# Laurent Series expansion without geometric series

There are several functions in complex analysis which I have not been able to get the Laurent expansion for, both of which are very different from the examples I see online and in the (4) textbooks I have checked out...:

I need to find the Laurent expansion about each singularity of the following function: $$f(z) = {1 \over z^6+1}$$

I had no issue with finding the singular points, but I don't see how to create a Laurent expansion from there---all of the online examples show something like: $$f(x) = {1 \over z(z-1)}$$ In which it is much more clear how to use a geometric series to find the Laurent series.

I also have the same issue for the following function: $$f(z) = {1 \over z^4+2z^2+1}$$ I can find the singularities, but where do I go from there? The examples found online are tough to map onto these problems.

Let $\alpha=e^{(2n+1)\pi i/6}$ be one of the roots of $z^6+1$ and $\alpha w=z-\alpha$. \begin{align} \frac1{z^6+1} &=\frac1{1-(1+w)^6}\tag{1}\\ &=-\frac1w\frac1{6+15w+20w^2+15w^3+6w^4+w^5}\tag{2}\\ &=\sum_{k=-1}^\infty b_kw^k\tag{3}\\ &=-\frac1{6w}+\frac5{12}-\frac{35}{72}w+\frac{35}{144}w^2+\frac{119}{864}w^3+\dots\tag{4} \end{align}

Explanation:
$(1)$: $z^6+1=1+\alpha^6(1+w)^6=1-(1+w)^6$
$(2)$: Binomial Theorem
$(3)$: label the powers of $w$ in the expansion of $(2)$
$(4)$: multiply both sides of $(2)$ and $(3)$ by $w\!\left(6+15w+20w^2+15w^3+6w^4+w^5\right)$:
$\phantom{(3)\,}$ $\color{#C00000}{-1}=\left(6+15w+20w^2+15w^3+6w^4+w^5\right)\sum\limits_{k=-1}^\infty b_kw^{k+1}$
$\phantom{(3)\,}$ $\phantom{-1}=\color{#C00000}{6b_{-1}}+\color{#00A000}{(15b_{-1}+6b_0)}w+\color{#00A000}{(20b_{-1}+15b_0+6b_1)}w^2$
$\phantom{(3)\,}$ $\phantom{-1}+\color{#00A000}{(15b_{-1}+20b_0+15b_1+6b_2)}w^3+\color{#00A000}{(6b_{-1}+15b_0+20b_1+15b_2+6b_3)}w^4$
$\phantom{(3)\,}$ $\phantom{-1}+\sum\limits_{k=4}^\infty\color{#0000F0}{(b_{k-5}+6b_{k-4}+15b_{k-3}+20b_{k-2}+15b_{k-1}+6b_k)}w^{k+1}$
$\phantom{(3)\,}$ The red term is $-1$ and gives $b_{-1}=-\frac16$
$\phantom{(3)\,}$ The green terms are $0$ and give the other coefficients in $(4)$.
$\phantom{(3)\,}$ The blue term in the sum is $0$ and gives the recursion in $(5)$.

where, for $k\ge4$, $$b_k=-\frac{15b_{k-1}+20b_{k-2}+15b_{k-3}+6b_{k-4}+b_{k-5}}6\tag{5}$$ Then substitute $w=\frac{z-\alpha}\alpha$ into $(3)$ to get $$\frac1{z^6+1}=\sum_{k=-1}^\infty\frac{b_k}{\alpha^k}(z-\alpha)^k\tag{6}$$

• I'm a bit confused about how you went from the first line to the second, and then from the second to the third. Where did the 1/w factor come from, and how did you get the series in the third line from the second? – Marko Bakić Feb 11 '16 at 22:28
• @MarkoBakić: sorry it took so long, but I have added some explanation. – robjohn Feb 12 '16 at 7:52

Remember that the coefficients of the Laurent series around, for example $\;z=i\;$ , are given by

$$a_n=\frac1{2\pi i}\int_C\frac{\frac1{z^6+1}}{(z-i)^{n+1}}dz=\frac1{2\pi i}\int_C\frac{\frac1{(z+i)(z^4-z^2+1)}}{(z-i)^{n+2}}dz$$

