Question

How would I find a cubic polynomial with
roots 1 and i? Thanks!

Answer 1

(x-i)(x-1)² or (x-1)(x-i)²

Answer 2

You need to know that complex roots come in complex conjugate pairs
also, a cubic (3rd power) has 3 roots.
for a quadratic with roots 2 and 3, the poylnomial is (x-2)(x-3)

Answer 3

y2, not repeated i, but complex conjugate pairs.

Answer 4

I am honestly trying to understand, but I am not sure how I would use that information to start the question. Is there a general process for this type of question?

Answer 5

First, can you identify the 3 roots? Do you know what complex conjugate pair is?

Answer 6

I'm not sure how to find the third root, but I realize that two of the roots are (1+0i) and (0+1i).

Answer 7

no. The roots are 1, +i , and -i
(the complex conjugate of a+bi is a-bi, here a=0, b=1)

Answer 8

Oh, I see. Okay.

Answer 9

Now we know the terms (x-i)(x- -i)(x-1)= p(x)
or (x-i)(x+i)(x-1) = p(x)
multiply out to get the full expression

Answer 10

Oh, i thought he wanted the roots to be only +i and 1
i didn't know that -i is also included

Answer 11

You always get imaginary roots in conjugate pairs (unless your polynomial has complex coeffs, which we typically do not have)

Answer 12

Okay, thank you very much! I understand it now. :)

Answer 13

Hmm, Ok, thanks for the information