= An exponential has no isolated pole
{title2=$\exp(g)\text{ has no pole}$}
If $g$ is <holomorphic> on a punctured disc about $a$, the function $e^g$ cannot have a <pole> at $a$. Indeed, $(e^g)'/e^g=g'$, whereas a <pole> of order $n\geq1$ would give <residue> $-n$ for this <logarithmic derivative>. The <Laurent series> derivative of $g$ has <residue> zero, since its only candidate term comes from differentiating the constant coefficient. This contradiction rules out a <pole>, regardless of the type of the <isolated singularity> of $g$.
Back to article page
