HELP! Specialist question

[itsangelan]

Can someone please help me work out this question? I've been stuck on it for awhile. Thanks in advance!

If z=1+i is a zero of the polynomial z^3+az^2+bz+10-6i, find constants a and b, given that they are real.

[Sine]

Can someone please help me work out this question? I've been stuck on it for awhile. Thanks in advance!

If z=1+i is a zero of the polynomial z^3+az^2+bz+10-6i, find constants a and b, given that they are real.

so the question has given you one solution and a "hidden" solution since z=1+i Using the Conjugate Root thereom z=1-i is also a solution. Use this to produce a quadractic factor of the cubic and then use long division to find the final linear factor. Then expand the quadractic factor and the linear factor to read off a and b.

[ellipse]

so the question has given you one solution and a "hidden" solution since z=1+i Using the Conjugate Root thereom z=1-i is also a solution. Use this to produce a quadractic factor of the cubic and then use long division to find the final linear factor. Then expand the quadractic factor and the linear factor to read off a and b.

wouldnt the complex conjugate root theorem not work in this situation? Cause all the coefficients must be real, but the constant term is a complex term. I'm thinking of subbing in 1+i, expanding it then equating the real and imaginary part

[itsangelan]

so the question has given you one solution and a "hidden" solution since z=1+i Using the Conjugate Root thereom z=1-i is also a solution. Use this to produce a quadractic factor of the cubic and then use long division to find the final linear factor. Then expand the quadractic factor and the linear factor to read off a and b.

Sorry for being a pain but I've produced a quadratic factor by multiplying (z-1-i) and (z-1+i) that gives me (z^2-2z+2).
However, I'm just unable to use long division to find the final factor. I'm currently stuck at that stage.

[itsangelan]

wouldnt the complex conjugate root theorem not work in this situation? Cause all the coefficients must be real, but the constant term is a complex term. I'm thinking of subbing in 1+i, expanding it then equating the real and imaginary part

I've tried subbing in 1+i into the equation, letting that equal 0 and 1-i into the equation letting that equal 0. I've then tried solving it simultaneously but it just hasnt worked out for me.

[ellipse]

I've tried subbing in 1+i into the equation, letting that equal 0 and 1-i into the equation letting that equal 0. I've then tried solving it simultaneously but it just hasnt worked out for me.

this is what i came up with:
sub 1+i into the equation and make it =0 since its a root
(1+i)^3+a(1+i)^2+b(1+i)+10-6i=0
now simplifying this, we get:
-2+2i+2ia+b+bi+10-6i=0
Now equating the real part:
-2+b+10=0
hence, b=-8
Now equating the imaginary part:
2+2a+b-6=0
subbing in b=-8 and solving for a, we get a=6

so a=6, b=-8

[itsangelan]

Oh gosh thats genius. Thank you so much!

[Sine]

sorry didn't see that 6i (just read that a and b are real), so CRT doesn't work in this case.

A lot of the time in spec you will see a question and not be able to do it in your head in a sense but only realise what to do once you start having a go at the question.

try subbing in the solution that you have and see what happens.

EDIT: ellipse got it