How to find a real function with that has certain imaginary roots

Question

I recently started learning about imaginary numbers, and a question asked to find a quadratic with roots of $2i$, $4i$. Solved simply by expanding $(x-2i)(x-4i)$ into $x^2 - 6ix - 8 = 0$.

However, how would I find a real function with the same roots, that is a function without $i$ in it? I attempted by saying:

$-b/2a = 3i$ (is this a correct assumption?) $\dfrac{\sqrt{b^2 - 4ac}}{2a} = i $

Then I made $-b = 6ai$, and thus $b^2 = -36a^2$. Then I solved for $c$. But I keep reaching a dead end, saying $b = 6ai$ and $c =8a$. Whilst this agrees with the initial function $x^2 - 6ix - 8$, it’s not what I’m looking for. Any help would be appreciated, thanks!

Answer 1

Let $f(x)=ax^2+bx+c$ with $a,b,c \in \mathbb R$ and $a \ne 0$.

If $z_0 \in \mathbb C$ is a root of $f$, then $ \overline{z_0}$ is also a root of $f$.

Consequence: $2i$ and $4i$ are never roots of $f$ if $a,b,c$ are real.

Answer 2

For quadratic no solution. If not restrict solution to quadratic, then $$(x-2i)(x+2i)(x-4i)(x+4i)$$

Answer 3

If you want a polynomial $f$ with real coefficients and roots $2i$ and $4i$, then $f$ factors as $$f=(x-2i)(x-4i)g,$$ for some polynomial $g$. Because $f$ is real it satisfies $\bar{f}=f$, i.e. it is invariant under complex conjugation. So $$(x-2i)(x-4i)g=f=\bar{f}=(x+2i)(x+4i)\bar{g},$$ which shows that $f$ also has factors $x+2i$ and $x+4i$. It follows that $f$ is divisible by $$(x-2i)(x+2i)(x-4i)(x+4i)=(x^2+4)(x^2+16)=x^4+20x^2+64.$$ In particular $f$ cannot be quadratic.