how to calculate $\int\limits_{|z| = 2} \frac{z^4dz}{z^4+1}$?

Question

The original integral is:

$$\int\limits_{|z| = 2} \frac{z^4dz}{z^4+1}$$

It is seen that integration area is a circle with radius 2, the integral has 4 different poles and can be solved as $z = \sqrt[4]{-1}$ (I can calculate it easily by the formula, know that $\arg = \pi$ ).

But, by theory, to calculate such integrals, the power of a denumerator has to be at least two degrees lower than numerator's one, but what to do if they are equal?

Answer 1

Since $\int_{|z|=2}1dz=0$ and $\frac{z^4}{z^4+1}=1-\frac1{z^4+1}$, $$\int_{|z|=2}\frac{z^4dz}{z^4+1}=-\int_{|z|=2}\frac{dz}{z^4+1}$$

Answer 2

Note that we can write our integral as: $$I = \int_{|z|=2} 1\, dz - \int_{|z|=2} \frac{1}{z^4+1}\, dz$$ All the poles of the function $z^4+1$ are in $|z|=2$, so if the roots are $z_1, z_2, z_3, z_4$, the residue at any one point can be found by: $$\text{ Res }_{z = z_i} f(z) = \frac{1}{(z^4+1)’}\bigg \lvert_{z=z_i} = \frac{1}{4z_i^3}$$

Hope you can take it from here.

Answer 3

Note that, for $R>2$, the function $$ f(x)=\frac{1}{z^4+1} $$ has no pole in $D=\{z|\ 2\le |z|\le R\}$. Therefore $$\int_{|z| = 2} \frac{z^4dz}{z^4+1}=-\int_{|z| = 2} \frac{dz}{z^4+1}=-\int_{|z| = R} \frac{dz}{z^4+1}.$$ But $$ \bigg|\int_{|z| = R} \frac{dz}{z^4+1}\bigg|\le\int_{|z| = R}\frac{1}{|z|^4-1}|dz|=\frac{1}{R^4-1}2\pi R\to0 $$ as $R\to\infty$. So $$\int_{|z| = 2} \frac{z^4dz}{z^4+1}=-\lim_{R\to\infty}\int_{|z| = R} \frac{dz}{z^4+1}=0.$$