Laurent's Theorem


by mathfied
Tags: laurent, theorem
mathfied
mathfied is offline
#1
Aug24-08, 05:42 PM
P: 16
Hi just a bit of help needed here as I don;t know where to start:

Part (A)
----------------------------
Suppose [latex]f(z) = u(x,y) + iv(x,y)\;and\;g(z) = v(x,y) + iu(x,y)[/latex] are analytic in some domain D. Show that both u and v are constant functions..?

I guess we have to use the CRE here but not really sure how to approach this..?

Part (B)
----------------------------
Let f be a holomorphic function on the punctured disk [latex]D'(0,R) = \left\{ {z \in C:0 < |z| < R} \right\}[/latex] where R>0 is fixed. What is the formulae for c_n in the Laurent expansion:
[latex]
f(z) = \sum\limits_{n = - \infty }^\infty {c_n z_n }[/latex].

Using these formulae, prove that if f is bounded on D'(0,R), it has a removable singularity at 0.

- Well I know that:
[latex]c_n = \frac{1}
{{2\pi i}}\int\limits_{\gamma _r }^{} {\frac{{f(s)}}
{{(s - z_0 )^{n + 1} }}} ds = \frac{{f^{(n)} (z_0 )}}
{{n!}}[/latex].
Any suggestions from here?


PART (C)
-------------------
Find the maximal radius R>0 for which the function [latex]
f(z) = (\sin z)^{ - 1}[/latex] is holomorphic in D'(0,R) and find the principal part of its Laurent expansion about z_0=0

??

Any help would be greatly appreciated.

Thanks a lot
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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Dick
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#2
Aug24-08, 10:29 PM
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I'll start you out with the first one. CRE's for the f(z) tell you u_x=v_y and u_y=-v_x. CRE's for g(z) tell you v_x=u_y and v_y=-u_x. What happens when you put both of these together?
Dick
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#3
Aug24-08, 11:25 PM
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For the second one, you might want to focus your efforts on proving that c_n=0 for n<0.

mathfied
mathfied is offline
#4
Aug25-08, 09:29 AM
P: 16

Laurent's Theorem


hmm so for part (1)
u_x = v_y = -u_x AND
u_y = -v_x = v_x

so u and v are constant because u_x = -u_x and -v_x = v_x

is that correct?
Dick
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#5
Aug25-08, 09:13 PM
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Yes. u_x=-u_x means u_x=0. The same for all of the other stuff. All of the partial derivatives are zero. Hence?


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