Find a Laurent expansion at \(z=0\) for \[ f(z) \da e^{1\over z}\cos\qty{1\over z} .\]
Solution
Let \(g(z) \da e^z\cos(z)\), an entire function with \(g(1/z ) = f(z)\) for \(z\ne0\). The Taylor series of \(g\) at \(0\), evaluated at \(1/z\), is the Laurent series of \(f\) on \(\abs z>0\): \[ g(z) &= e^{z}\cos(z)\\ &= {1\over 2}e^z\qty{e^{iz} + e^{-iz}} \\ &= {1\over 2}\qty{e^{(1+i)z} + e^{(1-i)z}} \\ &= {1\over 2} \sum_{k\geq 0}\qty{(1+i)^k + (1-i)^k} {z^k\over k!} \\ \implies f(z) &= {1\over 2} \sum_{k\geq 0}\qty{(1+i)^k + (1-i)^k} {1 \over k!z^k } \\ .\]