Reason for equality of $\frac{z^6+1}{z-z_k}=\sum\limits_{i=0}^{5} z_k^iz^{5-i}$

Question

I have this equation:

$$g_k(z)=\frac{z^6+1}{z-z_k}=\sum\limits_{i=0}^{5} z_k^iz^{5-i}$$

$z_k$ are the singularities of $f(z)=\frac{1}{z^6+1}$ with $k=0,..,5$. Afterwards I want to determine the residues of this function to compute the integral $$I=\frac{1}{2}\int\limits_{-\infty}^{+\infty} \frac{dx}{x^6+1}$$

I only need help to understand the equation above. Thanks for your help!

Answer 1

As $z_k^6=-1$, you get $$ z^6+1=z^6-z_k^6=(z-z_k)(z^5+z^4z_k+…+zz_k^4+z_k^5) $$ as in the binomial theorems or geometric sums.

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Note that also $$ z^6+1=(z^2+1)(z^4-z^2+1)=(z^2+1)((z^2+1)^2-3z^2) \\ =(z^2+1)(z^2+\sqrt3z+1)(z^2-\sqrt3z+1) $$ which allows to compute the integral using partial fraction decomposition and the arcus tangent.

Answer 2

Note that the following are equivalent in this case:

• $z_k$ is a singularity of $f(z).$
• $z_k$ is a pole of $f(z).$
• $z_k$ is a root of $z^6+1.$
• $z-z_k$ is a factor of $z^6+1.$

The expression you're curious about can be found by polynomial long division. It can be checked by multiplying both sides by $z-z_k.$ In both cases, we must bear in mind that $z_k^6=-1.$

Answer 3

First $(z_k^6+1) =0 $ $$ z^6+1=z^6 +1-(z_k^6+1)=z^6-z_k^6= (z-z_k)\sum_{i=1}^{5} z_k^i z^{5-i} $$ Then for $a>1$ $$\int_{-\infty}^{\infty} \frac{dx}{|x|^a +1} = 2\int_{0}^{\infty} \frac{dx}{x^a +1} = \frac{2}{a}\int_{0}^{\infty} \frac{u^{\frac{1}{a}-1}}{u +1} du =\frac{2}{a}B\left(\frac{1}{a}, 1-\frac{1}{a}\right) = \frac{2}{a}\Gamma\left(\frac{1}{a}\right)\Gamma\left(1-\frac{1}{a}\right) $$

take a= 6