Exploring Complex Numbers and Powers of -1

[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.

 

[CellCoree]

1- Yes, there is a complex number that satisfies x^x = i. This number is approximately 0.438283 + 0.360592i. However, it is important to note that this solution is not unique, as there are infinitely many complex numbers that satisfy this equation.

2- (-1)^(\sqrt{2}) is not a well-defined expression, as the power of a negative number must be a positive integer. However, if we consider the expression (-1)^{\sqrt{2}} instead, we can use the definition of complex powers to find that the result is approximately 0.20787 + 0.97814i. Again, this solution is not unique and there are infinitely many possible values for this expression.