# Quick way of solving the contour integral $\oint \frac{1}{1+z^5} dz$

Consider the contour integral in the complex plane: $$\oint \frac{1}{1+z^5} dz$$ Here the contour is a circle with radius $$3$$ with centre in the origin. If we look at the poles, they need to satisfy $$z^5 = -1$$. So the solutions of the poles are given by: \begin{align*} z_0 &= \cos(\frac{\pi}{5}) + i \sin(\frac{\pi}{5})\\ z_1 &= \cos(\frac{3\pi}{5}) + i \sin(\frac{3\pi}{5})\\ z_2 &= \cos(\pi) + i \sin(\pi) = -1\\ z_3 &= \cos(\frac{7\pi}{5}) + i \sin(\frac{7\pi}{5})\\ z_4 &= \cos(\frac{9\pi}{5}) + i \sin(\frac{9\pi}{5}) \end{align*} So one can use Cauchy's formula or the residue theorem to calculate for every solution the integral and then adding them up to get the full integral. But I have the feeling that there needs to be a more simple way of calculating the full contour integral. Can one just calculate the integral for one solution $$z_i$$ (like the simple solution $$-1$$) and then multiply by it $$5$$, suggesting that the others have the same value. This would make the calculation much efficienter.

EDIT: I now see that $$4$$ solutions are symmetric (the solutions except $$z=-1$$) in the complex plane. If one approximates the solutions of the poles in decimals, one finds: \begin{align*} z_0 &= 0.81 + 0.58i\\ z_1 &= -0.31 + 0.95i\\ z_2 &= -1\\ z_3 &= -0.31 -0.95i\\ z_4 &= 0.81 -0.58i \end{align*} So there are four symmetric solutions. For instance $$z_0$$ is symmetric with $$z_4$$, they are mirrored around the x-axis. Could this mean they cancell each other out so we only need to calculate the integral for $$z_2 = -1$$?

• No. You don't have any net residues at all. See below ... . – Oscar Lanzi Jan 30 '19 at 20:59

HINT:

All of the poles lie on the unit circle. Hence, for any values of $$R_1>1$$ and $$R_2>1$$, we have

$$\oint_{|z|=R_1}\frac1{1+z^5}\,dz=\oint_{|z|=R_2}\frac1{1+z^5}\,dz$$

Now, what happens for $$R_1=3$$ and $$R_2\to\infty$$? The answer is zero. This approach is tantamount to invoking the residue at infinity.

If one insists on summing the residues, then one may proceed in general as follows. First note that if $$z^n+1=0$$, then $$z=z_k=e^{i(2k-1)\pi/n}$$ for $$k=1,2,\dots,n$$.

Next, we calculate the residues of $$\frac{1}{z^n+1}$$ at $$z_k$$ using L'Hospital's Rule as follows:

\begin{align} \text{Res}\left(\frac1{z^n+1},z=z_k\right)&=\lim_{z\to z_k}\left(\frac{z-z_k}{z^n+1}\right)\\\\ &=\frac{1}{nz_k^{n-1}}\\\\ &=\frac1n e^{-i(2k-1)\pi(n-1)/n}\\\\ &=-\frac1n e^{-i\pi/n}\left(e^{i2\pi/n}\right)^k \end{align}

Summing all of the residues reveals for $$n>1$$

\begin{align} \sum_{k=1}^n \text{Res}\left(\frac1{1+z^n},z=z_k\right)&=-\frac1ne^{-i\pi/n}\sum_{k=1}^n\left(e^{i2\pi/n}\right)^k\\\\ &=-\frac1ne^{=i\pi/n} \left(\frac{e^{i2\pi/n}-\left(e^{i2\pi/n}\right)^{n+1}}{1-e^{i2\pi/n}}\right)\\\\ &=0 \end{align}

Hence, for $$n>1$$ the residues of $$\frac1{1+z^n}$$ add to zero. And we are done!

Do you know what the residue at the infinity is?

There is a relation between the residues at complex numbers and the residue at infinity when you have finitely many poles.

