Complex Numbers

[d0minicz]

Hi guys I need help with this to factorise over C



thanks

[Flaming_Arrow]

http://vcenotes.com/forum/index.php/topic,11821.msg142364.html#msg142364

[d0minicz]

Use the factor theorem to show that is a linear factor of

Do i divide through long division?

[Flaming_Arrow]

i would use remainder theorem

[d0minicz]

can you show me your working for that, thanks =]

[TrueTears]

z - 1 - i = 0
z = 1 + i

sub that in the equation and if it equals 0 then it is a factor.

[Flaming_Arrow]

[d0minicz]

and to find the other linear factors, how would i do that ?
do i divide through (z-1-i) ?

[/0]

You could but it could be messy.

From the conjugate root theorem, for a polynomial with real coefficients, every complex solution with has a conjugate solution .

So another solution would be , and so another factor would be

A polynomial of degree 3 has 3 solutions, so it can be factored as:



And since the only way to get term on the Left-hand-side is for the two constant terms on the Right-hand-side to be multiplied together, we must have , so .

So and those are our linear factors.

[TrueTears]

z - ( 1 + i ) is a factor therefore its conjugate will also be a factor since all the coefficients are real.

so z - (1-i) is also a factor.

now   yields an

long divide that to find the other linear factor.

[kamil9876]

z - ( 1 + i ) is a factor therefore its conjugate will also be a factor since all the coefficients are real.

so z - (1-i) is also a factor.

now   yields an

long divide that to find the other linear factor.

To save time, forget about long division but just realise that the final factor must have a real zero.

So 2c=12 by comparing coefficients
c=6.

[d0minicz]

thanks guys

when you multiplied the factors truetears, thats jus saying (z- (1+i)) x (z- (1-i)) x (z-r)  yeah ? since 3 factors

[TrueTears]

thanks guys

when you multiplied the factors truetears, thats jus saying (z- (1+i)) x (z- (1-i)) x (z-r)  yeah ? since 3 factors

yeap.

and (z-r) is the one you are trying to find out.

[d0minicz]

how do i know which ones to group ?
ie. (z-(1+i)) or ((z-1) +i)
thanks

[kamil9876]

how do i know which ones to group ?
ie. (z-(1+i)) or ((z-1) +i)
thanks

Both are correct, but the second one allows u to use difference of squares very nicely, so i recommend it.

Edit: the two are not equivalent, first one is z-1-i when expanded while the other is z-1+i. But the idea of having ((z-1)+i)((z-1)-i) grouped like this still stands.