Fill in the details of the following proof of the Laurent

Chapter 21, Problem 21.36

(choose chapter or problem)

Fill in the details of the following proof of the Laurent expansion theorem (21.5). Let z be in the annulus, and choose r1 and r2 such that 0 2 1 and 2 in the proof of Theorem 21.5 for 21.11, Section 21.2. line segments L1 and L2 between these circles (Figure 21.2), forming two closed paths 1 and 2 in the annulus (shown separately in Figure 21.3). Show that f (z) = 1 2i 1 f (w) w z dw and 1 2i 2 f (w) w z dw = 0. Add these to obtain f (w) = 1 2i 1 f (w) w z dw + 1 2i 2 f (w) w z dw with counterclockwise orientation on both paths. By noting that in these integrals, the parts of the integrals over the line segments vanish (these segments are traversed in both directions), show that f (z) = 1 2i 2 f (w) w z dz 1 2i 1 f (w) w z dw with both integrations counterclockwise over the closed path. On 2, show that |(z z0)/(w z0)|<1, and use the geometric series to write 1 w z = n=0 1 (w z0)n+1 (z z0) n . On 1, show that |(w z0)/(z z0)| < 1 to show that 1 w z = n=0 (w z0) n 1 (z z0)n+1 .Use these to show that f (z) = n=0 1 2i 2 f (w) (w z0)n+1 dw (z z0) n + n=0 1 2i 1f (w)(w z0) n dw 1 (z z0)n+1 . Finally, replace n =m 1 in the last summation, and then use the deformation theorem to replace 1 and 2 by any closed path about z0 and in the annulus.

Unfortunately, we don't have that question answered yet. But you can get it answered in just 5 hours by Logging in or Becoming a subscriber.

Becoming a subscriber
Or look for another answer

×

Login

Login or Sign up for access to all of our study tools and educational content!

Forgot password?
Register Now

×

Register

Sign up for access to all content on our site!

Or login if you already have an account

×

Reset password

If you have an active account we’ll send you an e-mail for password recovery

Or login if you have your password back