I'm revising complex numbers and having trouble with this question...(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Complex numbers, solving polynomial, signs of i

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