91Ó°ÊÓ

A complex conjugate of a complex number z is represented as \(\bar{z}\) and has the form \(a - bi\), where z = a + bi, and \(\bar{z}\) is the conjugate of z. The complex conjugate is obtained by changing the sign of the imaginary part of the complex number. A polynomial with real coefficients can be represented as: \[ p(z) = c_nz^n + c_{n-1}z^{n-1} + \cdots + c_1z + c_0 \] where \(c_n, c_{n-1}, \dots, c_1, c_0\) are real coefficients, and n is a non-negative integer.

Let's first find the value of p(z) for a complex number z = a + bi. We'll substitute z in the polynomial equation: \[ p(z) = c_n(a + bi)^n + c_{n-1}(a + bi)^{n-1} + \cdots + c_1(a + bi) + c_0 \]

Now, let's find the value of p(conjugate of z) by substituting \(\bar{z} = a - bi\) in the polynomial equation: \[ p(\bar{z}) = c_n(a - bi)^n + c_{n-1}(a - bi)^{n-1} + \cdots + c_1(a - bi) + c_0 \]

To show p(\(\bar{z}\)) = \(\overline{p(z)}\), let's compute the conjugate of the polynomial p(z): \[ \overline{p(z)} = \overline{c_n(a + bi)^n + c_{n-1}(a + bi)^{n-1} + \cdots + c_1(a + bi) + c_0} \] Since the conjugate distributes over addition and multiplication, and real coefficients are unaffected by taking the conjugate, we can write: \[ \overline{p(z)} = c_n \overline{(a + bi)^n} + c_{n-1} \overline{(a + bi)^{n-1}} + \cdots + c_1 \overline{(a + bi)} + c_0 \] Now, recall that \(\overline{z^n} = \overline{z}^n\) for any complex number z, thus: \[ \overline{p(z)} = c_n (a - bi)^n + c_{n-1} (a - bi)^{n-1} + \cdots + c_1 (a - bi) + c_0 = p(\bar{z}) \] Hence, we have shown that for a polynomial with real coefficients and any complex number z, it holds true that \(p(\bar{z}) = \overline{p(z)}\).