I need to develop Laurent series for $\frac{e^{z}}{z}-z_{0}$.
My attempt:
I have just developed a Taylor series for the exponent and got :$\frac{1}{z}\cdot \sum_{n=0}^{\infty }\frac{z^{n}}{n!}-z_{0}$
But in the solution, they said we need to change $z$ into $z'=z-z_{0}$ and then to develop Taylor series to the exponent.
Whay my solution is wrong?
More than surely, the problem is what Kavi Rama Murthy wrote in comment.
What they say in the solution is a very good trick : when you need a Taylor or Laurent series around $z=z_0$, start letting $z=x+z_0$ and work the expansion around $x=0$. This will save a lor of time and computation effort.
When finished, replace $x$ by $(z-z_0)$.
Just a small hint : for the coefficients in this problem, think about the complete and incomplete gamma functions.
Edit
To help you starting, as said, let $z=x+z_0$ to make $$\frac{e^{z}}{z}-z_{0}=\frac{e^{z_0+x}}{z_0+x}-z_0=e^{z_0}\frac{e^{x}}{z_0+x}-z_0$$