find the Laurent series centered at $z=1$ $$ f(z)=\frac{e^z}{(z-1)^2} $$ I thought that the denominator part is safe by our center and the expansion is just about the exponential which is a Taylor series but that doesn't match the calculator solution. Any help is appreciated.
Solution:
so we are good at $(z-1)^2$, then we just need to do Taylor expansion for $e^z$ at $z=1$.( that's the center for our Laurent series), which would be $$ f(z) = \frac{1}{(z-1)^2}\sum \frac{e}{k!}(z-1)^k $$ thanks for everyone making the hint!