didnt know where to post this sorry

how do we know that e^ix is a circle when plotted??

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- Dec 26th 2008, 10:30 AM #1

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- Dec 26th 2008, 10:39 AM #2

- Dec 26th 2008, 10:40 AM #3

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- Dec 26th 2008, 10:41 AM #4

- Dec 27th 2008, 06:32 AM #5

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- Dec 27th 2008, 08:36 AM #6

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- Dec 28th 2008, 03:46 AM #7
It may help to be able to derive Euler's formula from the polar form of a complex number $\displaystyle \cos{x} + i\sin{x}$. We can do this using derivatives.

Let $\displaystyle z = \cos{x} + i\sin{x}$.

$\displaystyle \frac{dz}{dx} = -\sin{x} + i\cos{x}$.

But since $\displaystyle i^2 = -1$

$\displaystyle \frac{dz}{dx} = i^2\sin{x} + i\cos{x}$

$\displaystyle \frac{dz}{dx} = i(\cos{x} + i\sin{x})$

$\displaystyle \frac{dz}{dx} = iz$.

If we divide both sides by z, we get

$\displaystyle \frac{1}{z}\frac{dz}{dx} = i$

and so integrating both sides with respect to x gives

$\displaystyle \int{\frac{1}{z}\frac{dz}{dx}\,dx} = \int{i\,dx}$

$\displaystyle \int{\frac{1}{z}\,dz} = ix + C$

$\displaystyle \ln|z| = ix + C$

$\displaystyle |z| = e^{ix + C}$

$\displaystyle |z| = e^Ce^{ix}$

$\displaystyle z = \pm e^Ce^{ix}$

$\displaystyle z = Ae^{ix}$.

$\displaystyle \cos{x} + i\sin{x} = Ae^{ix}$.

Substituting $\displaystyle x = 0$ into the equation gives $\displaystyle A = 1$.

So $\displaystyle z = \cos{x} + i\sin{x} = e^{ix}$.

And since we know $\displaystyle \cos{x} + i\sin{x}$ is a circle of radius 1, so must be $\displaystyle e^{ix}$.

- Dec 28th 2008, 04:17 AM #8
Proof by calculus

*does*require making some assumptions about complex derivatives. However, it can be proved that convenient properties (such as the chain rule) carry over from R to C.

Personally I prefer the proof by Maclaurin series.

You can also work with $\displaystyle e^{ix} = \lim_{n \rightarrow \infty} \left( 1 + \frac{ix}{n} \right)^n$. It's not too hard to show that the modulus is equal to 1 and that the argument is equal to x (I leave this for others who are interested).

- Dec 28th 2008, 04:23 AM #9

- Dec 29th 2008, 04:29 AM #10

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- Dec 29th 2008, 11:24 AM #11
haha, short answer, but probably not the answer you want, is to study complex analysis. a complex analysis text will go through all the proofs and show what things carry over from real analysis and what things don't. you will find that complex analysis is a lot like working in $\displaystyle \mathbb{R} \times \mathbb{R}$. and so derivatives are defined in the same way, but the limits are taken differently. this leaves a lot of familiar properties intact though. you can look these things up if you're interested.

- Dec 30th 2008, 02:44 AM #12

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