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(edited due to mooserman352's hint)

e^ix may have 2 results that are real if x should be real. This can easily seen from the fact that i*sin x must become zero to cancel out i. You could furthur determine the real values when setting that x into the cosine.

I would go into more detail if I were sure that this isn't some sort of homework.

[Edited by - haegarr on October 8, 2006 11:23:18 AM]

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ln(pi/4) = -0.2415644753.

so if x = 0.2415644753*i, then e^(ix) = e^(-0.2415644753) = pi/4

i'm not sure if that's what you were asking.

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Quote:
Original post by haegarr
e^ix may have 2 results that are real. This can easily seen from the fact that i*sin x must become zero.


i don't think that's true. there is no restriction that x must be real, so cos(x)+i*sin(x) being real does not imply that sin(x) is zero.

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Quote:
Original post by mooserman352
Quote:
Original post by haegarr
e^ix may have 2 results that are real. This can easily seen from the fact that i*sin x must become zero.


i don't think that's true. there is no restriction that x must be real, so cos(x)+i*sin(x) being real does not imply that sin(x) is zero.

That (i.e. your correction) has some thruth! I've narrowed my above reply accordingly.

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i'm not entirely sure. the original post is unclear. typically, a person gives an equation, for example, x^2+3x = 3, and asks how to solve for x. in this case, the op writes an identity e^(ix)=cos(x)+i*sin(x), and asks to solve for a real number, ie pi/4. this actually means to solve for a case when x is a real number, but I'm pretty sure given the context that that is not what he meant.

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Agreed - the question makes little sense.
The equation is Euler's Formula. It is an identity, necessarily true for all real values of x. If you're just looking for a real solution, you won't have a very hard time finding one [rolleyes].

Regards
Admiral

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