## Loser66 one year ago Find all value of $$i^i$$ Please, help.

1. anonymous

$i=e^{\frac{\pi}{2}i}$

2. Loser66

then?

3. freckles

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

4. freckles

hint i^2=-1

5. Loser66

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

6. freckles

i*i is i^2 not i+i

7. freckles

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

8. Loser66

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

9. Loser66

That is $$z = e^{ilogi}$$

10. freckles

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

11. Loser66

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

12. Loser66

Ok, got it

13. Zarkon

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

14. 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.

15. Loser66

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

16. freckles

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

17. freckles

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

18. Loser66

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

19. Loser66

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

20. anonymous

we can do it that way too if you like

21. anonymous

@Loser66