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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.
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.


1. Poles of 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
Let's analyze the functions f(z)f(z)f(z) and g(z)g(z)g(z) to
determine the order of the poles at z=iz = iz=i.
Function f(z)=z2(z−i)4f(z) = \frac{z^2}{(z-i)^4}f(z)=(z−i)4z2:
• The denominator of f(z)f(z)f(z) is (z−i)4(z-i)^4(z−i)4,
which clearly has a factor of (z−i)(z-i)(z−i) raised to the
power 4.
• The numerator z2z^2z2 is a polynomial that is analytic
(holomorphic) everywhere, including at z=iz = iz=i. At
z=iz = iz=i, the numerator z2z^2z2 evaluates to i2=−1i^2 =
-1i2=−1, which is non-zero.
Since the numerator does not vanish at z=iz = iz=i, the pole
is solely determined by the denominator (z−i)4(z-
i)^4(z−i)4.

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