# Complex analysis: Coefficients of Laurent series

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 $z=0$. – Alex R. Apr 10 '12 at 18:20
You should not embed images of text. You should simply write down the text!! – André Caldas Apr 10 '12 at 18:45
These are really great exam questions : crisp, covering much material, not too easy, not too difficult. Kudos to the examiner! – Georges Elencwajg Oct 23 '12 at 21:52

(a) It's given that $$f(z)=\sum_{n=-\infty}^\infty a_nz^n\,\,,\,\,\text{is analytic in}\,\,\Bbb C-\{0\}$$ and it has a pole of order 5 in $\,z=0\,$ , so the conditions here are $$a_n=0\,\,\,\forall n<-5\,\,,\,a_{-5}\neq 0$$

$$(b)\;\;\;\;\;\;\;\;f(z)-7e^{1/z}=\sum_{n=-\infty}^\infty a_nz^n-\sum_{n=0}^\infty\frac{7/n!}{z^n}=\frac{a_{-5}-\frac{7}{5!}}{z^5}+\frac{a_{-4}-\frac{7}{4!}}{z^4}+...$$ so the conditions here are $$a_{-5}\neq\frac{7}{5!}\,\,\,,\,\,a_{-n}=\frac{7}{n!}\,\,\forall n>5$$

$$(c)\;\;\;\;\;\;\,\,\,f(1/z)=\sum_{n=-\infty}^\infty \frac{a_n}{z^n}=...a_{-1}z+a_0+\frac{a_1}{z}+...\frac{a_5}{z^5}$$ so the conditions here are $$a_5\neq 0\,\,,\,\,a_n=0\,\,\,\forall n>5$$

$$(d)\;\;\;\;\;\;\;\,\,\,\frac{1}{f(z)}\,\,\text{has a pole at zero}\,\Longleftrightarrow f(z)\,\,\text{has a zero at zero}$$ so the conditions here are $$f(z)=a_5z^5+a_6z^6+...\Longrightarrow\,a_5\neq 0\,\,,\,\,a_n=0\,\,\forall n<5$$

(e) From what we did in (c) it follows that it must be that $\forall m\in\Bbb N\,\,\exists\,\,n\in\Bbb N\,\,,\,n>m\,\,\,s.t.\,\,a_n\neq 0$

(g) The only possible singularity we have to worry about within $\,|z|=1\,$ is $\,z=0\,$ , and then $$\oint_\Gamma f(z)dz=2\pi i\,Res_{z=0}(f)=2\pi i a_{-1}$$ so the above equals $$1\Longleftrightarrow a_{-1}=-\frac{i}{2\pi}\,\,\,and\,\,\,\exists k\in\Bbb N\,\,s.t.\,\,a_{-m}=0\,\,\,\forall m>k$$ The last part above means the singularity at $\,z=0\,$ is a pole.

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