$f\left( z\right) =\dfrac {e^{-z}}{\left( z-1\right) \left( z+2\right) ^{2}}$ function $0 <\left| z+2\right| <3$ open the Laurent series in the area

Question

$f\left( z\right) =\dfrac {e^{-z}}{\left( z-1\right) \left( z+2\right) ^{2}}$ function $0 <\left| z+2\right| <3$ open the Laurent series in the area

How do I edit a function?

Answer 1

$f\left( z\right) =\dfrac {e^{-z}}{\left( z-1\right) \left( z+2\right) ^{2}}$ 

$\dfrac {1}{z-1}=\dfrac {1}{z+2-3}=\dfrac {1}{3\left( \dfrac {z+2}{3}-1\right) }$

$=-\dfrac {1}{3}\dfrac {1}{1-\dfrac {z+2}{3}}$

$=-\displaystyle\sum ^{\infty }_{n=0}\dfrac {\left( z+2\right) ^{n}}{3^{n+1}}$

$f(z)=-e^{-z}\displaystyle\sum ^{\infty }_{n=0}\dfrac {z+2^{n-2}}{3^{n+1}}$   ,   $0<|z+2|<3$ 

what are wrongs in the solution?