I want to calculate $I=\int_{C(i,2)}^{}\frac{1}{(z^2+4)^2}\: dz$
if the denominator wasn't squared it would be easy, break the integral in 2 sums then consider 2 circles... ect
Now I am not so sure, I thought maybe doing somewhat the same would work.
$I=\int_{C(i,2)}\left [ \frac{i}{4(z+2i)}-\frac{i}{4(z-2i)} \right ]^2\: dz=\int_{C(i,2)} \frac{-1}{16(z+2i)^2}+\frac{1}{8(z-2i)(z+2i)} +\frac{-1}{16(z-2i)^2} \: dz$
I have calculated the middle one using the above method, but for the other 2 can i use Cauchy formula for $f(z)=-1/16$ ? And say they are equal to zero?
Your approach is almost correct except for one minor detail.
• $z=-2i$ lies outside $C$, so the first integral contributes nothing (due to Cauchy's integral theorem). The integral formula does not apply.
• For the second integral, you can use the formula with $f(z)=\frac1{8(z+2i)}$:
$$\oint_C\frac{\mathrm dz}{8(z+2i)(z-2i)}=2\pi i\cdot\frac1{8(z+2i)}\bigg|_{z=2i}=\frac\pi{16}$$
• For the third integral, use Cauchy's differentiation formula, indeed with $f(z)=-\frac1{16}$:
$$-\oint_C\frac{\mathrm dz}{16(z-2i)^2}=2\pi i\cdot\left(\frac{\mathrm d}{\mathrm dz}\left[-\frac1{16}\right]\right)\bigg|_{z=2i}=0$$
Then $I=\dfrac\pi{16}$.
You get the same result using the residue theorem:
$$I=2\pi i\operatorname{Res}\left(\frac1{(z^2+4)^2},z=2i\right)=2\pi i\lim_{z\to2i}\frac{\mathrm d}{\mathrm dz}\left[\frac{(z-2i)^2}{(z^2+4)^2}\right]=2\pi i\cdot\left(-\frac i{32}\right)=\frac\pi{16}$$