Compute Integrals: Integrate (z^3-6z^2+4)dz from -1+i to 1

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Discussion Overview

The discussion centers on the integration of the complex function (z^3-6z^2+4)dz along a curve from the point -1+i to 1. The scope includes mathematical reasoning and exploration of complex integration techniques.

Discussion Character

  • Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant proposes to integrate the function by separating it into individual components: z^3dz, -6z^2dz, and 4dz.
  • Another participant suggests that a parametric representation of the curve is unnecessary due to the existence of an anti-derivative for the polynomial function, indicating that the integral's value depends only on the endpoints.
  • A request is made for the original poster to share their progress to facilitate assistance from others.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of a parametric representation for the integration process, indicating a lack of consensus on the approach to take.

Contextual Notes

The discussion does not resolve the implications of path independence in complex integration, nor does it clarify the assumptions regarding the choice of curve for integration.

asqw121
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Integrate (z^3-6z^2+4)dz where the function is any curve joining -1+i to 1. Z is complex number
 
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Please post your progress so far so our helpers know exactly where you are stuck or what you may be doing wrong. (Nod)
 
I am trying to first separate them to integrate individual z^3dz, -6z^2dz and 4dz.

Line integral γ(t)= t*i+(1-t)*(-1-+i)
 
There is no need to find a parametric representation of the curve because the function you are integrating has an anti derivative , indeed it is a polynomial . By independence of the path , the integral only depends on the initial and final points .
 

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