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 interior of the closed curve.
My first problem is $\gamma$ as it seems that it is neither simple nor closed curve. At first I tried defining $\gamma$ as $$ \begin{align} \gamma(t)=\gamma_1(t)+\gamma_2(t), t\in[0,\frac{\pi}{2}] \\ \gamma(t)=\gamma_1(t)+\gamma_2(t)+\gamma_3(t), t\in[\frac{\pi}{2},2\pi] \\ \gamma(t)=\gamma_1(t)+\gamma_2(t)+\gamma_3(t), t\in[2\pi,\frac{9\pi}{2}] \\ \end{align} $$ but that didn't work cause $\gamma(\frac{\pi}{2})$ had 2 different values and so did $\gamma(2\pi)$.
My second attempt was to "move" $\gamma_3$ so it would be applied at $2\pi$ , but that doesn't yield any results.
I am not sure that I followed the correct approach so any help , advice or guidance towards the right solution would be welcome.