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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?

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    $\begingroup$ Seems correct to me. – Another simple approach is using the partial fraction decomposition of the integrand. $\endgroup$ – Christian Blatter Jan 5 '15 at 11:09
  • $\begingroup$ Thank you. And then, after decomposition I use Cauchy formula yes? $\endgroup$ – luka5z Jan 5 '15 at 11:15

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