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## Main Question or Discussion Point

Is this true:

[tex]e^{i(a+bx)}=cos(a+bx)+i sin(a+bx)[/tex] ?

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Is this true:

[tex]e^{i(a+bx)}=cos(a+bx)+i sin(a+bx)[/tex] ?

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- #2

chroot

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Thanks

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So if you have the e^xi expression and set x = pi, this will equal -1. Adding 2Pi equals 1. So far so good? But here comes what I don't understand. 1 can also be written e^0, so using logarithms we say: e^i2Pi = e^0 -> i2Pi = 0, but this can't be true as i2Pi = 6,28i. I got one possible solution from my math teacher but it did not quite tell me how it could equal 0. Anyone care to give an explenation a shot?

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e^x = e^y <=> x = y is true.

For complex numbers, it must be the case that

e^x = e^y <=> x = y is false.

This shouldn't be so surprising. For the real numbers, you can't say things like...

x^2 = y^2 <=> x = y

sin(x) = sin(y) <=> x = y

...

I suppose it's just an accident that you can't do that for the exponential function over complex numbers.

- #6

dx

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You mean injection.My best guess would be that the exponential function is not a bijection over the complex numbers, while it is so over the real numbers. That is, for real numbers,

e^x = e^y <=> x = y is true.

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I tend to talk in terms of bijection or not bijection. I usually don't delve into things like injectivity and surjectivity. I think that the exponential is neither bijective nor injective over the complex numbers... then again, what do I know?

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dx

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You said exp was a bijection over the reals which it isn't. It's an injection, but not a bijection.

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For the less mathematically versed, this is a little like taking the square root of a number and having to remember that there are actually two roots (+/-). When you take the log of a complex number, there are (potentially) multiple correct results, so you have to specify which one you're choosing.

(At least, I think that's a correct statement; someone please correct me if it's not.)

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Oops. You're right. I guess that's the price you pay when you get too used to using the general terminology too often.

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