Calculate $\int_{γ}\frac{\sin z}{z^4}dz$ where $γ$ is the unit circle.
I tried to consider $f(z)=\frac{\sin z}{z^3}$ so I can have $\int_{γ}\frac{f(z)}{z}dz$ and use Cauchy integral formula, but I realize I can't because $f(z)$ is not holomorphic on the disc centered at zero.
I am new to Cauchy formula trying to understand it (along with the course on complex analysis). I think a hint would be helpful because I am stuck.
Best Answer
use the residue theorem, $z=0$ is a singularity, then $$\int_{\gamma}\frac{\sin(z)}{z^4}dz=2\pi i Res(f;z=0)=2\pi i \left(\frac{-1}{6}\right)=-\frac{i\pi}{3}$$ By cauchy
$$\int_\gamma\frac{g(z)}{(z-z_0)^{n+1}}=\frac{2\pi i}{n!}g^{n}(z_0)$$ where $g(z)=\sin(z)$ so $z=0$ $g(z)$ is holomorphic at $z=0$ then:
$\frac{d^3}{dz^3}g(z_0)=-1$ hence $$\int_{\gamma}\frac{\sin(z)}{z^4}dz=\frac{2\pi i}{3!}(-1)=-\frac{i\pi}{3} $$