202301111401

Type :Note Tags : Complex Analysis

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Complex Polynomials

Fundamental Theorem of Algebra

Note

Every polynomial $p(z)\in \mathbb{C}[z]$ has a root. Hence, every polynomial factors into a product of linear polynomials. A polynomial has a Zero of order $m$ at $a\in \mathbb{C}$ if $p(z)=(z-a{)}^m\cdot q(z)$ where $q(a)\ne 0$.

Lucas Theorem

Proposition: If all zeroes of a polynomial lie in a half plane in $\mathbb{C}$, Then all zeroes of $p^{\prime }(z)$ in it too. Which implies that if you take the convex hull of the zeroes of $p(z)$, inclusive of the boundary line, then all the roots of $p^{\prime }(z)$ will all lie in size that. Proof: Suppose $p(z)=(z-{\alpha }_1)\cdots (z-{\alpha }_n)$ Then

$$\frac{p^{\prime }(z)}{p(z)}=\frac{1}{z-{\alpha }_1}+\cdots +\frac{1}{z-{\alpha }_n}$$

Suppose the half plane which contains the zeroes of $p(z)$ is

$$H:\frac{Im(z-a)}{b}<0$$

We have ${\alpha }_k\in H\forall k$, but suppose $z_0\in \mathbb{C}\setminus H$

$$\begin{array}{rlr}\frac{Im(z_0-{\alpha }_k)}{b}& =\frac{Im(z_0-a)}{b}-\frac{Im(z_0-{\alpha }_k)}{b}& \\ & ⟹b(z_0-{\alpha }_k{)}^{-1}<0& \forall k\\ ⟹Im(b)\frac{p^{\prime }(z_0)}{p(z_0)}& =Im\frac{b}{z-{\alpha }_1}+\cdots +Im\frac{b}{z-{\alpha }_n}<0& \\ & ⟹p^{\prime }(z_0)\ne 0& \end{array}$$

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Related Problems

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References

Analytic Function