# What is $\sqrt i^i$?

my answer would be $$i$$ because I used the definition $$i=\sqrt-1$$ and replaced the $$i$$ with that

I'm not sure if I got it right though

• Do you mean $(\sqrt{i})^i$ or $\sqrt{i^i}$? Mar 8 '19 at 15:22
• @Wojowu in either case, the result is the same. Mar 8 '19 at 15:26
• Try considering taking logarithm on both sides of $y=(\sqrt{i})^i$. Mar 8 '19 at 15:27
• You can see $i^i$. Mar 8 '19 at 15:27
• You can see square roots of complex numbers here, in particular $\sqrt{i^i}$. Mar 8 '19 at 15:28

$$\sqrt{i}^i=i^{i/2}=\exp \left(\left(\dfrac{\pi i}{2}+2\pi i n\right)\cdot \dfrac{i}{2}\right)=\exp \dfrac{-\pi}{4}\cdot\exp-\pi n, \ n\in \mathbb{Z}$$
• Complex exponentiation doesn't work like real exponentiation. There are infinitely many values of $i^i$ (and two square roots of each.) Mar 8 '19 at 15:36
• This uses one of the values of $\log i$. There are many more, which will result in a different value Mar 8 '19 at 15:37
• @RossMillikan Would adding $2\pi n, \ n\in \mathbb{Z}$ in the expression for $i$ generalize? Mar 8 '19 at 15:38