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)

• Welcome to the Mathematics StackExchange. Can you show what you have attempted so far? – George Coote Feb 20 '18 at 11:35
• $\int_\gamma z\mathrm dz$ that was part a but i was given parametarrization for z in terms of t, as far as i can understand I need to come up with paramatarization for both f(z) and |z-a| <- which is a circle but have no idea how to do anything about it, perhaps I did not understand the lecture on countour integration (and I suspect this is not holomorphic in this partcular region otherwise the answer would be simply 0) – user6396911 Feb 20 '18 at 11:40
• Do you know the residue theorem? Do you know which poles are contained in your contour? – bames Feb 20 '18 at 11:44
• no, we have not come across it yet, it's further down in the lectures – user6396911 Feb 20 '18 at 11:46
• Is this an Imperial College homework question? If so, in the lectures so far you have covered integrating round a circle centre the origin containing $1/z$. You have also covered that integrating round a path where the function is holomorphic everyone on and inside the path gives zero. That is enough to deal with this example. – almagest Feb 20 '18 at 12:05

4 Answers

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 $$1=\frac{1}{2\pi i}\int\limits_{|z-a|=a}\frac{1}{z-1}dz$$

• How did you come up with a magical $f(\omega)$? – user6396911 Feb 20 '18 at 21:13
• It's already in my DNA for 20 years :) ... 1st it was easy to observe that $z^4-1=(z-1)(z^3+z^2+z+1)$, then that $w=1$ is inside the circle and all the other roots of $z^4-1$ aren't (same way you did it) and ... everything else is mechanical application of Cauchy's integral formula. – rtybase Feb 20 '18 at 21:20

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 $\frac{i\pi}{2}$.

• is there a way to solve it without using this magic theorem? – user6396911 Feb 20 '18 at 11:49

$z^4-1=0\to z=\pm 1,\pm i$

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:

$\displaystyle\oint_\gamma\frac{z}{z^4 − 1}\mathrm dz=2\pi i Res_{z\to1}(f(z))$

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.