Complex Conjugate Root Theorem

Let p(x) be a polynomial with real coefficients c_k. P(x)= ∑_(k=0)^n c_kx^k Using the properties of conjugation, the conjugate of the polynomial evaluated at a+bi can be rewritten as expressed in the following table.

P(a+bi)=∑_(k=0)^n c_k (a+bi)^k	The conjugate of the polynomial evaluated at a+bi
P(a+bi)= ∑_(k=0)^n c_k (a+bi)^k	The conjugate of a sum is the sum of the conjugates.
P(a+bi)= ∑_(k=0)^n c_k (a+bi)^k	The conjugate of a product is the product of the conjugates.
P(a+bi)= ∑_(k=0)^n c_k (a+bi)^k	The conjugate of a power is the power of the conjugate.
P(a+bi)= ∑_(k=0)^n c_k(a+bi) ^k	The conjugate of a real number is itself.
P(a+bi)= ∑_(k=0)^n c_k(a-bi)^k	The conjugate of a+bi is a-bi.
P(a+bi)=P(a-bi)	The polynomial evaluated at a-bi

It is given that a+bi is a root of the polynomial p(x). P(a+bi)=0 The conjugate of the real number 0 is 0. Therefore, the conjugate of the polynomial evaluated at a+bi is 0. P(a+bi)=0 This equation, along with the last equality in the table, shows that a-bi, the conjugate of a+bi, is also a root of the polynomial. P(a+bi) = P(a-bi) [0.1cm] P(a+bi) = 0 ⇓ P(a-bi)=0