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Loser66

  • one year ago

Find all value of \(i^i\) Please, help.

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  1. anonymous
    • one year ago
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    \[i=e^{\frac{\pi}{2}i}\]

  2. Loser66
    • one year ago
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    then?

  3. freckles
    • one year ago
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    \[i^i=(e^{\frac{\pi}{2}i})^i\]

  4. freckles
    • one year ago
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    hint i^2=-1

  5. Loser66
    • one year ago
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    i*i in exponent is not i+i =2i, right?

  6. freckles
    • one year ago
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    i*i is i^2 not i+i

  7. freckles
    • one year ago
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    \[(a^b)^c=a^{ b \cdot c} \text{ not } a^{b+c}\]

  8. Loser66
    • one year ago
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    I don't know how to go on this way. To me, I let \(z = i^i , then~~ e^ {logz} = e^{logi^i} = e^{ilogi}\)

  9. Loser66
    • one year ago
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    That is \(z = e^{ilogi}\)

  10. freckles
    • one year ago
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    doing what satellite says gives us exp(-pi/2)

  11. Loser66
    • one year ago
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    Then let ilog i = w, hence \(z = e^w\) and solve it as usual to get some thing like i^i = {z : ......}

  12. Loser66
    • one year ago
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    Ok, got it

  13. Zarkon
    • one year ago
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    \[e^{i\theta}=\cos(\theta)+i\sin(\theta)\] \[e^{i\frac{\pi}{2}}=\cos\left(\frac{\pi}{2}\right)+i\sin\left(\frac{\pi}{2}\right)=i\]

  14. Loser66
    • one year ago
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    Yes, I got that part @Zarkon @satellite73 @freckles @Jhannybean But my Prof wants me to follow the way he taught in class. :) All I need is the steps to make sure that I don't miss any detail.

  15. Loser66
    • one year ago
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    If I go that way, then \(i^i = e^ {-\pi/2}\) , then how to put + 2kpi in?

  16. freckles
    • one year ago
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    \[e^{i \theta}=e^{i (\theta+2 n \pi)}\]

  17. freckles
    • one year ago
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    \[(e^{(\frac{\pi}{2}+2n \pi) i})^i=e^{-(\frac{\pi}{2}+2n \pi)}\]

  18. Loser66
    • one year ago
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    Actually, my Prof wants me to do something like logz = log|z| + iarg (z) + 2ikpi

  19. Loser66
    • one year ago
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    Hence, I tried to follow that way by making it so complicated like that :)

  20. anonymous
    • one year ago
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    we can do it that way too if you like

  21. anonymous
    • one year ago
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    @Loser66

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