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MAT3705 Assignment 4 (COMPLETE ANSWERS) 2024 - DUE 5 September 2024 ; 100% TRUSTED Complete, trusted solutions and explanations. R46,84   Add to cart
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MAT3705 Assignment 4 (COMPLETE ANSWERS) 2024 - DUE 5 September 2024 ; 100% TRUSTED Complete, trusted solutions and explanations.

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MAT3705 Assignment 4 (COMPLETE ANSWERS) 2024 - DUE 5 September 2024 ; 100% TRUSTED Complete, trusted solutions and explanations.

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  • August 22, 2024
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, MAT3705 Assignment 4 (COMPLETE ANSWERS) 2024 -
DUE 5 September 2024 ; 100% TRUSTED Complete,
trusted solutions and explanations.
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.


Question 1: Poles of f(z)f(z)f(z) and g(z)g(z)g(z) at z=iz = iz=i
Let's first examine the functions f(z)f(z)f(z) and g(z)g(z)g(z):
• f(z)=z2(z−i)4f(z) = z^2 (z - i)^4f(z)=z2(z−i)4
• g(z)=z2+1(z−i)4g(z) = \frac{z^2 + 1}{(z -
i)^4}g(z)=(z−i)4z2+1
f(z)=z2(z−i)4f(z) = z^2 (z - i)^4f(z)=z2(z−i)4
To determine the nature of the singularity at z=iz = iz=i for
f(z)f(z)f(z), let's expand f(z)f(z)f(z) around z=iz = iz=i:
1. Notice that (z−i)4(z - i)^4(z−i)4 has a factor of (z−i)(z -
i)(z−i) raised to the power of 4. This factor contributes to a
pole at z=iz = iz=i.
2. The z2z^2z2 term is analytic at z=iz = iz=i because it is a
polynomial and does not contribute to the pole’s order.

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