Question

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

Answer 1

\[i=e^{\frac{\pi}{2}i}\]

Answer 2

then?

Answer 3

\[i^i=(e^{\frac{\pi}{2}i})^i\]

Answer 4

hint i^2=-1

Answer 5

i*i in exponent is not i+i =2i, right?

Answer 6

i*i is i^2 not i+i

Answer 7

\[(a^b)^c=a^{ b \cdot c} \text{ not } a^{b+c}\]

Answer 8

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

Answer 9

That is \(z = e^{ilogi}\)

Answer 10

doing what satellite says gives us exp(-pi/2)

Answer 11

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

Answer 12

Ok, got it

Answer 13

\[e^{i\theta}=\cos(\theta)+i\sin(\theta)\]
\[e^{i\frac{\pi}{2}}=\cos\left(\frac{\pi}{2}\right)+i\sin\left(\frac{\pi}{2}\right)=i\]

Answer 14

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.

Answer 15

If I go that way, then \(i^i = e^ {-\pi/2}\) , then how to put + 2kpi in?

Answer 16

\[e^{i \theta}=e^{i (\theta+2 n \pi)}\]

Answer 17

\[(e^{(\frac{\pi}{2}+2n \pi) i})^i=e^{-(\frac{\pi}{2}+2n \pi)}\]

Answer 18

Actually, my Prof wants me to do something like logz = log|z| + iarg (z) + 2ikpi

Answer 19

Hence, I tried to follow that way by making it so complicated like that :)

Answer 20

we can do it that way too if you like

Answer 21

@Loser66