The relation is $$\sum_{k=0}^{n} \operatorname{Res}_{a_k}(f) = -\operatorname{Res}_{\infty}(f),$$ where $$\{ a_{k} : n(\Gamma,a_k)\ne 0\}_{k=0}^{n}$$ are the poles (inside the path $$\Gamma$$) of the function you want to compute the integral and $$f$$ is the function you are integrating. Here $$n(\gamma,z)$$ is the index of the complex number $$z$$ with respect to the path $$\gamma$$. I'm supposing that the curve only winds one time around the origin.

• This can help and I will try it, but in my edit I suggest making use of the symmetrie of the solutions, could that be working? – Belgium_Physics Jan 30 '19 at 19:26
• The symmetries betweem the complex poles let you to compute fastly that integral, but introducing the infinity you are studying one more symmetrie, a symmetrie in the Riemann Sphere. What I want to say you is that the symmetries between the poles/zeros can appear by more than one way. But yes, you will be correct if you use the symmetries avaliable at the complex plane. – DrinkingDonuts Jan 30 '19 at 19:36

Here is a subtle approach that completely gets around having to determine the residues.

Cut the plane along the positive real axis. Do your integration around a contour that goes like this:

Part 1: Counterclockwise from $$z=3$$ on the upper side of the cut to $$x=3$$ on the lower side of the cut.

Part 2: Follow the lower side of the cut out to $$z=R$$, where we will take the limit as $$R\to\infty$$.

Part 3: Clockwise around the circle $$|z|=R$$ ending back on the upper side of the cut.

Part 4: Back to $$z=3$$ along the upper side of the cut.

This contour encloses no singularities, therefore the integral around it is zero. But also:

Parts 2 and 4: The real integral converges between $$x=3$$ and $$x=\infty$$ by comparison with $$1/x^5$$ whose antiderivative $$(-1/(4x^4))+C$$ also converges. Therefore Part 2 and the additive inverse of Part 4 must have a common value, forcing these parts to cancel each other.

Part 3: The integrand is $$O(1/R^5)$$ and the length of the contour is $$2\pi R$$, so by the triangle inequality the Part 3 integral is $$O(1/R^4)\to 0$$.

So the Part 1 integral which is your original integral must also be zero.

In general: When you integrate a rational function around a contour that covers all of its poles and the function decreases faster than $$1/z$$ as $$z\to\infty$$, you must get zero. The residues inside the contour are forced to add up to zero, to conform.

• Is there a theorem or proof for this in general property? It looks something interesting that I have not encountered yet. – Belgium_Physics Jan 30 '19 at 22:17
• If you use the type of contour described here, for tge functions I include, you get this result. Parts 2 and 4 always cancel by convergence, and Part 3 always is $O(1/R)$ or less. – Oscar Lanzi Jan 30 '19 at 22:30

Another way to go about this (but this is actually closely related to the solutions above invoking the residue at infinity or arguing that extending the radius of the contour doesn't affect the integral and taking the limit of infinite radius), is to substitute $$z = \frac{1}{w}$$. The integral then becomes:

$$\oint_C \frac{w^3 dw}{1+w^5}$$

where $$C$$ is now a counterclockwise contour with radius $$\dfrac{1}{3}$$ (note that the transform makes the original contour a clockwise contour and we can then make that a counterclockwise one by using the minus sign in $$dz = -\frac{dw}{w^2}$$). Since there are now no poles inside the contour, the integral is zero.

You can use this transform to derive the result that a counterclockwise contour integral with winding number 1 is also given by minus the sum of all the residues outside the contour where you also need to include a suitably defined "residue at infinity".

Partial fractions can work. I think this approach is pretty much equivalent to Mark Viola's residue one, but it's nice to see the agreement.

Let $$\zeta$$ be a primitive fifth root of unity. If you want to be explicit, let $$\zeta = e^{2\pi/5}$$. But the important thing is that

• the roots of $$z^5+1$$ are $$-1$$, $$-\zeta$$, $$-\zeta^2$$, $$-\zeta^3$$, and $$-\zeta^4$$.
• Since $$(z^5 - 1) = (z-1)(z^4+z^3+z^2+z+1)$$, we know that $$1 + \zeta + \zeta^2 + \zeta^3 + \zeta^4 = 0$$.

