## ybarrap one year ago Challenge Question: What is i^i, where i = sqrt(1)

1. AbdullahM

But... $$\sf i = \sqrt{-1}$$

2. dan815

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3. dan815

maybe its better to rewrite in polar form

4. ybarrap

Yep! $$\Large{ i^i=\sqrt{-1}^{\sqrt{-1}}=? }$$

5. dan815

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6. dan815

feels like i cheated just using euler form hehe

7. ybarrap

That's it! Fantastic that an imaginary numbers like that make a real number!

8. dan815

yes pretty cool

9. dan815

however -.- i dont think imaginary numbers are any different form real numbers

10. dan815

the term "imaginary" is needlessly misleading

11. dan815

@ybarrap what is your interpretation of a number raised to the power of an imaginary number though

12. ybarrap

I'm with you, I think Euler -- I think rotations and this particular rotation keeps us on the real axis $$e^{nix}=(\cos x + i \sin x)^n\\ {e^{\pi i/2}}^i\text{ (i.e. a rotation)}=(\cos \pi/2 + i \sin \pi/2)^i=\text{ a real number}\\$$

13. dan815

mhm its just a bit weird to think about, i was just thinking, okay building up exponent laws for integers, fractions, then negative numbers get a bit weird, and imaginary numbers cant wrap my head around it