# Find the principal part of the Laurent expansion of $\frac{(z^{2}-2z-3)^{2}}{\cos(\pi z)+1}$ around $z_{0}=1$

Problem: The function

$f(z) = \frac{(z^{2}-2z-3)^{2}}{\cos(\pi z)+1}$

has an isolated singularity at $z_0=1$.

a) Find the principal (singular) part of the Laurent expansion of $f$ in a punctured neighbourhood of $z_0=1$.

b) In which region does the Laurent expansion converge?

My thoughts: Usually I am given the region in which the Laurent expansion converges but now I have to determine that myself. Will the Laurent series still be uniquely determined? If so, I have an idea about how to proceed: to manipulate the function so that we have $z-1$ in place of $z$, find Taylor series for every $z-1$ and then just divide when you have the expansion in the top and the bottom. I am not sure if this is a good idea or not and if so I do not know how to change $cos(\pi z)$ into something I can work with.

All input appreciated! I am studying for an exam in complex analysis, this is a problem from an old exam.

• Yes the Laurent series is uniquely determined (in fact can come from a suitable Cauchy integration formula on an annulus), and I suppose you could write $\cos(\pi(z-1)+\pi) = -\cos(\pi(z-1))$, similar trick so you can make use of expansion about $0$ and plugging in a value.
– Evan
Aug 26, 2013 at 19:08
• @Evan I didn't understand how that was helpful but then I realized you meant $cos(\pi(z-1)+\pi) = cos(\pi z))$ (or am I mistaken?). Smart, I suppose you can proceed from there with the method I mentioned?
– Sid
Aug 26, 2013 at 19:13
• @Sid I think I meant what I said. Use cosine of sums formula to get $\cos(y+\pi) = -\cos(y)$. Hmm.. I hope there's a good pattern when you do the division :)
– Evan
Aug 26, 2013 at 19:15
• I'm sorry for my confusion. @Potato Another singularity of the function will be in $z=3$ so the disk can be no larger than of radius 2. This does not tell me how large it is though, or does it?
– Sid
Aug 26, 2013 at 19:19
• @Sid See my answer below. Aug 26, 2013 at 19:27

The Laurent series in any given annulus is uniquely determined. (You should look up a proof of this. It's important.) In this case, you have convergence in a punctured disk. To determine this disk, look at the singularities of the function and note the distance from 1 to the nearest singularity. (If you return to the proof of the existence of Laurent series, you will see that the radius of convergence will be the largest possible punctured disk around with point that doesn't contain another singularity.)

The hard part about finding the Laurent series here is inverting the denominator. If you write out the power series for the denominator, you will see that it is

$$(z-1)^2(a+b(z-1)+c(z-1)^2+\dots)$$

where $a,b,c$ are some constants. So you can factor out the $\frac{1}{(z-1)^2}$ term and use the normal procedure for inverting power series with a nonzero constant term. You can use your idea (writing everything in terms of $z-1$) to take care of the numerator.

• Thanks, I did this and found that the principal part must equal $\frac{2! 16} {{\pi}^{2} z^{2}} = \frac{32}{\pi^{2}z^{2}}$ which matches the student solutions. Finding an expression for the entire Laurent series is much more difficult (I don't know how to invert the whole power series) but that was not a part of the problem.
– Sid
Aug 26, 2013 at 22:48
• @Sid You can invert a power series $f$ with nonzero constant term by considering some power series $g$ with coefficients $a_i$ and looking at the equations you get from equating coefficients in $fg=1$. So you must have $f_0a_0=1$ and so on. You can recursively solve these equations to get as many terms of the power series as you want. This method should be detailed in any text on complex analysis. (I know you don't need this right now, but you should be aware of it.) Aug 27, 2013 at 0:26
• According to the student solutions, the region of convergence is $0 < |z-1| < 4$. Do you have any idea how this can be, considering that the function is not analytic in the point $z=3$?
– Sid
Aug 27, 2013 at 23:22
• @Sid It is! The zero in the numerator cancels the pole in the denominator. Aug 28, 2013 at 0:24

The Maple commands $$f := (z^2-2*z-3)^2/(cos(Pi*z)+1): with(numapprox): laurent(f, z = 1, 7);$$ produce $$32\,{\pi }^{-2} \left( z-1 \right) ^{-2}+2\,{\frac {-8+4/3\,{\pi }^{2 }}{{\pi }^{2}}}+2\,{\frac {1-{\frac {2}{45}}\,{\pi }^{4}+\frac 1 9 \, \left( {\pi }^{2}-6 \right) {\pi }^{2}}{{\pi }^{2}}} \left( z-1 \right) ^{2}+ O \left( \left( z-1 \right) ^{4} \right) .$$

• The same output is given by $series(f, z = 1, 5).$ Aug 27, 2013 at 3:46
• @learner: The Maple syntax differs from the $TEX$ syntax. Aug 27, 2013 at 4:39
• I'm downvoting this because it does not provide a solution or proof, and because it does answer any of the asker's questions about finding Laurent series (which form over half of the body of the question). Aug 27, 2013 at 4:53
• @ Potato: This adds to your excellent answer, giving the result. Aug 27, 2013 at 5:07
• Ah, yes, my mistake. I thought this just duplicated an existing answer, but now I see my words were a bit too strong. Could you make a small edit to the answer so that I can remove the downvote? My vote is now "locked in" and cannot be changed unless you make an edit. Aug 27, 2013 at 5:09