Solutions for assignment 7

Solutions for assignment 7 - Solutions for assignment 7 1...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Section 5.3 Question 2: The limit of the ratios of the terms in \sum a_j (z-z_0)^j is L |z-z_0|, because of the given information that lim_j |a_{j+1} / a_j| = L. Thus, by the ratio test, the series converges when |z-z_0| < 1/L and diverges when |z-z_0| > 1/L. Thus the radius of convergence of the series is 1/L. Question 3 (b). L = 2, so the radius of convergence is 1/2, and the circle of convergence is |z - 1| = 1/2. (d) L = 1/3, so the radius of convergence is 3, and the circle of convergence is |z - i| = 3. Section 5.5 Question 3: Although it is not mentioned explicitly, it is assumed that we are computing the Laurent series aroud zero. The easiest way is to break the function up into partial fractions: z/(z+1)(z-2) = (1/3)/(z+1) + (2/3)/(z-2); You can either work out the constants 1/3, 2/3 by the usual method taught to you in lower-division, or you can compute the residues of z/(z+1)(z-2) at -1 and 2 respectively. (a) When |z|<1 we have
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 3

Solutions for assignment 7 - Solutions for assignment 7 1...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online