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
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?
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?
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?