Calculate $\oint_{\gamma}\frac {\sin z}{z(z-2i)}dz$ on $|z| =3$ in trigonometric sense and on the inverse trigonometric sense. $\oint_{\gamma}\frac {\sin z}{z(z-2i)}dz$ 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.
 A: $$\int_{\gamma}\dfrac{\frac{\sin z}{z}}{z-2i}\mathrm dz=2\pi i\mathrm{Res}_{z=2i}=2\pi i\cdot\lim_{z\to 2i}\left[(z-2i)\cdot\dfrac{\frac{\sin z}{z}}{z-2i}\right]$$

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:$$2\pi if(2i)=\int_\gamma\dfrac{{\frac{\sin z}{z}}}{z-2i}\mathrm dz \ : f(z)=\dfrac{\sin z}{z}$$
A: Another way for anticlockwise (or trigonometric) (yes, I read the question which mentions Residue theorem) which I believe is easier to grasp, you can rewrite 
$$\int\limits_{|z|=3}\frac {\sin z}{z(z-2i)}dz=
\frac{1}{2 i}\int\limits_{|z|=3}\sin{z} \left(\frac{1}{z-2i}-\frac{1}{z}\right)dz=\\
\frac{1}{2 i}\int\limits_{|z|=3}\frac{\sin{z}}{z-2i}dz-\frac{1}{2 i}\int\limits_{|z|=3}\frac{\sin{z}}{z}dz=...$$
and now use Cauchy's integral formula
$$f^{(n)}(a)=\frac{n!}{2\pi i}\int\limits_{\gamma}\frac{f(z)}{(z-a)^{n+1}}dz \tag{1}$$
In this case, both $0$ and $2i$ are "inside" $|z|=3$ and $f(z)=\sin{z}$, thus
$$...=\pi f(2i)-\pi f(0)=\pi\sin{(2i)}$$
For clockwise, just flip the sign of the result above. 
