100% satisfaction guarantee Immediately available after payment Both online and in PDF No strings attached
logo-home
Previously searched by you
Previously searched by you
logo
MAT3705 Assignment 4 (COMPLETE ANSWERS) 2024 - DUE 5 September 2024 ; 100% TRUSTED Complete, trusted solutions and explanations R46,84   Add to cart
Quickly navigate to
Preview
  2.  
  3. University Of South Africa
  4.  
  5. Complex Analysis

Exam (elaborations)

MAT3705 Assignment 4 (COMPLETE ANSWERS) 2024 - DUE 5 September 2024 ; 100% TRUSTED Complete, trusted solutions and explanations
 7 views  0 purchase

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

Preview 2 out of 11  pages

Document information

  • August 22, 2024
  • 11
  • 2024/2025
  • Exam (elaborations)
  • Questions & answers

Subjects

  • mat3705 assignment 4 complete answers 2024 due

Connected book

book image

Book Title:

Author(s):

  • Edition:
  • ISBN:
  • Edition:

More summaries for

All for this textbook (9)

Written for

All documents for this subject (14)

Seller

avatar-seller
Academy360

Reviews received

Content preview

, 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.

The benefits of buying summaries with Stuvia:

Guaranteed quality through customer reviews

Guaranteed quality through customer reviews

Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.

Quick and easy check-out

Quick and easy check-out

You can quickly pay through EFT, credit card or Stuvia-credit for the summaries. There is no membership needed.

Focus on what matters

Focus on what matters

Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!

Frequently asked questions

What do I get when I buy this document?

You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.

Satisfaction guarantee: how does it work?

Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.

Who am I buying this summary from?

Stuvia is a marketplace, so you are not buying this document from us, but from seller Academy360. Stuvia facilitates payment to the seller.

Will I be stuck with a subscription?

No, you only buy this summary for R46,84. You're not tied to anything after your purchase.

Can Stuvia be trusted?

4.6 stars on Google & Trustpilot (+1000 reviews)

67866 documents were sold in the last 30 days

Founded in 2010, the go-to place to buy summaries for 14 years now

Start selling
R46,84
  • (0)
  Buy now