Let p(z) = z³ + (1 + j) z² + (2 + j) z + 3, where z is a complex number.

Which one of the following is true?

Complex Polynomial Root Analysis

The given polynomial is $p(z)=z^3+(1+j)z^2+(2+j)z+3$, where $z$ is a complex number. This is a cubic polynomial, meaning it has exactly three roots (counting multiplicity). A key characteristic of this polynomial is that its coefficients are not all real numbers. Specifically, the coefficient of $z^2$ is $(1+j)$ and the coefficient of $z$ is $(2+j)$, both of which are complex numbers with non-zero imaginary parts.

Understanding the nature of polynomial roots, especially when coefficients are complex, is crucial for evaluating the given statements. Many common theorems about polynomial roots (like complex conjugate pairs) apply strictly to polynomials with real coefficients.

Analyzing Each Option for p(z)

- Option 1: All the roots cannot be real Consider a polynomial $P(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x+a_0$. If all the coefficients $(a_0,a_1,\dots ,a_n)$ are real numbers, then the roots can be real or complex (appearing in conjugate pairs). If all the roots of a polynomial were real numbers, say $r_1,r_2,\dots ,r_n$, then the polynomial could be written in factored form as $P(x)=C(x-r_1)(x-r_2)\dots (x-r_n)$. When this product is expanded, all the resulting coefficients $(a_0,a_1,\dots ,a_n)$ must necessarily be real numbers. Our given polynomial is $p(z)=z^3+(1+j)z^2+(2+j)z+3$. Here, the coefficients are $1$, $(1+j)$, $(2+j)$, and $3$. Since the coefficients $(1+j)$ and $(2+j)$ are complex numbers (not purely real), it is impossible for all the roots of $p(z)$ to be real. If all roots were real, the expanded polynomial would have only real coefficients, which contradicts the given $p(z)$. Therefore, this statement is true.
- **Option 2: $\overline{p(z)}=p(\overline{z})$ for all $z$ ** The property $\overline{p(z)}=p(\overline{z})$ holds true if and only if **all coefficients of the polynomial $p(z)$ are real numbers**. Let's demonstrate this: Let $p(z)=a_n{z}^n+a_{n-1}{z}^{n-1}+\cdots +a_{1}z+a_0$. - The conjugate of $p(z)$ is: $\overline{p(z)}=\overline{a_n{z}^n+a_{n-1}{z}^{n-1}+\cdots +a_{1}z+a_0}=\overline{a_n}\overline{z^n}+\overline{a_{n-1}}\overline{z^{n-1}}+\cdots +\overline{a_1}\overline{z}+\overline{a_0}$ - Substituting $\overline{z}$ into $p(z)$ gives: $p(\overline{z})=a_n(\overline{z}{)}^n+a_{n-1}(\overline{z}{)}^{n-1}+\cdots +a_1(\overline{z})+a_0$ For $\overline{p(z)}=p(\overline{z})$ to be true for all $z$, we must have $\overline{a_k}=a_k$ for every coefficient $a_k$. This means each coefficient $a_k$ must be a real number. In our polynomial $p(z)=z^3+(1+j)z^2+(2+j)z+3$, the coefficients are $a_3=1$, $a_2=(1+j)$, $a_1=(2+j)$, and $a_0=3$. Since $a_2=(1+j)$ is not a real number ($\overline{1+j}=1-j\ne 1+j$), and $a_1=(2+j)$ is not a real number ($\overline{2+j}=2-j\ne 2+j$), this condition does not hold for $p(z)$. Therefore, this statement is false.
- **Option 3: The sum of the roots of $p(z)=0$ is a real number** According to Vieta's formulas, for a polynomial equation $a_n{z}^n+a_{n-1}{z}^{n-1}+\cdots +a_{1}z+a_0=0$, the sum of its roots is given by the formula $-\frac{a_{n-1}}{a_n}$. For our cubic polynomial $p(z)=z^3+(1+j)z^2+(2+j)z+3$, we identify the coefficients: - $a_3=1$ (coefficient of $z^3$) - $a_2=(1+j)$ (coefficient of $z^2$) The sum of the roots is calculated as: $$\text{Sum of roots}=-\frac{a_2}{a_3}=-\frac{(1+j)}{1}=-1-j$$ The result $-1-j$ is a complex number (it has a non-zero imaginary part $-1$), not a real number. Therefore, this statement is false.
- **Option 4: The complex roots of the equation $p(z)=0$ come in conjugate pairs** The fundamental property that complex roots of a polynomial equation appear in conjugate pairs (i.e., if $z_0$ is a root, then its conjugate $\overline{z_0}$ is also a root) is valid if and only if all the coefficients of the polynomial are real numbers. As established, the polynomial $p(z)=z^3+(1+j)z^2+(2+j)z+3$ has complex coefficients (namely $1+j$ and $2+j$). Because not all coefficients are real, the conjugate root theorem does not apply. For example, consider a simple polynomial with a complex coefficient, $q(z)=z-j$. The root of this equation is $z=j$. The conjugate of this root is $\overline{j}=-j$. However, if we substitute $-j$ into $q(z)$, we get $q(-j)=-j-j=-2j$, which is not zero. Thus, $-j$ is not a root, and the complex root $j$ does not come in a conjugate pair. Therefore, this statement is false.

Conclusion

Based on the detailed analysis of each option, only the statement "All the roots cannot be real" is true for the given polynomial $p(z)=z^3+(1+j)z^2+(2+j)z+3$. The presence of non-real coefficients significantly changes the properties of its roots compared to polynomials with strictly real coefficients.