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Loser66
 one year ago
Find all value of \(i^i\)
Please, help.
Loser66
 one year ago
Find all value of \(i^i\) Please, help.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[i=e^{\frac{\pi}{2}i}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[i^i=(e^{\frac{\pi}{2}i})^i\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0i*i in exponent is not i+i =2i, right?

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[(a^b)^c=a^{ b \cdot c} \text{ not } a^{b+c}\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0That is \(z = e^{ilogi}\)

freckles
 one year ago
Best ResponseYou've already chosen the best response.0doing what satellite says gives us exp(pi/2)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Then let ilog i = w, hence \(z = e^w\) and solve it as usual to get some thing like i^i = {z : ......}

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1\[e^{i\theta}=\cos(\theta)+i\sin(\theta)\] \[e^{i\frac{\pi}{2}}=\cos\left(\frac{\pi}{2}\right)+i\sin\left(\frac{\pi}{2}\right)=i\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Yes, 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
 one year ago
Best ResponseYou've already chosen the best response.0If I go that way, then \(i^i = e^ {\pi/2}\) , then how to put + 2kpi in?

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[e^{i \theta}=e^{i (\theta+2 n \pi)}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[(e^{(\frac{\pi}{2}+2n \pi) i})^i=e^{(\frac{\pi}{2}+2n \pi)}\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Actually, my Prof wants me to do something like logz = logz + iarg (z) + 2ikpi

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Hence, I tried to follow that way by making it so complicated like that :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we can do it that way too if you like
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