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You could always use my historical proof that iⁱ = e^-π/2 ∈ R. I did it for a talent show and won, so I guess it could be used as good mathematical magic. It was a great day for mathematics. If you would like me to write out the problem, just ask.

Hopefully we all know Euler's formula:

For this problem, let 0 ≤ θ ≤ 2π. Now ask yourself, for what value of θ will eⁱθ = i? Why yes, it's π/2!

So now we know that eⁱπ/2 = i. Now let's take it to the next level:

Now if we remove the restriction 0 ≤ θ ≤ 2π, then we get the general solution

where n ∈ Z. I'm not sure why you have just n and not 2n, Ricky. Odd values of n in your solution would give

Now take a simple case of some 0 ≤ θ ≤ 2π that could give eⁱθ = -i. 3π/2 is our value. Then

So, given the same interval 0 ≤ θ ≤ 2π, (-i)ⁱ ≠ iⁱ, so I will conclude that your n should be 2n.

Last edited by Zhylliolom (2006-07-14 08:07:19)

Hey you guys are discussing something too difficult here!
Though I read it too.

Mathsisfun is good at fun maths and perhaps he is the best person you can consult to.

Or you could just try looking in the Exercises forum. We're adding some every now and then, so check back regularly!

Last edited by Devantè (2009-02-15 10:18:30)