the original text is from brown's textbook complex variables and applications, which said
we want to find out the series of "$ e^{1/z}$",and since we have already known the maclaurin series expansion $$e^z = \sum_{n=0}^{\infty}\frac{z^n}{n!}$$ we replace $z$ by $1/z$ to get $$e^{1/z}=\sum_{n=0}^{\infty}\frac{1}{n!z^n}$$ note that no positive powers of z appear hear, since the coefficient of the positive powers are zero.
why is that? And there is another text said that
observe that the integrand in expression (3) can be written $f(z)(z-z_0)^{n-1}$.Thus it is clear that when $f$ is actually analytic throughout the disk$|z-z_0|<R_2$, this integrand is too. Hence all of the coefficients $b_n$ are zero
and why is that? the $b_n$ coefficients of Laurent series are $$b_n = \frac{1}{2 \pi i}\int_C \frac{f(z)}{(z-z_0)^{-n+1}}dz$$ Thank you very much if anyone could give my any suggestions.
