How do I obtain the Laurent series for $f(z)=\frac 1{\cos(z^4)-1}$ about $0$? I know that $$\cos(z^4)-1=-\frac{z^8}{2!}+\frac{z^{16}}{4!}+...$$ but how do I take the reciprocal of this series (please do not use little-o notation)? Or are there better methods to obtain the required series?

Long Division ... still worth learning  

To get more terms in the quotient, use more terms in the divisor.
