Question

Show that i^i is a real number.

Answer 1

\[\Large i = e^{\frac{\pi i}{2}}\]

So \(\Large i^i = \left(e^{\frac{\pi i}{2}}\right)^i = e^{\frac{\pi i^2}{2}} = \boxed{e^{-\frac{\pi}{2}}}\)

Answer 2

\[i^i=e^{i\ln(i)}=e^{i[i(\pi/2+2k\pi)]}=e^{-[\pi/2+2k\pi]}, k\in\mathbb{Z}\]

Answer 3

all of which are real. using k=0 give geerky42's particular value (thought there are infinity many)

Answer 4

Ahhh, but can you show that i^i is real without using the identity? ;) \[\Large e^{i \theta} = \cos \theta + i \sin \theta\]

Answer 5

@geerky42 and @Zarkon I had to leave, I'm sorry I wasn't around to clarify this while you were here!

Answer 6

Ah, I remember it now... my friend proved it elegantly by finding out the complex conjugate of this.

Answer 7

Yeah exactly! If a number is equal to its complex conjugate then it must have 0 imaginary part. \[\large (i^i )^*= (-i)^{-i} = \left( \frac{-1}{i} \right)^i=\left( \frac{i^2}{i} \right)^i = i^i\]