Evaluate the integral using partial fractions $\oint_\gamma\frac{z}{z^4 − 1}\mathrm dz$ Evaluate the integral
$\oint_\gamma\frac{z}{z^4 − 1}\mathrm dz$,
where $\gamma = \{z : |z − a| = a\}$, $a > 1$.
So without using residue theorem and considering that $z = a + ae^{i*\omega}$ gives usa very hard integral formula (at least idk how to solve it) the proper solution includes partial fractions: $\frac{z}{z^4 − 1} = \frac{1}{4}*(\frac{1}{z+1}+\frac{1}{z-1}-\frac{1}{z+i}-\frac{1}{z-i})$, from here we get that the only point which lies in the circle $|z-a|=a$ is $z=1$. Then by Cauchy-Goursat theorem we get that all the other integrals are equal to zero except for $\frac{1}{z-1}$ which is for some reason equal to $2i\pi$. 
The question now is: why the other three ointegrals are 0 and why is the last one equal to $2i\pi$ (without deviding it by 4)
 A: Residue theorem is the standard tool for solving this kind of integrals. First off, factor the polynomial at the denominator:
$$z^4-1=(z^2+1)(z^2-1)=(z+i)(z-i)(z+1)(z-1) $$
Thus the function has $4$ singularities, which are the simple poles $i,-i,1,-1$, and is holomorphic elsewhere. Your curve $\gamma$ is a circle of radius $a$ centred in $a$. It crosses the imaginary axis only in $z=0$, and since $a>1$, the only singularity within the circle is $z=1$.
Hence by residue theorem
$$ \int_{\gamma}\frac{z}{z^4-1}dz=2\pi i \operatorname{Res}_1$$
with
$$\operatorname{Res}_1= \frac{1}{(1+i)(1-i)(1+1)}=\frac{1}{4}$$
Hence the integral is equal to $\frac{i\pi}{2}$.
A: $z^4-1=0\to z=\pm 1,\pm i$
We now need to check if these roots are in the circle. 
$|z-a|=a$ is a circle with radius $a$ and center $(a,0)$ and only $z=1$ lies in the circle. Therefore:
$\displaystyle\oint_\gamma\frac{z}{z^4 − 1}\mathrm dz=2\pi i Res_{z\to1}(f(z))$
A: If you're expected to use parameterisation, consider $z = a + ae^{i\theta} \implies |z-a|=|a+ae^{i\theta}-a| = a$. You then have, 
$$\int_{\gamma} \frac z {z^4 - 1} \mathrm dz = i\int_0^{2\pi} \frac{ae^{i\theta}(a+ae^{i\theta})}{(a+ae^{i\theta})^4 - 1} \mathrm d\theta$$
But typically I'd go with the residue theorem. 
A: Alternative way, without using residue theorem, but still using Cauchy's integral formula, stating:
$$f(w)=\frac{1}{2\pi i} \int\limits_{\gamma}\frac{f(z)}{z-w}dz$$
we replace $w=1$ (since you proved that only this point is in the circle) and $f(z)=\frac{z}{z^3+z^2+z+1}$ then we have
$$\frac{1}{4}=f(1)=\frac{1}{2\pi i}\int\limits_{|z-a|=a}\frac{f(z)}{z-1}dz=\\
\frac{1}{2\pi i}\int\limits_{|z-a|=a}\frac{z}{(z-1)(z^3+z^2+z+1)}dz=\\
\frac{1}{2\pi i}\int\limits_{|z-a|=a}\frac{z}{z^4-1}dz$$
and the result follows. 
You can also use this theorem to show 
$$1=\frac{1}{2\pi i}\int\limits_{|z-a|=a}\frac{1}{z-1}dz$$
