complex numbers?

Question

Find a polynomial expression of least possible degree having roots 3i and sqrt(2) if its coefficients are:

(i) complex numbers
(ii) real numbers
(iii) rational numbers

plz help, iam totally lost! Step by step plz!

Answer 1

(i)

Normally, when we have real coefficients, complex roots always come in conjugate pairs; in our case, the conjugate pair of 3i would be -3i. But since we don't care about real coefficients, it's simple enough to make a quadratic with roots 3i and sqrt(2). That is, if
(x - 3i) (x - sqrt(2)) = 0 has roots 3i and sqrt(2), then the corresponding function has "zeros" of 3i and sqrt(2):

f(x) = (x - 3i) (x - sqrt(2)).

Expanding this out (using FOIL):

f(x) = x^2 - x sqrt(2) - 3i(x) + [3sqrt(2)]i, or
f(x) = x^2 - (sqrt2 - 3i)x + [3sqrt(2)]i

(ii) As we stated earlier, for real coefficient, conjugate roots always come in pairs, so we merely create an equation with 3i, -3i and sqrt(2) as roots.

f(x) = (x - 3i) (x + 3i) (x - sqrt(2))

Note that (x - 3i)(x + 3i) = ( x^2 - (3i)^2)
= (x^2 - 9i^2), and noting i^2 = -1,
= (x^2 - 9(-1)) = (x^2 + 9), so we have

f(x) = (x^2 + 9) (x - sqrt(2))

Foiling this out makes it become a cubic.

f(x) = x^3 - sqrt(2)x^2 + 9x - 9sqrt(2)

(iii) In order to obtain *rational* numbers, not only do we have to pair complex roots with their conjugate, but we also have to pair radicals (square roots) with their conjugate as well. This should do it:

f(x) = (x - 3i) (x + 3i) (x - sqrt(2)) (x + sqrt(2))

Note that we have two pairs of conjugates; therefore, two difference of squares take place.

f(x) = (x^2 + 9) (x^2 - 2)

Which we just foil out, and it becomes a quartic.

f(x) = x^4 - 2x^2 + 9x^2 - 18
f(x) = x^4 + 7x^2 - 18

Answer 2

An equation with two roots must have degree of at least 2. We have

ax^2 + bx + c = 0

a(3i)^2 + 3ib + c = 0
-9a + 3ib + c = 0
and
2a + sqrt(2)b + c = 0