Proving Complex Number Equality

[Werg22]

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...

 

[ComputerGeek]

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.

 

[robert Ihnot]

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.

 

[neurocomp2003]

well what do you get when you multiple one C:P(x) to a R:P(x)

 

[Werg22]

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?