So, for example, using Cauchy's Integral formula, we get that

$$a_{-1}=\frac1{2\pi i}\int_C\frac{\frac1{(z+i)(z^4-z^2+1)}}{(z-i)}=\left.\left(\frac1{(z+i)(z^4-z^2+1)}\right)\right|_{z=i}=\frac1{2i(1+1+1)}=\frac1{6i}=-\frac i6$$

and etc. BTW, this already gives you the residue at $\;z=i\;$

As an option, you can try the following (disclaimer: it is going to be very ugly):

$$z^6+1=(z-i+i)^6+1=(z-i)^6+6i(z-i)^5-15(z-i)^4-20i(z-i)^3+15(z-i)^2+6i(z-i)\rlap{\;\;\,\color{red}/}{-1}+\rlap{\color{red}/\;}1\implies$$

$$\frac1{z^6+1}=\frac1{z-i}\,\frac1{6i+15(z-i)-20i(z-i)^2-15(z-i)^3+6i(z-i)^4+(z-i)^5}=$$

$$=\frac1{6i(z-i)}\,\frac1{1+\frac{15}{6i}(z-i)+\ldots+\frac1{6i}(z-i)^5}=$$

$$-\frac i{6(z-i)}\left[1-\left(\frac{15}{6i}(z-i)+\ldots+\frac1{6i}(z-i)^5\right)+\left(\frac{15}{6i}(z-i)+\ldots+\frac1{6i}(z-i)^5\right)^2-..\right]$$

based on the usual development for $\;\frac1{1+z}\;$

If you only need a few summands of the series the above is not too terrible:

$$\frac1{z^6+1}=-\frac i{6(z-i)}+\frac{15}2+\frac{35}{72}i(z-i)+\ldots$$

For the other singularities something similar can be done, but I'm not sure whether it can be made less awful.

Edition by request: for the other function, we have

$$z^4+2z^2+1=(z^2+1)^2=(z-i)^2(z+i)^2$$

Take for example $\;z=-i\;$ , a double pole:

$$\frac1{(z-i)^2(z+i)^2}=\frac i{4(z+i)-4i}\left(\frac1{(z-i)^2}-\frac1{(z+i)^2}\right)=$$

$$=-\frac14\frac1{1+i(z+i)}\left(\frac1{(z+i-(1+i))^2}-\frac1{(z+i)^2}\right)=$$

$$=\frac14\frac1{1+i(z+i)}\frac1{2i}\frac1{\left(1-\frac{z+i}{1+i}\right)^2}-\frac14\left(1-i(z+i)-(z+i)^2+\ldots\right)\frac1{(z+i)^2}=$$

$$=\frac1{8i}\left(1-i(z+i)-(z+i)^2+\ldots\right)\left(1+\frac{z+i}{1+i}+\frac{(z+i)^2}{(1+i)^2}\ldots\right)^2-\\$$

$$-\frac14\left(\frac1{(z+i)^2}-\frac i{z+i}-1+\ldots\right)=...\,\text{etc.}$$

and it gets very nasty. If you mainly require the principal part:

$$a_{-2}=\frac1{2\pi i}\int_C\frac{\frac1{(z-i)^2(z+i)^2}}{(z+i)^{-1}}dz=\frac1{2\pi i}\int_C\frac{\frac1{(z-i)^2}}{(z+i)}dz=\left.\frac1{(z-i)^2}\right|_{z=-i}=-\frac14$$

$$a_{-1}=\frac1{2\pi i}\int_C\frac{\frac1{(z-i)^2(z+i)^2}}{(z+i)^0}dz=\frac1{2\pi i}\int_C\frac{\frac1{(z-i)^2}}{(z+i)^2}dz=\left.\left(\frac1{(z-i)^2}\right)'\right|_{z=-i}=\\$$

$$=\left.-\frac2{(z-i)^3}\right|_{z=-i}=\frac i4$$

If you watch closely in the first part, those two are precisely the coefficients of $\;(z+i)^{-2},\,\,(z+i)^{-1}\;$ respectively, that we got there

• So, since each of the poles is of order one, I can use Cauchy's integral formula to find the principal part for each easily? Then for the rest of the terms in the series, can I use the normal derivative formula for the terms of the taylor series? – Marko Bakić Feb 11 '16 at 19:36
• Well, the principal part is pretty short in each case: all the poles are simple, so yes: it's an easy calculation in this case. Now, using derivatives is always possible, yet I wouldn't call that "easy", In fact, I think it may be very tough and lengthy. – DonAntonio Feb 11 '16 at 19:44
• In the other function from the original problem, the poles aren't simple---what is the procedure in that case? – Marko Bakić Feb 11 '16 at 22:15
• I have surprisingly never seen an example of laurent series with higher order poles... – Marko Bakić Feb 11 '16 at 22:16
• @MarkoBakić Check the part I added to my answer of the other function, with a pole of order two. – DonAntonio Feb 11 '16 at 23:22