# Complex Numbers course work

1. Dec 3, 2009

### zenite

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

2. Relevant 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 arguement I got from this approach is i$$\theta$$ which doesn't make sense. Tried using natural log, but didn't work out too.

2. Dec 3, 2009

### Codexus

Start by rewriting in exponential form and then use:

(eix)n = einx

That should do the trick

3. Dec 3, 2009

### zenite

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

4. Dec 3, 2009

### 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)

5. Dec 3, 2009

### zenite

thanks alot. 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, elny = y for my working, thats where the ln comes from. but yours is much more simplified.

6. Dec 3, 2009

### Codexus

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