Computing the integral using cauchy's theorem

Question

I need to integrate

$$\oint _{|z|=1} \frac {\sin z}{z}\, dz$$

I write $\sin z=z-\frac{z^3}{3!}+\frac{z^5}{5!}-...$, then I get by dividing by $x$, the series $1-\frac{z^2}{3!}+\frac{z^4}{5!}-...$. I am confused as to how to use this to integrate. What happens to $\frac{z^2}{3!}+\frac{z^4}{5!}-...$ now?

Thanks in advance!

Answer 1

The function $f(z)=sin(z)$ is entire. In particular, it is analytic on the unit disk. By Cauchy's Formula, we have $$ f(z_0)=\frac{1}{2\pi i}\int_{\|z\|=1}\frac{f(z)}{(z-z_0)}dz $$ for any $z_0$ in the interior of the unit disk. Now apply this for a suitable choice of $z_0$.

Answer 2

Note that $\lim_{z \to 0} \frac{\sin z}{z} = 1$, and that the function $$ f(z) = \begin{cases} \sin z & z \neq 0 \\ 1 & z = 0 \end{cases} $$ is entire (that is, $\sin z/z$ has a removable discontinuity at $z = 0$). Because $f$ is entire, we have $$ \oint \frac{\sin z}{z}\,dz = \oint f(z)\,dz = 0 $$

Answer 3

We manage write the function $\frac{\sin z}{z}$ as a power series $$ F(z) = 1-\frac{z^2}{3!}+\frac{z^4}{5!}-..., $$ which converges in $\mathbb C$. That means $F(z)$ is analytic and by Cauchy's theorem $$ \oint_{|z|=1}F(z)\, dz = 0. $$

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To be more formal note that $\frac{\sin z}{z} = F(z)$ when $z\in\mathbb C\setminus 0$, and, in particular when $|z|=1$. So $$ \oint_{|z|=1}\frac{\sin z}{z}\, dz = \oint_{|z|=1}F(z)\, dz = 0. $$