Q24E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Let $P\left(r_1\right)=a_n{r}_1^n+a_{n-1}{r}_1^{n-1}+...+a_1{r}_1+a_0$  be a polynomialwith real coefficients $a_n,....,a_0$  . Prove that if r₁ is azero of  $P\left(r\right)$ , then so is its complex conjugate r₁. [Hint:Show that $\overline{P\left(r\right)}=P\left(\overline{r}\right)$  , where the bar denotes complexconjugation.]

Step-by-Step Solution

Answer

 If r₁ is a zero of the given polynomial P(r) then so is its complex conjugate r₁.

Step 1: Complex Conjugation

A complex conjugate of a complex number is another complex number that has the same real part as the original complex number and the imaginary part has the same magnitude but opposite sign. The product of a complex number and its complex conjugate is a real number.

Step 2: Use of Complex Conjugation properties

Suppose complex number r₁ is a zero of the given polynomial P(r). Hence P(r₁)=0:

 $P\left(r_1\right)=0\Rightarrow \overline{P\left(r_1\right)}=\overline{0}=0$

 

Using properties of complex conjugation:

 $\begin{array}{l}\overline{P\left(r_1\right)}=\overline{(a_n{r}_1^n+a_{n-1}{r}_1^{n-1}+\mathrm{...}+a_1{r}_1+a_0)}\\ =a_n\overline{r_1^n}+a_{n-1}\overline{r_1^{n-1}}+\mathrm{...}+a_1\overline{r_1}+a_0\\ =P\left(\overline{r_1}\right)=0\end{array}$

Hence, the final answer is:

If r₁ is a zero of the given polynomial P(r) then so is its complex conjugate r₁.