Compute the integral for $C$ given by $e^{i\theta}$, $\theta \in [0, 2\pi]$

Question

Here is the problem :

$$\int_{\mathcal{C}} \frac{z^9+e^{iz}+(7654)^{\sin(z)}z}{z-1} \,\mathrm{d}z,$$

for $C$ given by $C(\theta) = e^{i\theta}/2, \theta \in [0, 2\pi]$

I think that I can use the Coushy-Gorsat Theorem for this problem. But, I'm not sure it is correct or not.

Since the singularity point is equal to 1, and, it does not live in the $C(\theta)$, is it possible to use the Coushy-Gorsat Theorem? If so, I know that this integral will be $0$.

If my answer and approach are correct, could anyone give me a detailed account?

Answer 1

Pretty sure you're in my Complex Class at UT! The path is a circle with radius $\frac{1}{2}$. From my observation the function is not holomorphic when $z=1$, from the denominator. Therefore, the integral along that curve is going to be zero from the CG theorem because our integrand is holomorphic on the interior of that path.

Hopefully my assumption is right, because all the numerator functions seem to be holo on C, though I'm not sure about the last one regarding powers of sine.