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- Dec 26th 2008, 11:30 AM #1

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- Dec 26th 2008, 11:39 AM #2

- Dec 26th 2008, 11:40 AM #3

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- Dec 26th 2008, 11:41 AM #4

- Dec 27th 2008, 07:32 AM #5

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- Dec 27th 2008, 09:36 AM #6

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- Dec 28th 2008, 04:46 AM #7
It may help to be able to derive Euler's formula from the polar form of a complex number . We can do this using derivatives.

Let .

.

But since

.

If we divide both sides by z, we get

and so integrating both sides with respect to x gives

.

.

Substituting into the equation gives .

So .

And since we know is a circle of radius 1, so must be .

- Dec 28th 2008, 05:17 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 . 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, 05:23 AM #9

- Dec 29th 2008, 05:29 AM #10

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- Dec 29th 2008, 12:24 PM #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 . 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, 03:44 AM #12

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