, and $\gamma_3$, then $n(\gamma,z)=0$. If $z$ is inside the area bounded by $\gamma_2$ but outside the area bounded by the other two or inside the zone bounded by $\gamma_3$ but outside the area bounded by the other two, then $n(\gamma,z)=1$. If $z$ is inside the zone boundeded by $\gamma_1$, but outside the area bounded by $\gamma_3$, then $n(\gamma,z)=2$. And if it is in the area bounded by all of them, then $n(\gamma,z)=3$.