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.
 A: 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
