Complex numbers, solving polynomial, signs of i

  • Thread starter Platypus26
  • Start date
  • #1
I'm revising complex numbers and having trouble with this question...

Question:

Verify that 2 of the roots of the equation:

z^3 +2z^2 + z + 2 = 0

are i and -2. Find any remaining roots

Attempt at solution:

i^3 +2 i^2 + i + 2 =
(-1)i + 2(-1) +i + 2 =
-i -2 + i +2 =0
therefore i is a root

(-2)^3 +2(-2)^2 + (-2) + 2 =
-8 + 8 = 0
therefore -2 is a root


Let y be the remaining root to be found...

(z+i) (z-2)(z+y) = 0
(z^2 - 2z + iz -2i) (z+y) = 0
z^3 - 2z^2 + iz^2 -2iz +z^2 y -2zy +izy -2iy = 0

z^3 + (i+y-2) z^2 + (iy-2y-2i)z + (-2iy) = 0

so from this I should be able to work out y by equating the coefficients...
i + y-2 = 2 -> i+y = 4??
iy-2-2i = 1 iy - 2i = 3?
-2iy = 2 from this i think y=i but it doesn't seem to agree with the other equations.

This is where i'm stuck.
Does i multiplied by -i equal 1 or -1?
Also what is is (-i)^2 ?

Thanks
 

Answers and Replies

  • #2
CompuChip
Science Advisor
Homework Helper
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I think the mistake is at the very beginning...
(z+i) (z-2)(z+y) = 0

The idea of factoring is that if you plug in a root, one of the factors will be zero. Therefore, you should always write (z - root). If +i and -2 are roots, then, you would get
(z-i)(z+2)(z-y) = 0

Does i multiplied by -i equal 1 or -1?
i * -i = -1 * i *i = -1 * -1 = 1

Also what is is (-i)^2 ?
(-i)^2 = i^2 = -1
just like
(-2)^2 = 2^2 = 4
 
  • #3
34,552
6,266
Following up on what Compuchip said, if i and -2 are roots, z - i and z + 2 are factors, which means also that (z - i)(z + 2) is a factor. If you multiply this out and divide your original polynomial by it, you will get your third factor.
 
  • #4
mathman
Science Advisor
7,891
460
One thing that you should always be aware of is that if all the coefficients of a polynomial are real, then the complex roots will come in conjugate pairs. The conjugate of i is -i.
 

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