Let P(z) = z^3 + az^2 + bz + c, where a, b and c E R. Two of the roots of P(z) = 0 are -2 and (-3+2i). Find the value of a, b and c
Thanks in Advance,
Adam
You might want to let z = x + iy so it's easier to distinguish between the real and complex parts.
To find the third root, remember that if a complex number is a root, then its conjugate is also a root.
Then P(Z) will equal the product of its factors, and (remembering to match real with real, imaginary with imaginary!) solve for a,b,c.
And note that a fast way to expand $\displaystyle (z - (\alpha + i \beta))(z - (\alpha - i \beta))$ is to note that it's equal to $\displaystyle (z - \alpha - i \beta)(z - \alpha + i \beta) = ([z - \alpha] - i \beta)([z - \alpha] + i \beta) = (z - \alpha)^2 + b^2 = \, ....$