Register to reply 
Laurent's Theorem 
Share this thread: 
#1
Aug2408, 05:42 PM

P: 16

Hi just a bit of help needed here as I don;t know where to start:
Part (A)  Suppose [itex]f(z) = u(x,y) + iv(x,y)\;and\;g(z) = v(x,y) + iu(x,y)[/itex] 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 [itex]D'(0,R) = \left\{ {z \in C:0 < z < R} \right\}[/itex] where R>0 is fixed. What is the formulae for c_n in the Laurent expansion: [itex] f(z) = \sum\limits_{n =  \infty }^\infty {c_n z_n }[/itex]. Using these formulae, prove that if f is bounded on D'(0,R), it has a removable singularity at 0.  Well I know that: [itex]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!}}[/itex]. Any suggestions from here? PART (C)  Find the maximal radius R>0 for which the function [itex] f(z) = (\sin z)^{  1}[/itex] 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 


#2
Aug2408, 10:29 PM

Sci Advisor
HW Helper
Thanks
P: 25,228

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?



#3
Aug2408, 11:25 PM

Sci Advisor
HW Helper
Thanks
P: 25,228

For the second one, you might want to focus your efforts on proving that c_n=0 for n<0.



#4
Aug2508, 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? 


#5
Aug2508, 09:13 PM

Sci Advisor
HW Helper
Thanks
P: 25,228

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?



Register to reply 
Related Discussions  
Mean Value Theorem  Calculus & Beyond Homework  9  
Mean Value Theorem?  Calculus & Beyond Homework  7  
Greens theorem and cauchy theorem help  Calculus & Beyond Homework  4  
Gauss' Theorem/Stokes' theorem  Advanced Physics Homework  2  
Gauss's Divergance Theorem and Stokes's Theorem  Classical Physics  1 