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JasonL220

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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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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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