Find a real polynomial of least degree having 2i and 1-3i as roots.?

Answer 1

According to the fundamental theory of Algebra the imaginary # must have its reciprocal.
so you actually need 2i,-2i, 1-3i, and 1+3i
set it up in the (x- #) form
(x-2i)(x+2i) (x - (1-3i)) (x - (1+3i))
(x^ 2+ 4)(x^2 -2x +10)
x^4 -2x^3 + 14x^2 - 8x + 40

Answer 2

Complex roots always come in conjugate pairs. So the polynomial needs to be of degree four.

The roots are ±2i and 1 ± 3i.

The polynomial then is:

(x - 2i)(x + 2i)(x - 1 - 3i)(x - 1 + 3i) = 0

x^4 - 2x³ + 14x² - 8x + 40 = 0

Answer 3

If 2i and 1 - 3i are roots then so are -2i and 1 + 3i

putting these in factors
(x - 2i)(x + 2i)(x -1-3i)(x - 1+3i)

(x^ 2+ 4)(x^2 -2x +10)
x^4 -2x^3 + 14x^2 - 8x + 40