Can one expand $(a + b)^{{i}}$?

Question

I'm solving a transcendental equation and I currently have: $$-i\ln(ix + \sqrt{1 - x^2}) = \frac{1}{(i\ln(x) + \sqrt{1 - \ln(x)^2})^i}$$ I don't know how to binomially expand the bottom denominator expression with $i$, you cannot expand it with the same rules applied to binomials with positive integer powers, because when you organize nCR rows for each term, R cannot increase to $i$ because it isn't a real number - is the following binomial expression possible to expand? $$(a + b)^i$$

Answer 1

Expand $\left( a + b \right)^{i}$

Expansions

There are several ways to expand $\left( a + b \right)^{i}$: $$ \begin{align*} \left( a + b \right)^{i} &= \operatorname{_{2}F_{1}}\left( -i,\, y;\, y; 1 - a - b \right),\, \quad \left( \forall y \right) \tag{1.}\\ \left( a + b \right)^{i} &= -i \cdot \left( 1 - a - b \right)^{i} \cdot \operatorname{B}\left( 1 - a - b;\, -i,\, i \right) \tag{2.}\\ \left( a + b \right)^{i} &= \exp\left( \ln\left( a + b \right) \cdot i \right) \tag{3.}\\ \left( a + b \right)^{i} &= \cos(\ln\left( a + b \right)) + \sin(\ln\left( a + b \right)) \cdot i \tag{4.}\\ \end{align*} $$

Where $\operatorname{_{2}F_{1}}\left( a,\, b;\, c; z \right)$ is the Gaussian hypergeometric function and $\operatorname{B}\left( x;\, p,\, q \right)$ or $\operatorname{B}_{x}\left( p,\, q \right)$ is the Euler's incomplete beta function.

Derivations

$\left( 1. \right)$: $$ \begin{align*} \left( 1 - z \right)^{-x} &= \operatorname{_{2}F_{1}}\left( x,\, y;\, y; z \right),\, \quad \left( \forall y \right)\\ \left( a + b \right)^{i} &= \left( 1 - z \right)^{-x}\\ i = -x &\wedge a + b = 1 - z\\ -i = x &\wedge 1 - a - b = z\\ \end{align*} $$

$\left( 2. \right)$: $$ \begin{align*} \operatorname{B}\left( x;\, p,\, q \right) &= \frac{x^{p}}{p} \cdot \operatorname{_{2}F_{1}}\left( p,\, 1 - q;\, 1 + p; x \right)\\ \operatorname{B}\left( x;\, -i,\, i \right) &= \frac{x^{-i}}{-i} \cdot \operatorname{_{2}F_{1}}\left( -i,\, 1 - i;\, 1 - i; x \right)\\ -i \cdot x^{i} \cdot \operatorname{B}\left( x;\, q,\, p \right) &= \operatorname{_{2}F_{1}}\left( p,\, 1 - p;\, 1 + q; x \right)\\ \end{align*} $$

$\left( 3. \right)$: $$ \begin{align*} \left( a + b \right)^{i} &= \exp\left( \ln\left( \left( a + b \right)^{i} \right) \right)\\ \left( a + b \right)^{i} &= \exp\left( \ln\left( a + b \right) \cdot i \right)\\ \end{align*} $$

$\left( 4. \right)$: $$ \begin{align*} \left( a + b \right)^{i} &= \exp\left( \ln\left( a + b \right) \cdot i \right) \quad\mid\quad \text{Euler's formula}\\ \left( a + b \right)^{i} &= \cos(\ln\left( a + b \right)) + \sin(\ln\left( a + b \right)) \cdot i\\ \end{align*} $$

Simplification

If you want to solve the equation, you might want to note that you can simplify it even further using the complex logarithmic forms of the inverse of trigonometric functions $\arcsin\left( z \right) = -i \cdot \ln\left( \sqrt{1 - z^{2}} + i \cdot z \right) = i \cdot \ln\left( \sqrt{1 - z^{2}} - i \cdot z \right)$:

$$ \begin{align*} -i \cdot \ln\left( i \cdot x + \sqrt{1 - x^{2}} \right) &= \frac{1}{\left( i \cdot \ln\left( x \right) + \sqrt{1 - \ln\left( x \right)^{2}} \right)^{i}}\\ \arcsin\left( x \right) &= \exp\left( \arcsin\left( -\ln\left( x \right) \right) \cdot i \right)^{-i}\\ \end{align*} $$

With this simplification, only $2\,$ $x$s remain.

Answer 2

See the comment of jjag

Newton's binomial theorem is $$ (1+x)^q = \sum_{k=0}^\infty \binom{q}{k} x^k $$ where $q,x$ are complex numbers, $|x|<1$, and the binomaial coefficient is $$ \binom{q}{k} = \frac{q(q-1)(q-2)\cdots(q-k+1)}{k!} . $$

Then we can do $$ (a+b)^q = a^q\;\left(1+\frac{b}{a}\right)^q = \sum_{k=0}^\infty \binom{q}{k} a^{q-k}b^k $$ for $|a|>|b|$.

For $q=i=\sqrt{-1}$ we get $$ {i\choose 0}=1,\\ {i\choose 1}=i,\\ {i\choose 2}=-{\frac{1}{2}}-{\frac {i}{2}},\\ {i\choose 3}={\frac{1}{2}}+{\frac {i}{6}},\\ {i\choose 4}=-{\frac{5}{12}},\\ {i\choose 5}={\frac{1}{3}}-{\frac {i}{12}} $$