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?


1 Answer 1


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:

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

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


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