MATH 417 Homework 6 Instructor: D. Cabrera Due July Find the radius of convergence for each power series below. c n+1 c n (n + 1) 2 (z 3) n+1

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1 MATH 47 Homework 6 Instructor: D. Cabrera Due Jul 2. Find the radius of convergence for each power series below. (a) (b) n 2 (z 3) n n=2 e n (z + i) n n=4 Solution: (a) Using the Ratio Test we have L = lim = lim c n+ c n (n + ) 2 (z 3) n+ n 2 (z 3) n (n + ) 2 = lim z 3 n 2 ( = z 3 lim + n = z 3 The series converges when L = z 3 <. Therefore, the radius of convergence is. (b) Using the Ratio Test we have L = lim = lim c n+ c n = lim e z + i = e z + i ) 2 e n+ (z + i) n+ e n (z + i) n The series converges when L = e z + i < = z + i <. Therefore, the radius e of convergence is e.

2 2. What is the radius of convergence of the Series of f(z) = z = 0? about z = 3i? z 2 3z + 2 about Solution: The singular points of f(z) = z 2 3z + 2 = are z = and (z )(z 2) z = 2. Therefore, since f(z) is analtic at z = 0, it has a Series representation for all z satisfing z < R where R is the distance between z = 0 and the nearest singular point which is z =. Therefore, R = 0 =. Since f(z) is analtic at z = 3i, it has a Series representation for all z satisfing z 3i < R where R is the distance between z = 3i and the nearest singular point which is z =. Therefore, R = 3i = 0. Region of convergence about z = 0. Region of convergence about z = 3i. R = 2 3i 2 3. Find the Series of f(z) = z about z = 0 and state the region of validit. + z2 Write our answer in summation form. Solution: The singular points of f(z) are z = i and z = i. Since f(z) is analtic at z = 0, it has a Series representation for all z satisfing z < R where R is the distance between z = 0 and the closest singular point. Both singular points are at a distance of from the origin. Therefore, the region of validit is z <. We are looking for a series representation in the form f(z) = c n z n = c 0 + c z + c 2 z 2 + n=0 To get the Series we will write f(z) as f(z) = z + z 2 = z + z 2 2

3 and then use the Maclaurin Series for z 2 to get + z = z + z2 z 3 + and replace z with f(z) = z + z 2 f(z) = z ( z 2 + (z 2 ) 2 (z 2 ) 3 + ) f(z) = z z 3 + z 5 z 7 + f(z) = ( ) n z 2n+ n=0 4. Find the Series of f(z) = z + z our answer in summation form. about z = 0 in the region < z <. Write Solution: We are looking for a series representation in the form f(z) = n= To obtain this series we will rewrite f(z) as and then use the Maclaurin Series for z to get c n z n = + c 2 z 2 + c z + c 0 + c z + c 2 z 2 + f(z) = f(z) = z + z f(z) = z ( ) z z + f(z) = + z + z f(z) = z + ( z) 2 f(z) = z + z 2 z z = z + z2 z 3 + and replace z with ( ) 3 + z 0 f(z) = ( ) n z n n= or ( ) n z n n=0 3

4 5. Determine all regions for which f(z) has a Series epansion about z = 2. Then determine all regions for which f(z) has a Series epansion about z = 2. DO NOT FIND THE SERIES EXPANSIONS! (a) f(z) = e z (b) f(z) = z 2 + (c) f(z) = z(z + )() Solution: (a) The function is entire so it has a Series epansion that is valid for z 2 <. 2 (b) The function has singular points at z = i and z = i. Since f(z) is analtic at z = 2 it has a Series epansion for all z satisfing z 2 < R where R is the distance between z = 2 and the nearest singular point. Both singular points are at a distance of R = 5 from z = 2. Therefore, f(z) has a Series epansion in the region z 2 < 5 and a Series epansion in the region 5 < z 2 <. i -i 2 4

5 (c) The function has singular points at z = 0, z =, and z = 2i. Since f(z) is analtic at z = 2 it has a Series epansion for all z satisfing z 2 < R where R is the distance between z = 2 and the nearest singular point which is z = 0. The distance between these points is R = 2 so f(z) has a Series epansion in the region z 2 < 2. The net closest singular point is z = 2i. The distance between z = 2 and z = 2i is R = 2i 2 = 2 2. Therefore, f(z) has a Series epansion in the region 2 < z 2 < 2 2. The distance between z = 2 and the last singular point z = is R = 2 = 3. Therefore, f(z) has another Series epansion in the region 2 2 < z 2 < 3. Finall, f(z) has a third Series epansion in the region 3 < z 2 < i If we were interested in finding the series epansions for f(z) = z(z + ))) about z = 2, we would perform a Partial Fraction Decomposition of f(z) to get f(z) = z(z + )() = c z + c 2 z + + c 3 where c, c 2, and c 3 are comple numbers. Then, on each interval we would write either a or Series for each function and it would go as follows: 5

6 z 2 < 2 : f(z) = c z + c 2 z + + c 3 2 < z 2 < 2 2 : f(z) = c z + c 2 z + + c < z 2 < 3 : f(z) = c z + c 2 z + + c 3 3 < z 2 < : f(z) = c z + c 2 z + + c 3 6. Find the Series of f(z) = about z = in the region < z + < 3. z 2 4 It is not necessar to write our answer in summation form. However, ou should write out sufficientl man terms so that the pattern is clear. Solution: First, we use the Method of Partial Fractions to rewrite the function as f(z) = z 2 4 = 4 z 2 4 z + 2 The function f (z) = z 2 has a singular point at z = 2. Since f (z) is analtic at z = and the distance between z = and z = 2 is 3, f (z) has a Series epansion in the region z + < 3. Since we are looking for a series epansion for f(z) in the annulus < z + < 3, we will write the Series for f (z) around z =. 6

7 f (z) = z 2 f (z) = (z + ) 3 f (z) = ( ) z f (z) = 3 z + ( 3 f (z) = + z + ( ) 2 ( ) 3 z + z ) f (z) = 3 z + (z + )2 (z + ) The function f 2 (z) = z + 2 has a singular point at z = 2. Since f 2(z) is analtic at z = and the distance between z = and z = is, f 2 (z) has a Series epansion in the region z + <. However, we are interested in the series epansion of f(z) in the annulus < z + < 3. Therefore, we want to write the Series of f 2 (z) around z =. f 2 (z) = z 2 f 2 (z) = (z + ) 3 f 2 (z) = ( (z + ) 3 ) z + f 2 (z) = z + 3 ( z + f 2 (z) = + 3 ( ) 2 ( ) z + z ) z + z + f 2 (z) = z (z + ) (z + ) (z + ) 4 + 7

8 Putting the series epansions for f (z) and f 2 (z) back into the formula for f(z) we get f(z) = 4 f (z) 4 f 2(z) f(z) = [ 4 3 z + ] (z + ) f(z) = 4 (z + ) 4 3 (z + ) (z + ) 3 [ ] z (z + ) (z + ) (z + ) (z + )2 8