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ybarrap

  • one year ago

Challenge Question: What is i^i, where i = sqrt(1)

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  1. AbdullahM
    • one year ago
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    But... \(\sf i = \sqrt{-1}\)

  2. dan815
    • one year ago
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    |dw:1434833839810:dw|

  3. dan815
    • one year ago
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    maybe its better to rewrite in polar form

  4. ybarrap
    • one year ago
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    Yep! $$ \Large{ i^i=\sqrt{-1}^{\sqrt{-1}}=? } $$

  5. dan815
    • one year ago
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    |dw:1434833909276:dw|

  6. dan815
    • one year ago
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    feels like i cheated just using euler form hehe

  7. ybarrap
    • one year ago
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    That's it! Fantastic that an imaginary numbers like that make a real number!

  8. dan815
    • one year ago
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    yes pretty cool

  9. dan815
    • one year ago
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    however -.- i dont think imaginary numbers are any different form real numbers

  10. dan815
    • one year ago
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    the term "imaginary" is needlessly misleading

  11. dan815
    • one year ago
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    @ybarrap what is your interpretation of a number raised to the power of an imaginary number though

  12. ybarrap
    • one year ago
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    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
    • one year ago
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    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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