MATH 3210 Class Notes 03-05-2025

notevector-spaces

Refresher on Polynomials

Definition 1

Suppose $z=a+bi\in \mathbb{C}$. Then,

$$a=\mathrm{Re}z=\text{ real part of }z$$ $$b=\mathrm{Im}z=\text{ imaginary part of }z$$

So $z=(\mathrm{Re}z)+(\mathrm{Im}z)i$.

Defined Complex Conjugate

Defined Modulus (Absolute Value)

Theorem 1

Let $w,z\in \mathbb{C}$. Then,

• $z+\bar{z}=2(\mathrm{Re}z)$.
• $z-\bar{z}=2(\mathrm{Im}z)i$.
• $z\bar{z}=|z{|}^2$.
• $\overline{w+z}=\bar{w}+\bar{z}$.
• $\overline{\stackrel{ˉ}{z}}=z$.
• $|\mathrm{Re}z|\le |Z|$ and $|\mathrm{Im}z|\le |z|$.
• $|\bar{z}|=|z|$.
• $|wz|=|w|\cdot |z|$.
• $|w+z|\le |w|+|z|$ (triangle inequality).

Defined Root (Zero) of a Polynomial

Theorem

Suppose $m$ is a positive integer and $p\in P(\mathbb{F})$ of degree $m$. Then $p$ has at most $m$ zeros in $\mathbb{F}$.

Proof (Sketch)

If $\lambda \in \mathbb{F}$ is a zero, then $\exists q\in P(\mathbb{F})$ with degree $m-1$ such that $p(z)=(z-\lambda )q(z)$. Use induction on $m$. $\square$

Division Algorithm (Polynomials)

Fundamental Theorem of Algebra

Theorem 2

Suppose $p\in P(\mathbb{C})$ is a polynomial with real coefficients. If $\lambda \in \mathbb{C}$ is a zero of $p$, then so is $\bar{\lambda }$.

Theorem 3

Suppose $p\in P(\mathbb{R})$ is a non-constant polynomial. Then, $p$ has a unique factorization (up to order) of the form,

$$p(x)=c(x-{\lambda }_1)\cdots (x-{\lambda }_m)(x^2+b_{1}x+c_1)\cdots (x^2+b_{M}x+c_M),$$

where $c,{\lambda }_1,\dots ,{\lambda }_m,b_1,\dots ,b_M,c_1,\dots ,c_M\in \mathbb{R}$ and $b_k^2<4c_k$ for all $k=1,\dots ,M$.

Proof (Sketch)

Imagine $p\in P(\mathbb{C})$. If all the zeros are real numbers, we get our factorization and $M=0$ by the Fundamental Theorem of Algebra. If $\lambda \in \mathbb{C}$ is a zero, so is $\bar{\lambda }$ by Theorem 2. So,

$$p(x)=(x-\lambda )(x-\bar{\lambda })q(x)=(x^2-2(\mathrm{Re}\lambda )x+|\lambda {|}^2)q(x).\square$$