Finding the Suitable Parametrization for Computing ∫Cr (z - z0)n dz

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Homework Help Overview

The discussion revolves around computing the integral ∫Cr (z - z0)n dz, where Cr represents a circle in the complex plane defined by │z - z0│= r, traversed counterclockwise. Participants are exploring how to find a suitable parametrization for the circle.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster questions how to derive the parametrization z(t) for the circle, expressing confusion about its appearance in a solution. Other participants discuss the general parametrization of circles in the xy-plane and relate it to complex numbers using Euler's formula.

Discussion Status

The discussion includes attempts to clarify the parametrization of circles in the complex plane, with some participants confirming their understanding of the parametrization process. There is an exchange of ideas, but no explicit consensus has been reached regarding the original poster's question.

Contextual Notes

Participants are operating under the assumption that the integral is to be computed over a specific circle in the complex plane, with some details about the circle's center and radius being discussed.

nickolas2730
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compute the integral ∫Cr (z - z0)n dz,
with an integer and Cr the circle │z - z0│= r traversed once in the counterclockwise direction

Solution:

A suitable parametrization for Cr is give by z(t)= z0 + reit 0≤t≤2π
...
...

My question is , how to find that suitable z(t)?
i have no idea how to find out the z(t), it just pop out in the solution.

Thanks
 
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It's a circle in the real-imaginary plane.
A circle in the xy-plane can be parametrized by
x=r*cos(t)
y=r*sin(t)
with 0<=t<=2pi
This comes down to the fact that cos^2(t)+sin^(t)=1 as a circle is defined by x^2+y^2=R^2.
Now bear in mind Euler's formula
e^{ix}=cos(x)+i sin(x)
Alright?
 
so if C is the circle of │z - 2i │= 4
z(t) = 2i + 4e^it ?
i am doing it right?
 
You got it ;)
 
thanks!
 

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