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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.

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    $\begingroup$ $\displaystyle{\large z^{a} = \left[1 + \left(z - 1\right)\right]^{\,\,\,a}}$. $\endgroup$ – Felix Marin Jan 9 '14 at 5:54
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The magic words are: the (generalized) binomial theorem.

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

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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$ z^{a} = \bracks{1 + \pars{z - 1}}^{a} = \sum_{n = 0}^{\infty}{a \choose n}\pars{z - 1}^{n}\quad\mbox{where}\quad{a \choose n} \equiv {\Gamma\pars{a + 1} \over n!\,\,\Gamma\pars{a + 1 - n}} $$

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