I have 2 functions. I have to express the function in terms of a Laurent series.
The first function is $f(z) = \frac{z^5}{z - 1}$ in the point $z_0 = 1$ for $1 < \parallel z \parallel < \infty $.
In this case: $$ z^5 = (z - 1 + 1)^5 = \sum_{k = 0}^5 \binom{n}{k}(z - 1)^k $$ so, the problem is solved?
The other function , in the same conditions, is $f(z) = z^5e^{\frac{1}{z}}$. In this case i dont know how to answer.
The problem is indeed solved for $z^5$ (don't forget to divide by the $(z-1)$).
For $z^5e^{z^{-1}}$, find the Taylor expansion of $e^{z^{-1}}$ first, and then multiply with $z^5$.