Calculating winding number Let
$$
\begin{align}
\gamma= \gamma_1 +\gamma_2+\gamma_3,\\
\gamma_1(t)=e^{it}, t\in[0,2\pi] \\
\gamma_2(t)=-1+2e^{-2it}, t\in [0,2\pi]\\
\gamma_3(t)=1-i+e^{it},t\in [\frac{\pi}{2},\frac{9\pi}{2}]
\end{align}
$$
Calculate the value of $n(\gamma,z)$ as $z$ takes its value in $\mathbb{C}\backslash\gamma.$
$\gamma$ is the image of the closed curve.
as $\gamma$ is a closed and smooth by parts, we have that 
$$
n(\gamma,z)= \frac{1}{2\pi i}\int\limits_\gamma\frac{dc}{c-z} = \frac{1}{2\pi i} \left[ \int\limits_{\gamma_1}\frac{dc}{c-z}+ \int\limits_{\gamma_2}\frac{dc}{c-z} + \int\limits_{\gamma_3}\frac{dc}{c-z}\right]
$$
as  $\gamma_1'(t) = ie^{it}$, $\gamma_2'(t) = -4ie^{-2it}$ and $\gamma_3'(t) = ie^{it}$
i can compute those integrals as 
$$
n(\gamma,z)=\frac{1}{2\pi i} \left[ \int\limits_{0}^{2\pi}\frac{ie^{it}dt}{e^{it}-z}+ \int\limits_{0}^{2\pi}\frac{-4ie^{-2it}dt}{-1+2e^{-2it}-z} + \int\limits_{\pi /2}^{9\pi /2}\frac{ie^{it}}{1-i+e^{it}-z}\right]
$$
but when i compute then for any point i am finding $0$, did i made any mistake or am i making one when computing those integrals?
 A: The value depends on $z$. Note that $\gamma_1$ runs once counterclockwise around the circle with center $0$ and radius $1$, $\gamma_2$ runs twice clockwise around the circle with center $-1$ and radius $2$, $\gamma_3$ runs twice counterclockwise around the circle with center $1-i$ and radius $1$. Let $D_i$ be the open disk bounded by $\gamma_i$. Note that $D_1 \subset D_2$. Drawing a picture is helpful. For a point $z \in \mathbb C \setminus \gamma$ we get $n(\gamma,z) = n(\gamma_1,z) + n(\gamma_2,z) + n(\gamma_3,z)$. We have $n(\gamma_1,z) = 1$ for $z \in D_1$, $n(\gamma_1,z) = 0$ for $z \notin D_1$, $n(\gamma_2,z) = -2$ for $z \in D_2$, $n(\gamma_2,z) = 0$ for $z \notin D_2$, $n(\gamma_3,z) = 2$ for $z \in D_3$, $n(\gamma_3,z) = 0$ for $z \notin D_3$. Thus:


*

*If $z \in D_1 \cap D_3$, then $n(\gamma,z) = 1 -2 + 2 = 1$.

*If $z \in D_1 \setminus D_3$, then $n(\gamma,z) = 1- 2 + 0 = -1$.

*If $z \in D_2 \setminus (D_1 \cup D_3)$, then $n(\gamma,z) = 0 -2 + 0 = -2$.

*If $z \in (D_2 \cap D_3) \setminus D_1 $, then $n(\gamma,z) = 0 -2 + 2 = 0$.

*If $z \in D_3 \setminus D_2$, then $n(\gamma,z) = 0 + 0 + 2 = 2$.

*If $z \notin D_1  \cup D_2 \cup D_3$, then $n(\gamma,z) = 0 + 0 + 0 = 0$.
