#### Answer

$f(x)=x^4-6x^3+10x^2-6x+9$

#### Work Step by Step

The Conjugate Pairs Theorem states that when a polynomial has real coefficients, then any complex zeros occur in conjugate pairs. This means that, when $(p +i \ q)$ is a zero of a polynomial function with a real number of the coefficients, then its conjugate $(p –i q)$, is also a zero of the function.
We are given that the zeros of the function are: $\pm i$ and $3$ with multiplicity 2.
We write the factors of the function as $(x-zero)^{multiplicity}$ and multiply them to get the polynomial.
Therefore, we can write the equation of a function as:
$f(x) = (x-i)[x-(-i)](x-3)(x-3)\\=(x-i)(x+i)(x^2-6x+9)\\= (x^2+1)(x^2-6x+9)\\=x^2(x^2-6x+9)+1(x^2-6x+9)\\= x^4-6x^3+10x^2-6x+9$