Mathematics 350: Problems to Study Solutions

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1 Mathematics 350: Problems to Study Solutions April 25, 206. A Laurent series for cot(z centered at z 0 i converges in the annulus {z : < z i < R}. What is the largest possible value of R? Solution: The function cot(z cos(z has a singularity at each root of sin(z. The sin(z Laurent series will converge in the annulus centered at i and between the singularities at 0 and ±π. Hence R can be as large as + π Let f(z 2z z 2 +. (a Find the Taylor series for f centered at z 0 0. (Hint: partial fractions. Solution: The series will converge within a radius of (why?. Therefore we will assume that z <. By partial fractions, some algebra, and the celebrated geometric series formula, we obtain 2z z 2 + z + i + ( z i ( + i + z/i i z/i ( z/i k + (z/i k i i i k ( + ( k z k ( k 2z 2k+.

2 Here is an alternative approach. The hint suggested using partial fractions. This is not a bad idea, but things are actually a bit easier in this particular case. 2z z 2 + 2z ( z 2 2z ( z 2 k ( k 2z 2k+. (b Find the Laurent series for f centered at z 0 i converging in the annulus {z : 0 < z i < 2}. Solution: As above f(z /(z i + /(z + i. The function /(z i is already a Laurent series centered at i. We will work on the second part in the customary way, noting that 0 < z i < 2. z + i 2i + (z i ( 2i + (z i/(2i The Laurent series for f in this annulus is ( k (2i k+ (z ik. f(z z i + ( k (2i (z k+ ik. (c Find the Laurent series for f centered at z 0 i converging in the annulus {z : 2 < z i < }. Solution: As above we will work on /(z + i in the customary way, this time Page 2

3 noting that 2 < z i <. z + i 2i + (z i ( (z i + (2i/(z i ( 2i k (z i k+. The Laurent series for f in this annulus is f(z 2 z i + ( 2i k (z i. k+ k 3. Find a Laurent series for f(z e 3z /z 3 centered at z 0 0 and converging in the annulus {z : 0 < z }. Find Res(f, 0. Solution: Using the well-known Taylor series for e w and simple division, we obtain f(z e3z z 3 3 k z k 3, k! which is the Laurent series for f in the given annulus. The residue is Res(f, /2! 9/2, the coefficient of z. 4. Find a Laurent series for f(z sin(z/(z π centered at z 0 π and converging in the annulus {z : 0 < z π }. Find Res(f, π. Solution: The first step is to get a Taylor series for sin(z centered at z 0 π. We use the standard method : sin(z sin((z π + π sin(z π cos(π + sin(π cos(z π sin(z π k (z π2k+ (. (2k +! Page 3

4 All of the heavy lifting has been done, and the Laurent series is easily obtained: sin(z z π k+ (z π2k ( (2k +!. By identifying the coefficient of (z π, we see that Res(f, π Let f(z e z /(cos(z. Find Res(f, 0 via long division. Solution: Observe that Thus long division yields The residue in question is 2. cos(z z2 2 + z4 24 z e z + z + z2 2 + z3 6 + z e z cos(z 2 z 2 2 z 7 6 z 2 7z In each case, use Theorem 6.5. or Theorem to compute Res(f, z 0. (a f(z e 3z /z 3, z 0 0. (ompare with Problem 3. Solution: The function f has a pole of order 3, since e 0 0 and z 3 has a root of order 3 at 0. Thus Res(f, 0 2! lim z 0 D 2z 3 f(z. However z 3 f(z e 3z ; thus, D 2 z 3 f(z 9e 3z. Thus Res(f, 0 2! lim z 0 9e3z 9 2, which is the same answer we obtained in Problem 3. (b f(z sin(z/(z π, z 0 π.(ompare with Problem 4. Page 4

5 Solution: Let P (z sin(z and Q(z (z π. Since Q has a root of order at z π, the residue is easily calculated as P (z 0 /Q (z 0. In this case, Res(f, π sin(π/ 0, which is the same answer we obtained in Problem 4. (c f(z e z /(cos(z, z 0 0. (ompare with Problem 5. Solution: The denominator has root of order 2 at z 0 0 and the numerator is nonzero at z 0 0; thus, f has a pole of order 2 at z 0 0. Thus Res(f, 0! lim z 0 D z 2 e z cos(z lim z 0 e z (cos(z (2z + z2 + z 2 sin(z (cos(z 2 2, which is the same answer we obtained in Problem 4. The limit is not easy to calculate! A good heuristic for calculating such limits is to replace the transcendental functions by their Taylor series approximations near 0. Thus e z, cos(z z 2 /2, and sin(z z. The main point of this exercise is to convince you that long division is your friend. (d f(z cos(z/(e z 2, z 0 0. Solution: We see that cos(0 0 and the denominator has a root of order 2 at 0. Thus f has a pole of order 2 at 0. Long division is the best option here. By multiplication of series, (e z 2 z 2 + z 3 + 7z4 2 + This must be divided into cos(z z 2 /2 + z 4 /24. The result is It is clear that Res(f, 0. (e f(z sin(z/(e z 2, z 0 0. cos(z (e z 2 z 2 z 2 + 5z 2 + Solution: This is very similar to the above, with sin(z z z 3 /6 + replacing cos(z. Long division leads to sin(z (e z 2 z + z 4 + z2 2 + Page 5

6 It is clear that Res(f, 0. csc(z 7. Evaluate dz, where is the rectangle with corners at (4,, (,, (,, z and (4, traversed in the counterclockwise direction. Solution: Within the contour, the function f(z /(z sin(z has poles at z 0 and z π. By the residue theorem, f(zdz 2πi (Res(f, 0 + Res(f, π. The pole at the origin has order 2. I will calculate the first few terms of the Laurent series by long division: f(z z z learly Res(f, 0 0. The pole at π is order, so we can use the P/Q trick: Res(f, π sin(π + π cos(π π. The final answer is: f(zdz 2i 8. Evaluate e /(z 3 dz, where is the positively oriented circle of radius 6 centered at the origin. Solution: The function f(z e /(z 3 has singularity, at z 3. Thus f(zdz 2πi Res(f, 3. The Laurent series centered at z 0 3 in the annulus z 3 > 0 can be obtained by plugging w /(z 3 into the Taylor series for e w. We obtain: f(z k!(z 3 k + z 3 + 2(z learly Res(f, 3 ; thus, f(zdz 2πi. Page 6