From 52bec6d3f844825d1ac1d24fc11e6e5ce76d7b12 Mon Sep 17 00:00:00 2001 From: Andrew Lorimer Date: Fri, 22 Feb 2019 09:47:25 +1100 Subject: [PATCH] solving complex polynomials --- spec/complex.md | 16 ++++++++++++++++ 1 file changed, 16 insertions(+) diff --git a/spec/complex.md b/spec/complex.md index b1688eb..871e1d4 100755 --- a/spec/complex.md +++ b/spec/complex.md @@ -108,6 +108,22 @@ $z^2+a^2=z^2-(ai)^2=(z+ai)(z-ai)$ *Must include $\pm$ in solutions* +## Solving complex polynomials + +#### Dividing complex polynomials + +Dividing $P(z)$ by $D(z)$ gives quotient $Q(z)$ and remainder $R(z)$ such that: + +$$P(z) = D(z)Q(z) + R(z)$$ + +#### Remainder theorem + +Let $\alpha \in \mathbb{C}$. Remainder of $P(z) \div (z - \alpha)$ is $P(\alpha)$ + +## Conjugate root theorem + +Let $P(z)$ be a polynomial with real coefficients. If $a+bi$ is a solution to $P(z)=0$, with $a, b \in \mathbb{R}$, the the conjugate $a-bi$ is also a solution. + ## Polar form $$\begin{equation}\begin{split}z & =r \operatorname{cis} \theta \\ & = r(\operatorname{cos}\theta+i \operatorname{sin}\theta) \\ & = a + bi \end{split}\end{equation}$$ -- 2.43.2