Quadratic equation with complex coefficients

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Homework Help Overview

The discussion revolves around solving a quadratic equation with complex coefficients, specifically the equation z^2 + 4(1 + i(3^0.5))z - 16 = 0. Participants are exploring methods to approach the solution and the implications of using different forms for the complex variable.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • One participant attempts to verify their solution using polar coordinates, while another questions the choice of this method, suggesting that expressing z as a + bi might be simpler. There is also a discussion about eliminating the square root of the complex number in the determinant.

Discussion Status

The conversation is active, with participants sharing their reasoning and questioning the methods being used. Some guidance is offered regarding the simplification of the complex number involved, and there is an exploration of different approaches without a clear consensus on the best method.

Contextual Notes

Participants are considering the implications of using polar coordinates versus Cartesian forms, and there is mention of simplifying the expression -96 + 32i, indicating that certain assumptions or simplifications are under discussion.

elimenohpee
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Homework Statement


Solve the quadratic equation

z^2 + 4(1 + i(3^0.5))z - 16 = 0


Homework Equations





The Attempt at a Solution


I think I've done this correctly, I just wanted to verify.
I've only done the solution for k=0

http://i129.photobucket.com/albums/p201/elimenohpee182/Capture-1.png
 
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I'm interested to know, why did you use polar coordinates? Would it not be easier to let z = a + b.i, then solve for a and b?
 
I thought using polar coordinates would be easiest to eliminate the square root of the complex number.

I don't know if its right or not, that's why I wanted someone to check it.
 
You want to get rid of the square root of the determinant, so let [tex]\sqrt{-96+32i}=a+ib[/tex] on squaring both sides, we solve [tex]-96+32i=a^2-b^2+2abi[/tex]

Thus you have two equations to solve, [tex]-96=a^2-b^2[/tex] and [tex]32=2ab[/tex] since the real and imaginary coefficients must be equal.

But first you may want to check if you can simplify -96+32i. Notice 96=32*3
 

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