Finding the real part of z=i^i using De Moivre's formula

[zenite]

1. Find the real part of z=iⁱ by using De Moivre's formula.

Homework Equations

z= r(cos$$\theta$$ + i sin$$\theta$$)
zⁿ= rⁿ(cos(n$$\theta$$) + i sin(n$$\theta$$))


I tried using n=i to solve and got the ans 1ⁱ, but somehow feel that its not that simple. And the resultant argument I got from this approach is i$$\theta$$ which doesn't make sense. Tried using natural log, but didn't work out too.

 

[Codexus]

Start by rewriting in exponential form and then use:

(e^ix)ⁿ = e^inx

That should do the trick

 

[zenite]

z = iⁱ = e^i(lni)
so n=lni and the real part is cos(lni). is this correct?

 

[Codexus]

I'm not sure where your ln(i) comes from but that part is correct since ln(i) = $$i\pi/2$$. However it can be simplified further.

I would have just written:
$$i^{i} = (e^{i\pi/2})^{i} = e^{i i\pi/2} = e^{- \pi/2}$$ and that's your answer since this is a real number already. (Wolfram Alpha confirms it)

 

[zenite]

thanks a lot. I couldn't get the part where lni = i(PI)/2, tried googling but couldn't find anything. but I could understand your working, you make it look so simple.

I used the formula, e^lny = y for my working, that's where the ln comes from. but yours is much more simplified.

 

[Codexus]

I couldn't get the part where lni = i(PI)/2

Well, actually I just used Wolfram Alpha to find that, but if we combine our formulas, we have just proved it's true.