z is a complex number. find the taylor series expansion for g(z)=1/(z^3) about z0= 2.in what domain does the taylor series of g converge. z0 is z subscript 0
The Attempt at a Solution
I wrote g(z)=1/(z^3) = 1/(2+(z^3)-2) = (1/2)*1/(1+(z^3 -2)/2) then i was thinking I could use the identity 1/(1+z)=summation( (z^n) / (n!) ) when |z|<1 so then the expansion would be the summation (z^3 -2)^n / 2^n+1 in the domain |z^3 - 2| <2 I think this is right but it's not the taylor series expansion because I have the z^3 in there. the domain needs to be in the form |z-2| < R and in the summation I need to have (z-2)^n not (z^3 - 2)^n How do I get rid of the z^3 and have the answer in the right form?