Simplifying $x^i$ to real numbers

Question

I have been studying power functions, and started to think about imaginary powers. Take the function $x^i$. Because I don't know how to multiply a number $i$ times, I tried to simplify the equation

$x^i = x^{\sqrt{-1}} = x^{(-1)^{1/2}}$

Then, using the property of exponents that states an exponent to an exponent is the two multiplied, I get

$x^{(-1)^{1/2}} = x^{-1/2} = (\sqrt{x})^{-1} = \frac{1}{\sqrt{x}}$

However, plugging in any number shows that

$x^i \neq \frac{1}{\sqrt{x}}$

Where did I go wrong?

Answer 1

Then, using the property of exponents that states an exponent to an exponent is the two multiplied,

This is where you go wrong. You are thinking of $\left(a^b\right)^c=a^{(bc)}$, but you have $a^{(b)^c}$. That's not the same thing at all.

Thus $x^{(-1)^{1/2}}\neq x^{-1/2}$ because: $(-1)^{1/2}\neq -1/2$

Answer 2

Your definition works for Real numbers. But $(-1)^{1/2}$ is not a Real number. You need to use the complex logz, meaning you need to choose a branch of logz, so that you can define: $$x^i=e^{ilogx} $$;let $x=a+ib$, then this is equal to: $$e^{i(a+ib)}=e^{-b+ia} $$. If you do not select a branch of logz, you have a many-valued "function", i.e., you have many candidates for the logz.

once you have a branch, this is well-defined and single-valued.