I was reading today about this single variable binomial expansion
$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\ldots$
For example,
$(1+x)^2 = 1+2x+x^2$
However, is it also valid when exponents are imaginary?
$\begin{align*}
(1+x)^i &= 1+ix+\frac{i(i-1)}{2!}x^2+\frac{i(i-1)(i-2)}{3!}x^3+\ldots\\\
&= 1+ix+(-1/2 – i/2)x^2+(1/2+i/6)x^3+(-5/12)x^4+(1/3-i/12)x^5+\ldots
\end{align*}
$
Best Answer
This is just like any other Taylor series; the successive coefficients are just successive derivatives evaluated at $x=1$ with $1^i=1$ divided by the appropriate factorials. Since there is a branch cut for $(1+x)^i$ or with any noninteger exponent, you actually get one branch of the function, specifically the branch that equals $1$ at $x=0$ and is continuous within the convergence region of the series (see below).
With any exponent at all other than a nonnegative integer, convergence is limited by hitting a singularity, which is at $1+x=0$ or $x=-1$. So you converge in $|x|<1$.