# Need proof

1. Oct 2, 2005

### 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...

Last edited: Oct 2, 2005
2. Oct 2, 2005

### ComputerGeek

well, to start off, the first thing you need to do is to start it on your own.

3. Oct 2, 2005

### 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.

4. Oct 2, 2005

### neurocomp2003

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

5. Oct 3, 2005

### 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?