0
$\begingroup$

How can I find the power series for something like $z^{3/2}$?

I'm totally stuck in it and don't know how to start.

Thanks

1 Answers 1

1

First, you can't develop $z\mapsto z^{3/2}$ in a power series around $0$ because you have a branch cut $]-\infty,0]$. However, it is usual to develop such a function around $1$. (Precisely the disk of convergence is $D(1,1).$)

Writing $$z^{3/2} = (1+(z-1))^{3/2}$$ and using the general Newton's binomial formula, you find $$z^{3/2}=\sum_{n=0}^{+\infty} \binom{3/2}{n}(z-1)^n.$$

  • 0
    is there any way to find the coefficients without using the binomial theorem? anything related directly to complex, maybe? – 2017-02-25
  • 0
    Well you can directly compute the derivatives of $z^{3/2}$ and then evaluate them in $z=1$ to find the coefficients of the series, but you will see that the binomial's coefficients appear. – 2017-02-25
  • 0
    thanks a lot for your help. – 2017-02-25