Can complex numbers solve x^x = i and what is (-1)^{\sqrt{2}}?

[limitkiller]

1- is there any complex number, x ,such that x^x=i?

2- (-1)^($\sqrt{2}$)=?

 

[HallsofIvy]

1- is there any complex number, x ,such that x^x=i?

Yes, but finding it is non-trivial, involving, I think, the Lambert W function.

2- (-1)^($\sqrt{2}$)=?

We can write -1 in "polar form" as $e^{i\pi}$ and then $(-1)^{\sqrt{2}}= (e^{i\pi})^{\sqrt{2}}= e^{i\pi\sqrt{2}}= cos(\pi\sqrt{2})+ i sin(\pi\sqrt{2})$
or about .99+ .077i.

 

[limitkiller]

Yes, but finding it is non-trivial, involving, I think, the Lambert W function.


We can write -1 in "polar form" as $e^{i\pi}$ and then $(-1)^{\sqrt{2}}= (e^{i\pi})^{\sqrt{2}}= e^{i\pi\sqrt{2}}= cos(\pi\sqrt{2})+ i sin(\pi\sqrt{2})$
or about .99+ .077i.

Thanks.

 

[haruspex]

1- is there any complex number, x ,such that x^x=i?

Writing z = re^iθ, z^z = i gives θ sec(θ) e^{θ tan(θ)} = π/2 + 2πn and r = e^{θ tan(θ)}. For n = 0, θ has a solution in (π/6, π/4), and probably infinitely many for each n.