Find a real cubic polynomial whose has roots $1$ and $i$.

Question

I don't know how to show "real" in the question. In other words,how to eliminate $i$, because it's not real.

Answer 1

hint: $p(x) = (x-1)(x^2+1)$. Can you take it from here?

Answer 2

It seems $(x-1)(x^2+1)$ has roots $1, \pm i$. There's no way to make one with just roots $1, i$, but I assume that's not the intent.

Answer 3

HINT: For any polynomial with real coefficient $f(z) = 0 \implies f(\overline{z}) = 0$, where $z \in \mathbb{C}$

Answer 4

I think your point of confusion has to do with conjugate pairs. Saying there is a root i requires another root -i, always, so your source probably meant it to be implied or that's the trick they're getting at. But to explain why ...

The way I think about it, is the i component is a distance (well, root of the distance) above the y=0 if you start with a normal quadratic opening up. If you take (x-2)^2, you've got the regular x^2 shifted 2 right. Shift it down 1, your roots spread out, (x-2)^2 - 1 becomes [(x-2) -1]·[(x-2) +1] (so that when you multiply it out, the (x-2)·(1-1) cancels and the -1 remains) and your roots come from the resulting (x-3)·(x-1) (2 spreads out to 1 and 3).

For the rest to make sense, I need to point out that adding an x term is the same as starting with a left shifted parabola and shifting it down. Starting from x^2 and adding an x, it becomes x^2 +x = (x+0.5)^2 -0.25 = (x+0.5+0.5)·(x+0.5-0.5) =(x+1)·(x). So to be clear, my argument is about the effects of shifting a given parabola (that starts off just touching the y=0) either straight up or down, so (x-a)^2 +c, only because messing with the x term is just starting with a different parabola and shifting it up/down before applying the same arguments.

So if you shift (x-2)^2 up one, you need to get (x-2)^2 +1 into the form [(x-2) +b]·[(x-2) +c] so that the effect of b and c results in a +1. To do this there are two requirements on b and c. If b and c are real they can't be adding any x's, and if they are complex they still can't be adding x's and their i's need to cancel out. Real values of x need to correspond to real values of y.

If you consider (a+b)·(a+c) = a^2 + a(b+c) + bc, if a=(x-2) then it must be that b=-c or else (x-2)(b+c) will have the effect of adding x terms, and the goal was to just shift (x-2)^2 straight up. So you need to have an [(x-2)-c]·[(x-2)+c] so that your -(c^2) leaves a +1. Well that needs i! (-i)(+i)= -i^2 =1. So you have [(x-2)-i]·[(x-2)+i]=[(x-2)^2 +1]. Your roots didn't spread out, they went "up", and you MUST have two of them that are equal and opposite because you need a (-ci)·(+ci) to result in a +c^2 shift without any other effect on the parabola (no imaginary terms after it's all multiplied out). This is why the imaginary roots come in conjugate pairs with the same real part, you started with an (x-a)^2 and then put in oppositely signed i's in each (x-a) to shift it up. (or at least away from the y=0 line in the direction the quadratic is spreading, -1·(x-2)·(x-2) would still need the same conjugate pair -1·[(x-2)-i]·[(x-2)+i] to shift it down)

So if you want a root of i, with no real part, you must be starting with an x^2 (or (x-0)^2) and shifting it up. If you also want a root of 1, you'll need to multiply that upshifted x^2 by (x-1). This is easier to do if it's not in root form (as in (x-1+2i)·(x-1-2i)).

If we had say a quadratic x^2, shifted right 1 and up 4, we probably wouldn't write that as (x-1+2i)·(x-1-2i), maybe we like the look of [(x-1)^2 +4], or [x^2 - 2x +5]. Because the imaginary roots are in conjugate pairs, they'll always cancel out if you work out the terms.

Other responses seem to think we shouldn't give you the explicit answer (homework problem?) I hope I managed to (correctly) explain all the parts while not going over the finish line.