Let p(z) = z³ + (1 + j) z² + (2 + j) z + 3, where z is a complex number. Which one of the following is true?

The correct option is: All the roots cannot be real

Let the given polynomial be:
$$p(z)=z^3+(1+j)z^2+(2+j)z+3$$

Let us analyze the roots of the equation $p(z)=0$ by assuming, for the sake of contradiction, that all three roots are real. Let these real roots be $r_1$, $r_2$, and $r_3$ where $r_1,r_2,r_3\in ℝ$.

According to Vieta's formulas, for a cubic polynomial of the form $a{z}^3+b{z}^2+cz+d=0$ with roots $r_1,r_2,r_3$:
The sum of the roots is given by:
$$r_1+r_2+r_3=-\frac{b}{a}$$

For the given polynomial, we have $a=1$ and $b=1+j$. Therefore:
$$r_1+r_2+r_3=-(1+j)=-1-j$$

Since we assumed that $r_1,r_2,$ and $r_3$ are all real numbers, their sum $r_1+r_2+r_3$ must also be a real number. However, the sum of the roots is $-1-j$, which has a non-zero imaginary component ($-1j$). This is a direct contradiction.

Consequently, our initial assumption must be false. Hence, all the roots of $p(z)=0$ cannot be real numbers. At least one root must be strictly complex (non-real).