From: Andrew Lorimer Date: Thu, 28 Feb 2019 23:00:51 +0000 (+1100) Subject: update planner, create complex reference X-Git-Tag: yr12~233 X-Git-Url: https://git.lorimer.id.au/notes.git/diff_plain/8cf574c0abfcc83eb077cc9a2c5684ae8cf54a63?ds=sidebyside update planner, create complex reference --- diff --git a/planner.xlsx b/planner.xlsx index 62283a5..b0c6861 100644 Binary files a/planner.xlsx and b/planner.xlsx differ diff --git a/spec/complex-ref.pdf b/spec/complex-ref.pdf new file mode 100644 index 0000000..96ab8bf Binary files /dev/null and b/spec/complex-ref.pdf differ diff --git a/spec/complex.md b/spec/complex.md index c5ad6be..b70737f 100755 --- a/spec/complex.md +++ b/spec/complex.md @@ -1,81 +1,90 @@ +--- +geometry: margin=2cm + +graphics: yes +tables: yes +author: Andrew Lorimer +classoption: twocolumn +header-includes: +- \usepackage{harpoon} +- \usepackage{amsmath} +- \pagenumbering{gobble} + +--- + + # Complex & Imaginary Numbers ## Imaginary numbers -$i^2 = -1 \quad \therefore i = \sqrt {-1}$ +$$i^2 = -1 \quad \therefore i = \sqrt {-1}$$ ### Simplifying negative surds -$\sqrt{-2} = \sqrt{-1 \times 2}$ -$= \sqrt{2}i$ +\begin{equation}\begin{split}\sqrt{-2} & = \sqrt{-1 \times 2} \\ & = \sqrt{2}i\end{split}\end{equation} ## Complex numbers -$\mathbb{C} = \{a+bi : a, b \in \mathbb{R} \}$ +$$\mathbb{C} = \{a+bi : a, b \in \mathbb{R} \}$$ General form: $z=a+bi$ $\operatorname{Re}(z) = a, \quad \operatorname{Im}(z) = b$ ### Addition -If $z_1 = a+bi$ and $z_2=c+di$, then -$z_1+z_2 = (a+c)+(b+d)i$ +If $z_1 = a+bi$ and $z_2=c+di$, then + +$$z_1+z_2 = (a+c)+(b+d)i$$ ### Subtraction -If $z_1=a+bi$ and $z_2=c+di$, then $z_1−z_2=(a−c)+(b−d)i$ +If $z_1=a+bi$ and $z_2=c+di$, then + +$$z_1−z_2=(a−c)+(b−d)i$$ ### Multiplication by a real constant -If $z=a+bi$ and $k \in \mathbb{R}$, then $kz=ka+kbi$ +If $z=a+bi$ and $k \in \mathbb{R}$, then -### Powers of $i$ -$i^0=1$ -$i^1=i$ -$i^2=-1$ -$i^3=-i$ -$i^4=1$ -$\dots$ +$$kz=ka+kbi$$ -Therefore.. +### Powers of $i$ - $i^{4n} = 1$ - $i^{4n+1} = i$ - $i^{4n+2} = -1$ - $i^{4n+3} = -i$ -For $i^n$, divide $n$ by 4 and let remainder $= r$. Then $i^n = i^r$. +For $i^n$, find remainder $r$ when $n \div 4$. Then $i^n = i^r$. ### Multiplying complex expressions -If $z_1 = a+bi$ and $z_2=c+di$, then -$z_1 \times z_2 = (ac-bd)+(ad+bc)i$ +If $z_1 = a+bi$ and $z_2=c+di$, then -### Conjugates +$$z_1 \times z_2 = (ac-bd)+(ad+bc)i$$ -If $z=a+bi$, conjugate of $z$ is $\overline{z} = a-bi$ (flipped operator) +### Conjugates -Also, $z \overline{z} = (a+bi)(a-bi) = a^2+b^2 = |z|^2$ +If $z=a+bi$, conjugate is -- Multiplication and addition are associative +$$\overline{z} = a-bi$$ -#### Properties +##### Properties - $\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}$ - $\overline{z_1 z_2} = \overline{z_1} \cdot \overline{z_2}$ - $\overline{kz} = k \overline{z}, \text{ for } k \in \mathbb{R}$ -- $z \overline{z} = |z|^2$ +- $z \overline{z} = = (a+bi)(a-bi) = a^2+b^2 = |z|^2$ - $z + \overline{z} = 2 \operatorname{Re}(z)$ ### Modulus Distance from origin. -$|{z}|=\sqrt{a^2+b^2}$ -$\therefore z \overline{z} = |z|^2$ +$$|{z}|=\sqrt{a^2+b^2} \quad \therefore z \overline{z} = |z|^2$$ -#### Properties +###### Properties - $|z_1 z_2| = |z_1| |z_2|$ - $|{z_1 \over z_2}| = {|z_1| \over |z_2|}$ @@ -83,11 +92,11 @@ $\therefore z \overline{z} = |z|^2$ ### Multiplicative inverse -$z^{-1} = {1 \over z} = {{a-bi} \over {a^2+B^2}} = {\overline{z} \over {|z|^2}}$ +\begin{equation}\begin{split}z^{-1} & = {1 \over z} \\ & = {{a-bi} \over {a^2+B^2}} \\ & = {\overline{z} \over {|z|^2}}\end{split}\end{equation} ### Dividing complex numbers -${{z_1} \over {z_2}} = {{z_1\ {z_2}^{-1}}} = {{z_1 \overline{z_2}} \over {{|z_2|}^2}}$ +$${{z_1} \over {z_2}} = {{z_1\ {z_2}^{-1}}} = {{z_1 \overline{z_2}} \over {{|z_2|}^2}} \quad \text{multiplicative inverse}$$ (using multiplicative inverse) @@ -97,20 +106,18 @@ ${z_1 \over z_2} = {{(a+bi)(c-di)} \over {c^2+d^2}}$ ## Argand planes - Geometric representation of $\mathbb{C}$ -- Horizontal $= \operatorname{Re}(z)$; vertical $= \operatorname{Im}(z)$ +- horizontal $= \operatorname{Re}(z)$; vertical $= \operatorname{Im}(z)$ - Multiplication by $i$ results in an anticlockwise rotation of $\pi \over 2$ -## Solving complex quadratics - -To solve $z^2+a^2=0$ (sum of two squares): +## Solving complex polynomials -$z^2+a^2=z^2-(ai)^2=(z+ai)(z-ai)$ +**Include $\pm$ for all solutions, including imaginary** -*Must include $\pm$ in solutions* +## Solving complex quadratics -## Solving complex polynomials +To solve $z^2+a^2=0$ (sum of two squares): -Include $\pm$ for all solutions, including imaginary. +$$z^2+a^2=z^2-(ai)^2=(z+ai)(z-ai)$$ #### Dividing complex polynomials @@ -124,13 +131,13 @@ Let $\alpha \in \mathbb{C}$. Remainder of $P(z) \div (z - \alpha)$ is $P(\alpha) ## Conjugate root theorem -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. +If $a+bi$ is a solution to $P(z)=0$, with $a, b \in \mathbb{R}$, then the conjugate $\overline{z}=a-bi$ is also a solution. ## Polar form -$$\begin{equation}\begin{split}z & =r \operatorname{cis} \theta \\ & = r(\operatorname{cos}\theta+i \operatorname{sin}\theta) \\ & = a + bi \end{split}\end{equation}$$ +\begin{equation}\begin{split}z & =r \operatorname{cis} \theta \\ & = r(\operatorname{cos}\theta+i \operatorname{sin}\theta) \\ & = a + bi \end{split}\end{equation} -- $r=|z|$, given by Pythagoras ($r=\sqrt{\operatorname{Re}(z)^2 + \operatorname{Im}(z)^2}$) +- $r=|z|=\sqrt{\operatorname{Re}(z)^2 + \operatorname{Im}(z)^2}$ - $\theta=\operatorname{arg}(z)$ (on CAS: `arg(a+bi)`) - **principal argument** is $\operatorname{Arg}(z) \in (-\pi, \pi]$ (note capital $\operatorname{Arg}$) @@ -151,17 +158,22 @@ ${z_1 \over z_2} = {r_1 \over r_2} \operatorname{cis}(\theta_1-\theta_2)$ (divid ## de Moivres' Theorem -$(r\operatorname{cis}\theta)^n=r^n\operatorname{cis}(n\theta)$ where $n \in \mathbb{Z}$ +$$(r\operatorname{cis}\theta)^n=r^n\operatorname{cis}(n\theta) \text{ where } n \in \mathbb{Z}$$ ## Roots of complex numbers -$n$th roots of $r \operatorname{cis} \theta$ are: -$z={r^{1 \over n}} \cdot (\cos ({{\theta + 2k \pi} \over n}) + i \sin ({{\theta + 2 k \pi} \over n}))$ +$n$th roots of $z = r \operatorname{cis} \theta$ are + +$$z={r^{1 \over n}} \operatorname{cis}({{\theta + 2 k \pi} \over n})$$ Same modulus for all solutions. Arguments are separated by ${2 \pi} \over n$ +The solutions of $z^n=a \text{ where } a \in \mathbb{C}$ lie on circle + +$$x^2 + y^2 = (|a|^{1 \over n})^2$$ + ## Sketching complex graphs - **Straight line:** $\operatorname{Re}(z) = c$ or $\operatorname{Im}(z) = c$ (perpendicular bisector) or $\operatorname{Arg}(z) = \theta$ - **Circle:** $|z-z_1|^2 = c^2 |z_2+2|^2$ or $|z-(a + bi)| = c$ -- **Locus:** $\operatorname{Arg}(z) \lt \theta$ +- **Locus:** $\operatorname{Arg}(z) < \theta$