$e^{1/z}$ and Laurent expansion

Question

$e^\frac1z$ is not holomorphic at $z=0$, but it is known that it can be expanded as $$e^\frac1z=1+\frac1z+\frac1{2!z^2}+\frac1{3!z^3}+\cdots$$ The coefficients of this Laurent expansion are computed the same way as Taylor's. The question is how is that possible? If function is not holomoprhic at $z=0$, then it's not true that it is holomophic at $|z|<R$ and Taylor's coefficients can not be used. Please someone explain.

Answer 1

If $f$ is an entire function then $$ f(z)=a_0 +a_1z+a_2z^2+\ldots. $$ From Cauchy's integral formula $a_k=\frac{1}{2 \pi i} \oint_\gamma\frac{f(z)}{z^{k+1}}dz, \ k\in \mathbb{N}$ where $\gamma$ is the unit circle centered at zero. This means that for all $z\in\mathbb C$ $$ f(z)=a_0 +a_1z+a_2z^2+\ldots. $$ In particular the above equality holds for all non-zero complex numbers.

In our case the function $f(z)=e^z$ is entire with $$ f(z)=\sum_{n=0}^\infty\frac{z^n}{n!}. $$ Since the above equality holds for all non-zero complex numbers it follows that $$ e^{\frac{1}{z}}=f\left(\frac{1}{z}\right)=\sum_{n=0}^\infty\frac{1}{n!z^n},\qquad \forall z\in\mathbb C. $$ The above formula for (the Laurent series for) $f(1/z)$ was derived from the Taylor series of $f(z)$ by substituting $1/z$ for $z$ since the Taylor series formula holds for all non-zero complex numbers.