Loser66
  • Loser66
Find all value of \(i^i\) Please, help.
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
\[i=e^{\frac{\pi}{2}i}\]
Loser66
  • Loser66
then?
freckles
  • freckles
\[i^i=(e^{\frac{\pi}{2}i})^i\]

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freckles
  • freckles
hint i^2=-1
Loser66
  • Loser66
i*i in exponent is not i+i =2i, right?
freckles
  • freckles
i*i is i^2 not i+i
freckles
  • freckles
\[(a^b)^c=a^{ b \cdot c} \text{ not } a^{b+c}\]
Loser66
  • Loser66
I don't know how to go on this way. To me, I let \(z = i^i , then~~ e^ {logz} = e^{logi^i} = e^{ilogi}\)
Loser66
  • Loser66
That is \(z = e^{ilogi}\)
freckles
  • freckles
doing what satellite says gives us exp(-pi/2)
Loser66
  • Loser66
Then let ilog i = w, hence \(z = e^w\) and solve it as usual to get some thing like i^i = {z : ......}
Loser66
  • Loser66
Ok, got it
Zarkon
  • Zarkon
\[e^{i\theta}=\cos(\theta)+i\sin(\theta)\] \[e^{i\frac{\pi}{2}}=\cos\left(\frac{\pi}{2}\right)+i\sin\left(\frac{\pi}{2}\right)=i\]
Loser66
  • Loser66
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.
Loser66
  • Loser66
If I go that way, then \(i^i = e^ {-\pi/2}\) , then how to put + 2kpi in?
freckles
  • freckles
\[e^{i \theta}=e^{i (\theta+2 n \pi)}\]
freckles
  • freckles
\[(e^{(\frac{\pi}{2}+2n \pi) i})^i=e^{-(\frac{\pi}{2}+2n \pi)}\]
Loser66
  • Loser66
Actually, my Prof wants me to do something like logz = log|z| + iarg (z) + 2ikpi
Loser66
  • Loser66
Hence, I tried to follow that way by making it so complicated like that :)
anonymous
  • anonymous
we can do it that way too if you like
anonymous
  • anonymous

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