Value of $i^\sqrt3$

Question

Find all values of $i^\sqrt3$.

I am trying to apply de Moivre's formula here but cannot find a way to do so. I am not sure if i am approaching this wrong.

Answer 1

Hint: by definition, $b^c = \exp(c \log b)$. Use all values of $\log b$.

Answer 2

The de Moivre's formula is useful when you deal with a complex number of the form $e^{ix}$ ($x\in\mathbb{R}$).

When you have exponentiation of a complex number, what you should look for first is the definition, particularly when you have $z^w$ where $z$ is a complex number and $w$ is not an integer.

In your case, $$ i^{\sqrt{3}}:=\exp(\sqrt{3}\log(i)) $$ where $\log$ denotes the complex logarithm, which is a multi-valued function.

Answer 3

Consider: from deMoivre's

$e^{i \theta} = \cos \theta + i \sin \theta \tag 1$

with

$\theta = 2n \pi + \dfrac{\pi}{2}, \; n \in \Bbb Z, \tag 2$

we have

$e^{i(2n\pi + \pi/2)} = i; \tag 3$

thus

$i^{\sqrt 3} = e^{i(2n\pi + \pi/2)\sqrt 3}$ $= \cos ((2n\pi + \pi/2)\sqrt 3) + i \sin ((2n\pi + \pi/2)\sqrt 3), \ n \in \Bbb Z. \tag 4$

We may check for consistency: (4) yields

$-i = i^3 = (i^{\sqrt 3})^{\sqrt 3} $ $= (e^{i(2n\pi + \pi/2)\sqrt 3})^{\sqrt 3} = e^{i(2n\pi + \pi/2)(3)}$ $= e^{6\pi i}e^{3\pi i /2} = -i. ✓ \tag 5$