Integration of complex function? I have to find  $$\int_C \frac{dz}{(z^2+9)^2} \,dz$$ if 
C is the circle with the radius 3 and with the center at the point 2i. 
Now,I know how to find the above integral. $$\int_C \frac{dz}{(z-a)^n} \,dz$$ But I have no clue how to solve and find this one.HELP :/
 A: Just consider $C$ to be the circle centered at $a$ with radius $r$, $z-a=re^{i\theta},$ then
$$\int_C \frac{dz}{(z-a)^n} \,dz= \int_{0}^{2\pi} \frac{ire^{i\theta}d\theta}{re^{in\theta}}. $$
Now, I let you to finish it. The integral should be $0$ for every $n$ except for $n=1$ where it is $2\pi i$. 
Added: Note that, $n=1$ is a special case, now, for $n\neq 1$, we have
$$ \int_{0}^{2\pi} \frac{ire^{i\theta}d\theta}{re^{in\theta}}=i\int_{0}^{2\pi} {e^{i\theta(1-n)}d\theta} = \frac{e^{i\theta(1-n)}}{1-n}\Big|_{0}^{2\pi}= \frac{1}{1-n}( e^{i2\pi(1-n)}-1 )=0 ,$$
since $e^{i2m\pi} = 1,\, \forall m\in \mathbb{Z}. $
I think you can do it for $n=1$. Just subs $n=1$ in the integral and the result follows.
A: See Cauchy's differentiation formula.
http://en.wikipedia.org/wiki/Cauchy's_integral_formula
A: Hint: There are several ways to find the residue of $f(z)$ at a pole $z=p$.
One of my favourites: substitute $z = p + w$, write $f(z) = g(w) w^{-d}$ where
$d$ is the order of the pole (so $g$ is analytic at $w=0$), and find the
 Maclaurin series of $g$ up to the $w^{d-1}$ term. 
A: You can find the value of an integral within a closed curve using Residue Theory:
http://en.wikipedia.org/wiki/Residue_theorem
If you want to find the residue at a at an n'th order pole you can rewrite the inegrand as:
$$f(z)=\frac{\phi(z)}{(z-a)}$$
The residue is then found as:
$$Res{(z=a)}=\frac{\phi^{n-1}(a)}{(n-1)!}$$
So for example in your first integral there is a second order pole at 3i so: $$\int_{C} \frac{dz}{(z^2+9)^2}=2\pi iRes{(z=3i)}=2\pi i\phi^{'}(3i) $$
Where $$\phi(z)=\frac{1}{(z+3i)^2}$$
Note $\phi(z)$ was obtained by rewritting:
$$\frac{1}{(z^2+9)^2}=\frac{1}{(z+3i)^2(z-3i)^2} $$
