# power series expansion of $z^a$ at $z = 1$

I'm working through some problems in a complex analysis book, and one asks to compute the power series expansion of $f(z) = z^a$ at $z = 1$, where $a$ is a positive real number. The series should represent the branch that maps $(0,\infty) \rightarrow (0,\infty)$.

How do I compute this power series? I tried calculating several of the derivatives of $f$, and they seemed to get messy without an easily identifiable pattern. I also tried examining the Cauchy integral representation for the $n^{th}$ derivative of $f$, but that didn't get me any further. [edit:] I was representing the function as $e^{a*Log(z)}$, where $Log$ is the principal branch. I guess this answers my second question about different branches.

Secondly, how would I go about calculating the power series for a different branch? The choice of branch didn't figure into any of the computations I tried (and I think that's a problem).

Thirdly, how would I calculate the power series when $a$ is not necessarily real? Does this differ significantly from the real case?

I've already read some about branch cuts and phase factors of functions like $f(z) = z^a$ for complex $a$, but I was hoping this problem on power series might give me another perspective on the matter. From an adjacent problem I suspect that factorials are involved in the power series, but I don't see the connection.

• $\displaystyle{\large z^{a} = \left[1 + \left(z - 1\right)\right]^{\,\,\,a}}$. – Felix Marin Jan 9 '14 at 5:54

Perhaps you might use the binomial formula for complex numbers: $(z_1+z_2)^a=z_1^a+\frac {a}{1!}z_1^{(a-1)}z_2+\frac {a(a-1)}{2!}z_1^{(a-2)}z_2^2+...+\frac{a(a-1)(a-2)...(a-k+1)}{k!}z_1^{(a-k)}z_2^k+...+z_2^a$