Find a Laurent expansion about \(z=0\) of \[ f(z) \da \cos\qty{1- {1\over z}} ,\] and compute the “residue” coefficient \(c_{-1}\).
Solution
Write \(g(z) \da \cos(1-z)\), so \(g(1/z) = f(z)\), and expand: \[ g(z) &= \cos(1-z) \\ &= {1\over 2}\qty{e^{i(1-z)} + e^{-i(1-z)}} \\ &= {1\over 2}\qty{e^i e^{-iz} + e^{-i} e^{iz}}\\ &= {1\over 2}\sum_{k\geq 0} \qty{ (-i)^k e^i + i^k e^{-i} } {z^k \over k!} \\ \implies f(z) &= {1\over 2}\sum_{k\geq 0} \qty{ (-i)^k e^i + i^k e^{-i} } {1 \over k!z^k} .\]
Taking \(k=1\) yields \[ c_{-1} = {-ie^i + ie^{-i} \over 2} = -i\cdot {e^i - e^{-i}\over 2} = {e^i - e^{-i}\over 2i} = \sin(1) .\]