Classify the singularities at \(z=0\) of the following \[ f_1(z) &= {\Log(1+z) \sin(z) \over z^2} \\ f_2(z) &= e^{\sin\qty{1\over z}} \\ f_3(z) &= {1+z \over e^z-1} .\]
Solution
\(f_1\): removable, evident from Laurent expansion at \(z=0\): \[ z^{-2}\Log(1+z)\sin(z) &= \inverseof{z} \qty{ \sum_{k\geq 1} { (-z)^k \over k} } \qty{z - { z^3 \over 3!} + { z^5 \over 5!} - \cdots} \\ &= z^{-2} \qty{ -z+ { z^2 \over 2} - { z^3 \over 3} + \cdots} \qty{z - { z^3 \over 3!} + { z^5 \over 5!} - \cdots} \\ &= z^{-2}\qty{ z^2(-1) + z^3\qty{1\over 2} + z^2\qty{ -{ 1 \over 3!} - { 1 \over 3} } + \cdots } \\ &= 1 + {z \over 2} + \cdots .\]
\(f_2\): essential, evident from a sequence like \(z_k \da \inverseof{\qty{k\cdot {\pi\over 2} }}\) which makes \(\sin(z_k)\) oscillate between 0 and 1.
\(f_3\): pole of order 1 with residue 1, evident after some slightly clever Laurent manipulations: \[ {1\over e^z-1} &= {1 \over z + {1\over 2}z^2 + \cdots} \\ &= {1 \over z\qty{1 + {1\over 2}z + \cdots} } \\ &\da {1\over z \qty{1 + p(z)}} && p(z) \da {1\over 2} z + {1\over 3!}z^2 + \cdots \\ &= \inverseof{z} \sum_{k\geq 0}(-p(z))^k z^k \\ &= \inverseof{z} \qty{1 - zp(z) + z^2p(z)^2 - z^3p(z)^3 + \bigo(z^4)} \\ &= \inverseof{z}\qty{1- \bigo(z^2) + \bigo(z^4) - \bigo(z^6) } \\ &={1\over z} - \bigo(z) + \bigo(z^3) .\]