teclo said:
okay yah i have no clue. taylor series i might be able to manage, i don't know. i'll check the calc 2 book tommorow when I'm not drunk and have electrodynamics in six hours. i have no clue about imaginary numbers, it's been almost ten years since i was in high school working with them. where does z come from, and where can i get some clue as to this i business. i can follow the differentiation (almost). there may very well be no negative, i have had calculus 1 & 2 and that's it. working on 3 now, and this book on modern mathematical physics is mildly terrifying.
thanks so much, i really appreciate the effort afforded.
cheers
Worry not those into math books for science usually only assume one year of calculus. z is just the variable commonly used for complex numbers. Complex numbers are just a field extension of reals. That means your familar field axioms hold and you have an element i such that i^2=-1. The formal manipulation such books like is something like
[tex]e^{i\phi}=\sum_{n=0}^\infty \frac{(i\phi)^n}{n!}=\sum_{n=0}^\infty \left((-1)^n\frac{\phi^{2n}}{(2n)!}+(-1)^n\frac{\phi^{2n+1}}{(2n+1)!}i\right)=\cos(\phi)+i\sin(\phi)[/tex]
Another way to do the same thing is to note that is two analytic functions (those that can be represented by taylor series) have derivatives of all orders (0,1,2,...) equal at a point, then they are equal.
[tex]f(x):=e^{i\phi}[/tex]
[tex]f^{(n)}(x)=i^n e^{i\phi}[/tex]
[tex]f^{(n)}(0)=i^n[/tex]
[tex]g(x):=\cos(\phi)+i\sin(\phi)[/tex]
[tex]g^{(n)}(x)=\cos(\phi+n\pi/2)+i\sin(\phi+n\pi/2)[/tex]
[tex]g^{(n)}(0)=\cos(n\pi/2)+i\sin(n\pi/2)[/tex]
[tex]g^{(n)}(0)=i^n[/tex]
so they are equal
For any of that to have meaning as I mentioned before we need
-define exp(z) for complex z (or at least imaginary)
-show that the power series for exp(z) holds when z is complex (or at least imaginary)
-show that the series converges absolutely so that rearangment is allowed