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




ADMIN
[COMPANY NAME]

, 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
Given the functions f(z)=z2(z−i)4f(z) = \frac{z^2}{(z -
i)^4}f(z)=(z−i)4z2 and g(z)=z2+1(z−i)4g(z) = \frac{z^2 + 1}{(z
- i)^4}g(z)=(z−i)4z2+1:
1. Function f(z)f(z)f(z):
o Numerator Analysis: The numerator z2z^2z2 is a

polynomial that is analytic (holomorphic) everywhere
in the complex plane, including at z=iz = iz=i.
o Denominator Analysis: The denominator (z−i)4(z -

i)^4(z−i)4 has a zero of order 4 at z=iz = iz=i.
o Conclusion: Since the numerator does not vanish at

z=iz = iz=i and the denominator has a zero of order 4
at z=iz = iz=i, f(z)f(z)f(z) has a pole of order 4 at z=iz
= iz=i.
2. Function g(z)g(z)g(z):
o Numerator Analysis: The numerator z2+1z^2 +

1z2+1 is also a polynomial that is analytic

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