I have to calculate (without refering to residue theorem)
$$\int_{\partial B(2,3)} \frac{dz}{z^4-16}$$
My attempt:
First, I need to find singularities of $f(z)=\frac{1}{z^4-16}$.
$$z^4-16=(z^2-4)(z^2+4)=(z-2)(z+2)(z-2i)(z+2i)$$
Notice that point $z=-2$ doesnt lie inside our circle. Let $z_1:=2, z_2:=-2i, z_3:=2i$. They all lie inside our circle.
Let
$C_1:=\partial B(z_1,\frac{1}{1000})$ $C_2:=\partial B(z_2,\frac{1}{1000})$ $C_3:=\partial B(z_3,\frac{1}{1000})$
From Cauchy-Goursat theorem we have:
$$\int_{\partial B(2,3)} \frac{dz}{z^4-16}=\int_{C_1} \frac{dz}{z^4-16}+\int_{C_2} \frac{dz}{z^4-16}+\int_{C_3} \frac{dz}{z^4-16}$$
Now let's define functions:
$$f_1:=\frac{1}{(z+2)(z-2i)(z+2i)}$$ $$f_2:=\frac{1}{(z+2)(z-2)(z-2i)}$$ $$f_3:=\frac{1}{(z+2)(z-2)(z+2i)}$$
Note, that
$f_1$ is holomorphic in a region containing $C_1$
$f_2$ is holomorphic in a region containing $C_2$
$f_3$ is holomorphic in a region containing $C_3$
Now using Cachy integral formula to the above functions and circles we have:
$$\int_{\partial B(2,3)} \frac{dz}{z^4-16}=\int_{C_1} \frac{dz}{z^4-16}+\int_{C_2} \frac{dz}{z^4-16}+\int_{C_3} \frac{dz}{z^4-16}=$$ $$2\pi i(f_1(2)+f_2(-2i)+f_3(2i))$$
Is this correct reasoning?