- #1
mick25
- 13
- 0
Homework Statement
I have some past exam questions that I am confused with
Homework Equations
[itex]a_{n} = \frac{1}{2\pi i} \oint_\gamma \frac{f(z)}{z-a}\, dz[/itex]
The Attempt at a Solution
I'm not sure how to approach this, I'm completely lost and just attempted to solve a few:
a) it says [itex]f(z)[/itex] has a pole of order 5, so [itex] f(z) = \frac{g(z)}{z^5}, g(z)\neq0 [/itex]
so then I guess the condition is [itex]a_{4} = \frac{g^{(4)}(0)}{4!} [/itex]? But that's just applying the formula for the coefficients...
c) [itex] f(\frac{1}{z}) = \frac{g(\frac{1}{z})}{z^5} => f(z) = z^5g(z) [/itex]
so the coefficients are [itex] a_{n} = \frac{1}{2\pi i} \oint_\gamma z^5g(z) dz [/itex]?
d) [itex] \frac{1}{f(z)} = \frac{g(z)}{z^5} => f(z) = \frac{z^5}{g(z)} [/itex]
so, [itex] a_{n} = \frac{1}{2\pi i} \oint_\gamma \frac{z^5}{g(z)} dz [/itex]
g) [itex]a_{-1} = \frac{1}{2\pi i} \oint_ \gamma f(z) dz = \frac{1}{2\pi i} = Res(f; c)*I(\gamma; c) = -Res(f; c)[/itex]
h) [itex] \frac{a_{n}}{16} = 4^{n}a_{n} => 0 = a_{n}(4^{n} - 4^{-2}) => a_{n} = 0 [/itex] or [itex] n = -2 [/itex]for e) and f), I'm not sure what the relevance of the essential singularity is
Well, I think you can see I'm clearly lost, would appreciate if you could help me out.