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I have some past exam questions that I am confused with

http://i39.tinypic.com/vuwxl.png sorry, can't embed images yet

I'm not sure how to approach this, I'm completely lost and just attempted to solve a few:

a) it says $f(z)$ has a pole of order 5, so $ f(z) = \frac{g(z)}{z^5}, g(z)\neq0 $

so then I guess the condition is $a_{4} = \frac{g^{(4)}(0)}{4!}$?

c) $f(\frac{1}{z}) = \frac{g(\frac{1}{z})}{z^5} => f(z) = z^5g(z)$

so the coefficients are $a_{n} = \frac{1}{2\pi i} \oint_\gamma z^5g(z) dz$?

d) $\frac{1}{f(z)} = \frac{g(z)}{z^5} => f(z) = \frac{z^5}{g(z)}$

so, $a_{n} = \frac{1}{2\pi i} \oint_\gamma \frac{z^5}{g(z)} dz$

g) $a_{-1} = \frac{1}{2\pi i} \oint_ \gamma f(z) dz = \frac{1}{2\pi i} = Res(f; c)*I(\gamma; c) = -Res(f; c)$

h) $\frac{a_{n}}{16} = 4^{n}a_{n} => 0 = a_{n}(4^{n} - 4^{-2}) => a_{n} = 0$ or $n = -2$

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.

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    A few comments: for your part a), it should be $g(0)\neq 0$. As far as essential singularities go, one way to define "$f$ has an essential singularity at $z_0$" is to say that the Laurant expansion of $f$ around $z_0$ has infinitely many negative degree terms. Notice that $\exp(1/z)$ has an essential singularity at $z=0$.2012-04-10
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    You should not embed images of text. You should simply write down the text!!2012-04-10
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    These are really great exam questions : crisp, covering much material, not too easy, not too difficult. Kudos to the examiner!2012-10-23

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