ybarrap
  • ybarrap
Challenge Question: What is i^i, where i = sqrt(1)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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AbdullahM
  • AbdullahM
But... \(\sf i = \sqrt{-1}\)
dan815
  • dan815
|dw:1434833839810:dw|
dan815
  • dan815
maybe its better to rewrite in polar form

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ybarrap
  • ybarrap
Yep! $$ \Large{ i^i=\sqrt{-1}^{\sqrt{-1}}=? } $$
dan815
  • dan815
|dw:1434833909276:dw|
dan815
  • dan815
feels like i cheated just using euler form hehe
ybarrap
  • ybarrap
That's it! Fantastic that an imaginary numbers like that make a real number!
dan815
  • dan815
yes pretty cool
dan815
  • dan815
however -.- i dont think imaginary numbers are any different form real numbers
dan815
  • dan815
the term "imaginary" is needlessly misleading
dan815
  • dan815
@ybarrap what is your interpretation of a number raised to the power of an imaginary number though
ybarrap
  • ybarrap
I'm with you, I think Euler -- I think rotations and this particular rotation keeps us on the real axis $$ e^{nix}=(\cos x + i \sin x)^n\\ {e^{\pi i/2}}^i\text{ (i.e. a rotation)}=(\cos \pi/2 + i \sin \pi/2)^i=\text{ a real number}\\ $$
dan815
  • 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

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