Is $f$ continuous on a contour $\Gamma$? I am practicing for my midterm exam. I would appreciate it if someone could point me to any mistakes or flaws in my answer to this question I tried:

Given a contour $\Gamma : |z-\pi|=\frac{\pi}{2}$ traversed once counterclockwise, and define for all $z \in \mathbb{C}$ that are not on $\Gamma$ $$f(z) := \frac{1}{2\pi i} \int_\Gamma \frac{\cos \zeta}{\zeta - z} d\zeta $$
  and for $z$ that are on $\Gamma$, $f(z) := 0$. Is $f$ continuous on $\Gamma$?

My answer: We take any point $z_0$ on $\Gamma$ such that $\cos z_0 \neq 0$. The function $f(\zeta)=\cos \zeta$ is analytic in the right half of the complex plane. Cauchy's integral formula then says that for any $z_0$ in the interior of $\Gamma$, $f(z)=\cos z$. For continuity, we must have that by whichever path we go, $\lim_{z \rightarrow z_0} f(z) = f(z_0)$. If $z$ approaches $z_0$ along a path in the interior of $\Gamma$, $f(z)=\cos z \rightarrow \cos z_0 \neq 0$. However, $f(z_0)=0$, as $z_0$ lies on $\Gamma$. This means $\lim_{z \rightarrow z_0} f(z) \neq f(z_0)$, so $f$ is not continuous on all of $\Gamma$. 

I am especially unsure about my notation and way of writing everything down clearly. For example, I want to name $f(\zeta)=\cos \zeta$ since it fits perfectly in the Cauchy formula "form", but is that OK, or do I call it something else, for example $g(\zeta)=\cos \zeta$?
I feel a bit silly asking this sort of question, but I hope someone can help me out. Thanks in advance.

My answer:
We take any point $z_0$ on $\Gamma$ such that $\cos z_0\ne 0$. 
Cauchy's integral formula says that 
$$f(z)=\cos z$$
for any $z$ in the interior of $\Gamma$. 
If $z$ approaches $z_0$ along a path in the interior of $\Gamma$, $$f(z)=\cos z \to \cos z_0 \ne 0$$
since $\cos z$ is continous on $\mathbb{C}$.
However $f(z_0 )=0$ by it's definition, as $z_0$ lies on $\Gamma$. This means $$\lim_{z\to z_0} f(z)\ne f(z_0),$$ so $f$
is not continuous at $z_0\in \Gamma$. Therefore our conclusion is that the function $f(z)$ is not continous on $\Gamma$.
