Exam (elaborations)
MAT3705 Assignment 4 2024 - DUE 5 September 2024
- Course
- Complex Analysis (MAT3705)
- Institution
- University Of South Africa
MAT3705 Assignment 4 2024 - DUE 5 September 2024 QUESTIONS WITH COMPLETE ANSWERS
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[Company name]MAT3705 Assignment
4 2024 - DUE 5
September 2024
QUESTIONS WITH DETAILED ANSWERS
, MAT3705 Assignment 4 2024 - DUE 5 September 2024
1. Let f(z) = z2 (z−i)4 and g(z) = z2+1 (z−i)4 . Explain why f has a pole of order 4 at z = i, but g
has a pole of order 3 at z = i.
2. Let f(z) = sin z (z − π)2(z + π/2) and let C denote the positively oriented contour C = {z = 4eiθ
∈ C : 0 ≤ θ ≤ 2π}. (a) Identify the types of isolated singularities of f and calculate the residues of
f at these points. Provide reasons for your answers. (b) Use Cauchy’s Residue Theorem to
calculate Z C f(z) dz.
3. Let f(z) = (z + 1)2 z(z + 3i)(z + i/3)
(a) What type of isolated singularity is z = −i/3 of the function f? Provide reasons for your
answer.
(b) Calculate Resz=−i/3f(z). 1
(c) Calculate the value of k such that Z 2π 0 1 + cos θ 5 + 3 sin θ dθ = k Z C f(z) dz, where C is
the positively oriented contour C = {z = eit : 0 ≤ t ≤ 2π}.
(d) You are told (and do not have to calculate) that Resz=0f(z) = −1 and Resz=−3if(z) = 12+3i 4 .
Calculate the value of Z 2π 0 1 + cos θ 5 + 3 sin θ dθ.
4. Use Residue Theory to calculate Z ∞ −∞ x2 (x2 + 9)2 dx.
5. Let f(z) = z2 (z + 4)(z2 − 9) . Show that lim R→∞ Re Z CR f(z)ei5z dz = 0, where
CR denotes the positively oriented contour {Reiθ : 0 ≤ θ ≤ π}. Justify all steps.
6. Use Rouche’s Theorem to determine the number of roots of h(z) = 3z3 + 2z2 + 2z − 8 = 0
inside the disc {z ∈ C : |z| < 2}. Provide reasons for your answer.
Problem 1
Given functions:
• f(z)=z2(z−i)4f(z) = \frac{z^2}{(z - i)^4}f(z)=(z−i)4z2
• g(z)=z2+1(z−i)4g(z) = \frac{z^2 + 1}{(z - i)^4}g(z)=(z−i)4z2+1
You are asked to explain why fff has a pole of order 4 at z=iz = iz=i, but ggg has a pole of order
3 at z=iz = iz=i.
Solution:
1. For f(z)=z2(z−i)4f(z) = \frac{z^2}{(z - i)^4}f(z)=(z−i)4z2:
The function f(z)f(z)f(z) has a denominator with a factor (z−i)4(z - i)^4(z−i)4. Since the
numerator z2z^2z2 is analytic and non-zero at z=iz = iz=i, the pole at z=iz = iz=i is determined
solely by the factor in the denominator. Hence, f(z)f(z)f(z) has a pole of order 4 at z=iz = iz=i.
2. For g(z)=z2+1(z−i)4g(z) = \frac{z^2 + 1}{(z - i)^4}g(z)=(z−i)4z2+1:
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