$\oint_{\gamma}\frac {\sin z}{z(z-2i)}dz$ on $|z| =3$ in trigonometric sense and on the inverse trigonometric sense.
On the trigonometric sense, we can apply the Residue theorem, we have :
$z_1 = 0$ apparent singular point because $\lim_{z\to0} f(z)= -\frac 1{2i}.$ Which is different from infinity and it clearly exists. And also we have: $z_0=2i$ simple pole.
If we check that $|z_0|<3$ is true then $z_0\in Int\gamma.$ Then by the Residue theorem:
$$I = 2\pi i Res(f, 2i)$$
But $Res(f, 2i)= \frac {\sin(2i)}{2i} \implies I=\pi\sin(2i)=i\pi\frac {e^2-e^{-2}}{2}$.
And on the inverse trigonometric sense, I thought that i have to check if $|z_0|>3$ so then that is false then $I=0.$ Am i correct?
Ofcourse, with the notations: $I$ being the integral, and $f(z)$ the function under the integration sign.
$$\int_{\gamma}\dfrac{\frac{\sin z}{z}}{z-2i}\mathrm dz=2\pi i\mathrm{Res}_{z=2i}=2\pi i\cdot\lim_{z\to 2i}\left[(z-2i)\cdot\dfrac{\frac{\sin z}{z}}{z-2i}\right]$$
Aliter:
Another way to tackle this is to use Cauchy Integral formula that gives you the following thereby confirming what we obtained in the first part:$$2\pi if(2i)=\int_\gamma\dfrac{{\frac{\sin z}{z}}}{z-2i}\mathrm dz \ : f(z)=\dfrac{\sin z}{z}$$