Determine all roots of $i^{1/4}$

Question

trying to find all algebraic expressions for ${i}^{1/4}$.

Using. Le Moivre formula , I managed to get this :

${i}^{1/4}=\cos(\frac{\pi}{8})+i \sin(\frac{\pi}{8})=\sqrt{\frac{1+\frac{1}{\sqrt{2}}}{2}} + i \sqrt{\frac{1-\frac{1}{\sqrt{2}}}{2}}$

What's about other expressions.

Answer 1

Hint:. If $\exp(i\theta)^4=\exp(i4\theta)=i$, then $4\theta+2\pi n=\dfrac {\pi}2$, where $n\in\Bbb Z$. If you solve for $\theta, $ you should get the other solutions besides $\theta=\dfrac{\pi}8$.

Answer 2

The fourth roots are spaced on a circle, equally partitioning it. Thus, if you know one, you can find the rest by rotating by $π/2,$ or (which is the same thing) by multiplying by $i.$

Thus, since one of them as you found is $\cos(π/8)+i\sin(π/8),$ the others are $-\sin(π/8)+i\cos(π/8),$ etc.

Answer 3

Let this first root be $z$, and the other ones $zw$. Then

$$i=(zw)^4=z^4w^4=iw^4$$ and $w^4=1$. Hence you multiply $z$ by the fourth roots of unity, $1,i,-1,-i$.

Answer 4

If $(a+bi)^2=c+di$, then:

$$a=\pm\sqrt{\frac{c\pm |c+di|}{2}},b=\pm\sqrt{\frac{-c\pm |c+di|}{2}}$$

First we get the result: $$\sqrt{\lambda i}=\pm\sqrt{\frac\lambda 2}(1+i)$$

and then $$\sqrt[4]{\lambda i}=\sqrt[4]{\frac\lambda 2}\bigg(\pm\sqrt{\frac{1\pm\sqrt2}{2}}\pm i\sqrt{\frac{-1\pm\sqrt 2}{2}}\bigg)$$

What you seek is when $\lambda=1$ (the solutions are more exhaustive than necessary because this set of solutions considers $(p)+(q)i$ and $(qi)+(-pi)i$ as distinct)