Laurent's Theorem

**mathfied** · August 24th 2008, 02:40 PM

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:
$ 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:
$c_n = \frac{1} {{2\pi i}}\int\limits_{\gamma _r }^{} {\frac{{f(s)}} {{(s - z_0 )^{n + 1} }}} ds = \frac{{f^{(n)} (z_0 )}} {{n!}}$ .
Any suggestions from here?

PART (C)
-------------------
Find the maximal radius R>0 for which the function $ 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

**wingless** · August 24th 2008, 03:26 PM

A) Write CRE for both functions.

B) See here.

**mathfied** · August 25th 2008, 06:30 AM

PART (A)
CRE's for the f(z) :
u_x=v_y and u_y=-v_x.

CRE's for g(z) :
v_x=u_y and v_y=-u_x.

SO:
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?

**shawsend** · August 25th 2008, 08:49 AM

If $v_x=-v_x$ then $2v_x=0\Rightarrow v(x,y)=f(y)$

But you can also use the CR equations to show $2v_y=0\Rightarrow v(x,y)=f(x)$

Only way both statements are true is if $v(x,y)=k$

Then do these two steps for $u(x,y)$ as well.

**shawsend** · August 25th 2008, 09:50 AM

For part (c), the Laurent series for $\csc(z)$ is kind of messy. Could certainly use long division to get the first few terms. Here's one way to get the rest of the terms: First look at Mathworld under Bernoulli numbers. There they show how to derive:

$z\cot(z)=\sum_{n=0}^{\infty}\frac{(-1)^n B_{2n}(2z)^{2n}}{(2n)!}$

Now, if:

$\csc(z)=\sum_{n=0}^{\infty}\frac{(-1)^{n+1}2(2^{2n-1}-1)B_{2n}}{(2n)!}z^{2n-1}$

then split up the series for $\csc(z)$ into two sums, and figure out how the series for $z\cot(z)$ could be modified to represent those two series. Here's a start: what happens when I substitute $z\to z/2$ in the series for $z\cot(z)$ ?

Probably an easier way to get the terms though. Also, the radius of convergence extends to the nearest singularity which is $\pi$

**Opalg** · August 25th 2008, 12:17 PM

Originally Posted by mathfied

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

As shawsend has said, R = π. To find the principal part of the Laurent expansion, you might guess that since sin(z) is close to z when z is small, the principal part of 1/sin(z) ought to be 1/z. You can then justify that guess as follows.

Let $g(z) = \frac1{\sin z} - \frac1z = \frac{z-\sin z}{z\sin z}$ . Since z-sin(z) has degree 3 and z*sin(z) has degree 2, it follows that g(z) has a removable singularity at the origin and is therefore holomorphic in a neighbourhood of the origin. Therefore $\frac1{\sin z} = \frac1z + g(z)$ has the same principal part as 1/z (namely 1/z).

**shawsend** · August 25th 2008, 02:38 PM

I just realized I over-killed it: You only wanted the singular part. Do what Opalg said or just use long division:
$\csc(z)=\frac{1}{z-\frac{z^3}{3!}+\cdots}$ and the first term when you do that is $1/z$ then all the rest are positive powers of $z$ . But hey Mathfied, thanks a bunch. I hadn't known about all that stuff I posted above until I researched it and worked it out myself today. The Bernoulli stuff is interesting I think.

Thread: Laurent's Theorem

LinkBack

Thread Tools

Display

Laurent's Theorem

Similar Math Help Forum Discussions

Help with Laurent Expansions

Laurent Series/ Laurent Series Expansion

Laurent Series

Laurent

Laurent series

Search Tags