Complex numbers, solving polynomial, signs of i

[Platypus26]

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

 

[CompuChip]

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

 

[Mark44]

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.

 

[mathman]

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.

 

[CellCoree]

I understand your confusion and difficulties with complex numbers and solving polynomials. Let me provide some clarification and guidance.

Firstly, complex numbers are numbers that can be expressed in the form a + bi, where a and b are real numbers and i is the imaginary unit, defined as the square root of -1. In your equation, z^3 +2z^2 + z + 2 = 0, the coefficients (2, 1, 1, and 2) are all real numbers, so the roots of the equation can be either real or complex.

To verify that i and -2 are roots of the equation, you can simply substitute them into the equation and see if the result is equal to 0. In your attempt at solution, you made a small mistake in the first step where you wrote i^3 +2 i^2 + i + 2 = (-1)i + 2(-1) +i + 2. Remember that i^2 = -1, so the correct equation should be i^3 +2 i^2 + i + 2 = (-1)i + 2(-1) +i + 2 = 0.

Now, for finding the remaining root, you can use the fact that the sum of the roots of a cubic equation is equal to the negative of the coefficient of the second term (in this case, -2). Therefore, the sum of the three roots is -2. We already know that i and -2 are two of the roots, so the third root must be -2 - i.

To answer your question about (-i)^2, remember that i^2 = -1, so (-i)^2 = (-1)(-1) = 1. This is because when you multiply two negative numbers, the result is a positive number.

I hope this helps clarify your understanding of complex numbers and solving polynomials. Keep practicing and don't hesitate to ask for help when needed. Good luck!