Solving for |a| in a Complex Number Equation

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Discussion Overview

The discussion revolves around solving for the magnitude of a complex number \( |a| \) that satisfies the equation \( ia^3 + a^2 - a + 1 = 0 \). Participants explore different methods to approach the problem, including the potential use of Cardano's formula and the implications of reformulating the problem to find the maximum value of \( |a - 3 - 4i| \).

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant suggests substituting \( a = x + iy \) as a method to find \( |a| \).
  • Another participant references Cardano's formula as a potential tool for finding roots, noting that the roots appear to be complex and irrational.
  • A later post shifts the focus to finding the maximum value of \( |a - 3 - 4i| \), questioning whether this reformulation simplifies the solution process.
  • One participant expresses skepticism about whether the reformulation aids in finding a shorter solution, indicating that it does not seem to help.
  • There is a suggestion to verify the equations involved in the reformulation, implying uncertainty about their correctness.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the effectiveness of the proposed methods or whether the reformulation simplifies the problem. Multiple competing views remain regarding the best approach to solve for \( |a| \).

Contextual Notes

There are limitations related to the complexity of the roots and the potential irrationality involved in the solutions. The discussion does not resolve these complexities or provide a definitive method for solving the equation.

aviravir1
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if a is a complex number which satisfy ia^3+a^2-a+1=0

then find \left | a \right | ?

one way is to put a=x+iy

any other short way ?
 
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ok the real question was find the maximum value of \left | a -3-4i\right |<br />

now will that help in any way making the soln shorter...i guess not..but still will it?
 

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