Find a Laurent expansion for \[ f(z) = {1\over 1 + e^z} \] about \(z_0 = 0\) and \(z_1 = i\pi\).
Solution
At \(z=0\), we can use a geometric series approach since \(\abs{e^z} = e^{\Re(z)} \leq 1\) near \(0\). However, we still have to get rid of the leading 1 in the expansion of \(e^z\) in order to get a constant coefficient. \[ {1\over 1 + e^z} &= {1\over 1 + 1 + z + {1\over 2!}z^2 + {1\over 3!} z^3 + \bigo(z^4)} \\ &= {1\over 2 + z + {1\over 2!}z^2 + {1\over 3!} z^3 + \bigo(z^4) } \\ &= {1\over 2} {1\over 1 + {1\over 2} z + {1\over 2\cdot 2!}z^2 + {1\over 2\cdot 3!} z^3 + \bigo(z^4) } \\ &= {1\over 2}{1\over 1 - (-p(z)) } \qquad p(z) \da {1\over 2}z + {1\over 2\cdot 2!}z^2 + {1\over 2\cdot 3!}z^3 + \bigo(z^4) \\ &= {1\over 2} \sum_{k\geq 0} (-p(z))^k \\ &= {1\over 2}\Big[ 1 - \qty{{1\over 2}z + {1\over 2\cdot 2!}z^2 + {1\over 2\cdot 3!}z^3 + \bigo\qty{z^4} } \\ &\qquad + \qty{ {1\over 2}z + {1\over 2\cdot 2!}z^2 + {1\over 2\cdot 3!}z^3 + \bigo\qty{z^4} }^2 \\ &\qquad - \qty{{1\over 2} z + {1\over 2\cdot 2!}z^2 + {1\over 2\cdot 3!}z^3 + \bigo\qty{z^4} }^3 \\ &\qquad - \bigo(z^4) \Big]\\ &= {1\over 2} \Big[ 1 + z\qty{- {1\over 2}} + z^2\qty{- {1\over 2\cdot 2!} + \qty{1\over 2}^2}\\ &\qquad + z^3 \Big( -{1\over 2\cdot 3! } +\left[ \qty{1\over 2}^3 + {1\over 2}{1\over 2\cdot 2!}\right] - \qty{1\over 2}^3 \Big) \\ &\qquad + \bigo(z^4) \Big]\\ &= {1\over 2} - {1\over 4}z + 0z^2 + {1\over 48}z^3 + \bigo(z^4) .\]
Expanding at \(z-i\pi\): quite a bit easier. Let \(\omega \da z-i\pi\), then \[ {1\over 1 + e^z} &= {1\over 1 + e^{z-i\pi}e^{i\pi}} \\ &= {1\over 1 - e^{\omega} } \\ &= {1 \over -\omega - {1\over 2!}\omega^2 - {1\over 3!}\omega^3 - \bigo(\omega^4) } \\ &= -{1\over \omega} {1 \over 1 + {1\over 2!}\omega + {1\over 3!}\omega^2 + \bigo(\omega^3) } \\ &= -{1\over \omega}{1\over 1-(- p(z) ) } \qquad p(z) \da \sum_{k\geq 2}{\omega^{k-1}\over k!} \\ &= -{1\over \omega} \sum_{k\geq 0} (-p(z))^k \\ &= -{1\over \omega} \left[ 1 - \qty{{1\over 2!}\omega + {1\over 3!}\omega^2 + \bigo(\omega^3)} + \qty{{1\over 2!}\omega + {1\over 3!}\omega^2 + \bigo(\omega^3)}^2 - \bigo(\omega^3) \right] \\ &= -{1\over \omega} \left[ 1 + \omega\qty{-{1\over 2!}} + \omega^2\qty{-{1\over 3!} + \qty{1\over 2!}^2} + \bigo(\omega^3) \right] \\ &= -{1\over w} + {1\over 2} - {1\over 12}\omega + \bigo(\omega^2) .\]