Now for something about partial fractions that I didn't know until just now. If $$f(z) = (z-\alpha_1) \cdots (z-\alpha_n)$$, and all the roots are distinct, then $$\frac{1}{f(z)} = \frac{1}{f'(\alpha_1)(z-\alpha_1)} + \frac{1}{f'(\alpha_2)(z-\alpha_2)} + \dots + \frac{1}{f'(\alpha_n)(z-\alpha_n)}$$ To show this, start from $$\frac{1}{(z-\alpha_1) \cdots (z-\alpha_n)} = \sum_{i=1}^n \frac{A_i}{z-\alpha_i}$$ for some constants $$A_1, \dots, A_n$$. Clear out the denominators, and you get $$1 = \sum_{i=1}^n \frac{A_i f(z)}{z-\alpha_i} = \sum_{i=1}^n A_i \frac{f(z)-f(\alpha_i)}{z-\alpha_i} \tag{*}$$ Now for any $$i$$ and $$j$$, $$\lim_{z\to \alpha_j} \frac{f(z)-f(\alpha_i)}{z-\alpha_i} = \begin{cases} 0 & i \neq j \\ f'(\alpha_i) & i = j \end{cases}$$ So if we take the limit of both sides of ($$*$$) as $$z\to z_j$$ we get $$1 = A_j f'(\alpha_j)$$ for each $$j$$.

This means that \begin{align*} \frac{1}{z^5+1} &= \frac{1}{5(z+1)} + \frac{1}{5(-\zeta)^4(z+\zeta)} + \frac{1}{5(-\zeta^2)^4(z+\zeta^2)} + \frac{1}{5(-\zeta^3)^4(z+\zeta^3)} + \frac{1}{5(-\zeta^4)^4(z+\zeta^4)} \\&= \frac{1}{5}\left(\frac{1}{z+1} + \frac{1}{\zeta^4(z + \zeta)} + \frac{1}{\zeta^3(z + \zeta^2)} + \frac{1}{\zeta^2(z + \zeta^3)} + \frac{1}{\zeta(z + \zeta^4)}\right) \\&= \frac{1}{5}\left(\frac{1}{z+1} + \frac{\zeta}{z + \zeta} + \frac{\zeta^2}{z + \zeta^2} + \frac{\zeta^3}{z + \zeta^3} + \frac{\zeta^4}{z + \zeta^4}\right) \end{align*} You wanted an argument involving symmetry; what could be more symmetric than that equation?

Now $$C$$ encloses all those roots, so by the Cauchy Integral Formula, $$\oint_C \frac{dz}{z^5+1} = \frac{2\pi i}{5}\left(1 + \zeta + \zeta^2 + \zeta^3 + \zeta^4\right)$$ and as we showed above, the latter factor is zero.

• One can check this, but this would be hard. I've made an edit to my question where I make it simpler by using symmetries. – Belgium_Physics Jan 30 '19 at 19:25
• Unfortunately, if $P(X)=(X-a_1)....(X-a_n)$ has distinct roots, then $$\sum_{k=1}^n \frac{1}{X-a_k}=\frac{P'(X)}{P(X)}$$ This follows immediately from the product rule:$$P'(X)=((X-a_1)....(X-a_n))'=(X-a_2)(X-a_3)....(X-a_n)+(X-a_1)(X-a_3)....(X-a_n)+...+(X-a_1)(X-a_2)....(X-a_{n_1})$$ Now divide both sides by $(X -a_1)....(X-a_n)$. – N. S. Jan 31 '19 at 3:44
• Therefore $$\frac{1}{z-z_0} + \frac{1}{z-z_1} + \frac{1}{z-z_2} + \frac{1}{z-z_3} + \frac{1}{z-z_4}=\frac{5z^4}{z^5+1} \neq \frac{1}{z^5+1}$$ – N. S. Jan 31 '19 at 3:45
• @NS you’re right, of course. Like I said, it was a clue to try partial fractions. And although the numerators aren’t one, there is a decomposition of that form with the “1”s replaces by constants. – Matthew Leingang Jan 31 '19 at 11:35
• @N.S. I've replaced my hint with a full answer. – Matthew Leingang Feb 1 '19 at 22:06