Find a Laurent expansion about \(z_0=0\) for \(f(z) = \tan(z)\) by using polynomial long division on the series expansions for \(\sin(z)\) and \(\cos(z)\).
Solution
Computing the Laurent series for \(\tan(z)\) at \(z=0\):
\[ {\sin(z) \over \cos(z)} &= {z - {1\over 3!}z^3 + {1\over 5!} z^5 + \bigo(z^7) \over 1 - {1\over 2!} z^2 + {1\over 4!}z^4 + \bigo(z^6) } \\ \\ z - {1\over 3!}z^3 + {1\over 5!}z^5 &= \qty{1 - {1\over 2!}z^2 + {1\over 4!}z^4}(z) \ +\qty{ \qty{{1\over 2!} - {1\over 3!}}z^3 + \qty{-{1\over 4!} + {1\over 5!} }z^5 } \\ &= \qty{1 - {1\over 2!}z^2 + {1\over 4!}z^4}(z) \ +\qty{ {1\over 3}z^3 -{1\over 30}z^5 } \\ \\ {1\over 3}z^3 -{1\over 30}z^5 &= \qty{1 - {1\over 2!}z^2 + {1\over 4!}z^4}\qty{{1\over 3} z^3} \ +\qty{ \qty{ {1\over 6} - {1\over 30} } z^5 + \qty{ -{1\over 3\cdot 24} }z^7 } \\ &= \qty{1 - {1\over 2!}z^2 + {1\over 4!}z^4}\qty{{1\over 3} z^3} \ + \qty{ {2\over 15}z^5 - {1\over 72}z^7 } \\ \\ { {2\over 15}z^5 - {1\over 72}z^7 } &= \qty{1 - {1\over 2!}z^2 + {1\over 4!}z^4}\qty{{2\over 15} z^5} + \cdot \\ \\ \implies \tan(z) &= z + {1\over 3}z^3 + {2\over 15 }z^5 + \bigo(z^7) .\]