# Complex integral over curve example calculations

Given $$\gamma$$ is positively orientated circle with radius $$\sqrt2$$ and center at $$2 + i$$ calculate following integrals: $$\int_\gamma \frac{e^z \cos z}{(1+z^2)\sin z}dz \\ \int_\gamma \frac{\sin(z - 2)}{z - 2}dz$$

I wish I could add my attempt on this but I don't even know how to start. Should I parametrize my circle as $$\gamma(t) = 2 + i + \sqrt2e^{it}$$? Work with Cauchy's integral formula? Notice antyhing about those functions?

Let's deal with the first integral. Let us make some observations.

Observation $$(1)$$: We can write $$z^2 + 1 = (z + i)(z - i)$$. So, it is zero when $$z = \pm i$$.

Observation $$(2)$$: Polynomials, $$e^z, \cos z$$ and $$\sin z$$ are all entire functions. Moreover, $$1/ \sin z$$ is entire whenever it is defined.

Observation $$(3)$$: $$\sin z = 0 \iff z = k \pi$$ for some $$k \in \mathbb{Z}$$.

We have the positively-oriented circle $$\gamma(t) = (2 + i) + \sqrt{2}e^{it}$$ for $$t \in [0, 2 \pi]$$.

Now, $$(1)$$ tells us that $$z^2 + 1$$ is holomorphic inside and on $$\gamma$$ (as $$\pm i$$ is not contained inside $$\gamma(t)$$).

$$(2)$$ and $$(3)$$ tells us that $$1/\sin z$$ is entire whenever $$z \neq k \pi$$ for some $$k \in \mathbb{Z}$$, so it is holomorphic inside and on $$\gamma$$.

Combining all these observations we deduce that:

$$$$\frac{e^z \cos z}{(1 + z^2) \sin z} \; \text{ is holomorphic inside and on} \; \gamma(t).$$$$

So, Cauchy's Theorem tells us that the integral is zero. I'll leave you to do the other one.