Find a monic polynomial f(x) of least degree over C that has the given numbers...

Step 1
Given:
The zeros of the polynomial 3-i, i and 2
Step 2
The complex roots occur in pair, so conjugate of 3-i and i also be the zeros of the same polynomial
So, the polynomial has zeros: 3-i, 3+i, i , -i and 2
If ${\displaystyle a_1,a_2,a_3,a_4,a_5}$ are the zeros of the polynomial then the polynomial in factored form is written as:
${\displaystyle f\left(x\right)=a(x-a_1)(x-a_2)(x-a_3)(x-a_4)(x-a_5)}$
Where a is the leading coefficient since the given polynomial is monic so the leading coefficient is 1.
${\displaystyle \Rightarrow f\left(x\right)=1(x-(3-i))(x-(3+i))(x-\left(i\right))(x-(-i))(x-2)}$
${\displaystyle \Rightarrow f\left(x\right)=1(x-3+i)(x-3-i)(x-i)(x+i)(x-2)}$
Simplifying we get ${\displaystyle (x-3+i)(x-3-i)=x^2-6x+10}$
${\displaystyle (x-i)(x+i)=x^2+1}$
${\displaystyle \Rightarrow f\left(x\right)=(x^2-6x+10)(x^2+1)(x-2)}$
${\displaystyle \Rightarrow f\left(x\right)=x^5-8x^4+23x^3-28x^2+22x-20}$
Therefore, the required monic polynomial is ${\displaystyle f\left(x\right)=x^5-8x^4+23x^3-28x^2+22x-20}$