Thread: complex numbers.pt2

Problem:

Is it possible to determine the real constant a so that

is a root for the equation


I started out like so that if p(b)=0, where p(z) is a polynomial of a degree n,
then p(z)=(z-b)*f(z) where f(z) is a polynomial of a degree n-1

meaning that



but apparently that was wrong..

Sorry for all the questions, but I have an exam coming up this monday

Originally Posted by Greenb

Problem:

Is it possible to determine the real constant a so that

is a root for the equation


I started out like so that if p(b)=0, where p(z) is a polynomial of a degree n,
then p(z)=(z-b)*f(z) where f(z) is a polynomial of a degree n-1

meaning that



but apparently that was wrong..

Sorry for all the questions, but I have an exam coming up this monday

Do you mean is the root?

Just substitute into the cubic, simplify and equate the real and imaginary parts of the result to zero. Then solve for .

This is what I got:




somehow I think I need the values for d and e to proceed..

How do I "equate the real and imaginary parts of the result to zero" seeing that I should leave the left side intact?

Please check your OP.
It must contain some mistakes.
You want to be a root of .
If I have guessed correctly, there is a simple answer for .

Ah didn't see that I had typed it wrong Anyway, yes a=-6 but I really can't see how.

Originally Posted by Greenb

a=-6 but I really can't see how.

Indeed, that is what I got.

I thought way more complicated and was actually totally off the track hehe. Thanks for clearing it up for me, I now see that mr fantastic also meant this solution, I was trying to fit in his advice into my solution ^^