Binomial expansion of $(1+x)^i$ where $i^2 = -1$.

Question

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*} $

Answer 1

$(1+x)^i = e^{i\times\ln(1+x)}$ so it is also equals to $\cos(ln(1+x)+i\times\sin(\ln(1+x))$, where defined. So i left it to you to find the corresponding Taylor series.

Answer 2

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 $|x|<1$.

Answer 3

The result was found around 1665 by Isaac Newton in the form of the Binomial Series

$$ (1+x)^{\alpha }=\sum _{k=0}^{\infty }\;{\binom {\alpha }{k}}\;x^{k}=1+\alpha x+{\frac {\alpha (\alpha -1)}{2!}}x^{2}+{\frac {\alpha (\alpha -1)(\alpha -2)}{3!}}x^{3}+\cdots $$

where $ {\binom {\alpha }{k}}:=\frac{1}{k!} {\alpha (\alpha -1)(\alpha -2)\cdots (\alpha -k+1)}$ is the generalized binomial coefficient. This formula holds for any complex $α$ and $|x|<1$. Note that if $α∈ℕ$, then $\binom {\alpha }{k} = 0$ for $k>α$, so in this case it reduces to the regular binomial theorem.