Proving Complex Number Equality

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Discussion Overview

The discussion revolves around proving that the product of linear factors involving complex numbers results in a complex polynomial, particularly when none of the complex numbers are conjugates of each other. Participants explore various aspects of this topic, including conditions for real results and implications of complex roots.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions how to prove that the product of linear factors, given that the roots are complex and not conjugates, will always yield a complex result.
  • Another participant suggests that if the coefficients of the polynomial are real and a complex number is a root, then its conjugate must also be a root.
  • A participant presents a specific case with two factors and derives conditions under which the result can be real, indicating that certain relationships between the imaginary parts of the roots must hold.
  • There is a challenge regarding how to generalize the proof for any number of factors beyond the two presented.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus, as there are multiple competing views regarding the conditions under which the product of complex factors results in a complex or real polynomial. The discussion remains unresolved with various hypotheses and challenges presented.

Contextual Notes

Limitations include the dependence on specific assumptions about the coefficients and the nature of the roots. The discussion does not resolve the mathematical steps necessary for a general proof.

Werg22
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How to proove that

(x - a)(x - b)(x - c)...

If a, b, c... are complex numbers, and none is conjugent to another the result will always be complex? Complex as is not real for those who like to complicate things...
 
Last edited:
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Werg22 said:
How to proove that

(x - a)(x - b)(x - c)...

If a, b, c... are complex numbers, and none is conjugent to another the result will always be complex? Complex as is not real for those who like to complicate things...

well, to start off, the first thing you need to do is to start it on your own.
 
Well, if the equation's coefficients are not complex, but x+iY is a root, then x-iY is also a root, since we collect imaginary parts separate from real.
 
well what do you get when you multiple one C:P(x) to a R:P(x)
 
Well this the only thing i could proove; with two factors;

(x - (a + bi))(x - (c + di))
x^2 - x(a + bi + c+di) + (c+di)(a+bi)
x?2 -x(a+c + i(b+d)) + (ca -bd + i(da + cb))

In order for the result to be real;

b = -d
-da = cb

ba=cb

a=c

so it is actually

(x - (a + bi))(x - (a -bi)), otherwise the result is not real. But how to proove that it is also applicable for any amount of factors?
 

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