MathsGee Answers is Zero-Rated (You do not need data to access) on: Telkom | Dimension Data | Rain | MWEB
First time here? Checkout the FAQs!
x
MathsGee is Zero-Rated (You do not need data to access) on: Telkom |Dimension Data | Rain | MWEB

1 like 0 dislike
16 views
Suppose $p(z)$ is a polynomial with real coefficients. Prove that
(a) $\overline{p(z)}=p(\bar{z})$;
(b) $p(z)=0$ if and only if $p(\bar{z})=0$;
(c) the roots of $p(z)=0$ appear in conjugate pairs, i.e., if $z_{0}$ is a root of $p(z)=0$, so is $\bar{z}_{0}$.
in Mathematics by Gold Status (10,269 points) | 16 views

1 Answer

0 like 0 dislike
Best answer
Proof. Let $p(z)=a_{0}+a_{1} z+\ldots+a_{n} z^{n}$ for $a_{0}, a_{1}, \ldots, a_{n} \in \mathbb{R} .$ Then
$$
\begin{aligned}
\overline{p(z)} &=\overline{a_{0}+a_{1} z+\cdots+a_{n} z^{n}} \\
&=\overline{a_{0}}+\overline{a_{1} z}+\cdots+\overline{a_{n} z^{n}} \\
&=\overline{a_{0}}+\left(\overline{a_{1}}\right) \bar{z}+\cdots+\left(\overline{a_{n}}\right) \overline{z^{n}} \\
&=a_{0}+a_{1} \bar{z}+\cdots+a_{n} \bar{z}^{n}=p(\bar{z}) .
\end{aligned}
$$
If $p(z)=0$, then $\overline{p(z)}=0$ and hence $p(\bar{z})=\overline{p(z)}=0 ;$ on the other hand, if $p(\bar{z})=0$, then $\overline{p(z)}=p(\bar{z})=0$ and hence $p(z)=0$ By the above, $p\left(z_{0}\right)=0$ if and only if $p\left(\bar{z}_{0}\right)=0 .$ Therefore, $z_{0}$ is a root of $p(z)=0$ if and only if $\bar{z}_{0}$ is.
by Gold Status (10,269 points)

MathsGee provides answers to subject-specific educational questions for improved outcomes.



On MathsGee Answers, you can:


1. Ask questions
2. Answer questions
3. Comment on Answers
4. Vote on Questions and Answers
5. Donate to your favourite users

MathsGee Tools

Math Worksheet Generator

Math Algebra Solver

Trigonometry Simulations

Vectors Simulations

Matrix Arithmetic Simulations

Matrix Transformations Simulations

Quadratic Equations Simulations

Probability & Statistics Simulations

PHET Simulations

Visual Statistics

Other Tools

MathsGee ZOOM | eBook

16,915 questions
12,339 answers
122 comments
2,430 users