Part (A)----------------------------Solving Laurent's Theorem with CRE

[mathfied]

Hi just a bit of help needed here as I don;t know where to start:

Part (A)
----------------------------
Suppose f(z) = u(x,y) + iv(x,y)\;and\;g(z) = v(x,y) + iu(x,y) 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 D'(0,R) = \left\{ {z \in C:0 < |z| < R} \right\} where R>0 is fixed. What is the formulae for c_n in the Laurent expansion:
<br /> f(z) = \sum\limits_{n = - \infty }^\infty {c_n z_n }.

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

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


PART (C)
-------------------
Find the maximal radius R>0 for which the function <br /> f(z) = (\sin z)^{ - 1} 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

 

[Dick]

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]

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

 

[mathfied]

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]

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?