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FANTASMA DE GAVILAN
Lv 7
FANTASMA DE GAVILAN asked in Science & MathematicsMathematics · 1 decade ago

I want to know (i)^i?

z= (0 + 1i) ^i ; i = √ -1

3 Answers

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  • GoldenFibonacci
    Lv 5
    1 decade ago
    Favorite Answer

    Use Euler's theorem:

    e^(ix) = (e^x)^i

    By Euler's Theorem, e^(ipi/2+2pi*k)=i

    So i^i = e^(-pi/2)*e^(2ipi*k) = e^(-pi/2)

  • Anonymous
    1 decade ago

    In polar form:

    i = cos(π/2) + i sin(π/2) ==> i = e^(iπ/2).

    Applying exponential rules:

    i^i

    => [e^(iπ/2)]^i

    = e^(iπ/2 * i)

    = e^(i * i * π/2)

    = e^(i^2 * π/2)

    = e^(-π/2).

    I hope this helps!

  • Time Splinters
    Lv 6
    1 decade ago

    i^i = e^(-pi/2)

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