Complex numbers

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

2- (-1)^([itex]\sqrt{2}[/itex])=?
 

Answers and Replies

  • #2
HallsofIvy
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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)^([itex]\sqrt{2}[/itex])=?
We can write -1 in "polar form" as [itex]e^{i\pi}[/itex] and then [itex](-1)^{\sqrt{2}}= (e^{i\pi})^{\sqrt{2}}= e^{i\pi\sqrt{2}}= cos(\pi\sqrt{2})+ i sin(\pi\sqrt{2})[/itex]
or about .99+ .077i.
 
  • #3
Yes, but finding it is non-trivial, involving, I think, the Lambert W function.


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

Thanks.
 
  • #4
haruspex
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1- is there any complex number, x ,such that x^x=i?
Writing z = re, zz = i gives θ sec(θ) eθ tan(θ) = π/2 + 2πn and r = eθ tan(θ). For n = 0, θ has a solution in (π/6, π/4), and probably infinitely many for each n.
 
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