why is $\cos x + i \sin x = i^\frac{x}{90^\circ}$

Question

I was calculating $\sqrt{i}$ and $\sqrt[3]{i}$,

and I found out that in both of those cases,
$\cos x + i \sin x = i^\frac{x}{90}$
this formula is right.
and this formula is right on x=0,30,45,90,360
But I can't prove it. Can anybody prove this?

Answer 1

EDIT it seems I cannot divide fractions correctly, I have updated my answer

We first start with what you have (I switched to radians though)

$$ i^{\frac{2x}{\pi}} = e^{\frac{2x}{\pi}\log i} $$

using the complex log definition we know, assuming we are working on just one rotation of the unit circle $[0,2\pi)$

$$ \log i = \log(1) + i\left(\dfrac{\pi}{2}\right) $$

so we get that

$$ i^{\frac{2x}{\pi}} = e^{i\frac{2x}{\pi}\frac{\pi}{2}} = e^{ix} $$

which by Euler's identity is

$$ \cos(x)+i\sin(x) $$

thus proving your identity Q.E.D

Answer 2

Powers of complex numbers are typically defined using the main branch of $\ln$ with complex argument in the range $[0, 2\pi)$, as $$i^\theta = \exp (\theta \ln i) = \exp \left(\frac{i\pi \theta}{2}\right) = \cos \left(\frac{\pi \theta}{2}\right) + i \sin \left(\frac{\pi \theta}{2}\right)$$ where the argument to the trigonometric functions is in radians. This is equivalent to your formula with arguments in degrees.