separate complex numbers and matrices
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mechanical/mmme1026_maths_for_engineering/complex_numbers.md
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mechanical/mmme1026_maths_for_engineering/complex_numbers.md
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---
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author: Alvie Rahman
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date: \today
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title: MMME1026 // Complex Numbers
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tags: [ uni, nottingham, mechanical, engineering, mmme1026, maths, complex_numbers ]
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---
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# Complex Numbers
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## What is a Complex Number?
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- $i$ is the unit imaginary number, which is defined by:
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$$ i^2 = -1 $$
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- An arbritary complex number is written in the form
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$$z = x + iy$$
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Where:
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- $x$ is the real part of $z$ (which you may seen written as $\Re(z) = x$ or Re$(z) = x$)
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- $y$ is the imaginary part of $z$ (which you may seen written as $\Im(z) = y$ or Im$(z) = y$)
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- Two complex numbers are equal if both their real and imaginary parts are equal
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e.g. $$(3 + 4i) + (1 -2i) = (3+1) + i(4-2) = 2 + 2i$$
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### The Complex Conjugate
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Given complex number $z$:
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$$z = z + iy$$
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The complex conjugate of z, $\bar z$ is:
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$$\bar{z} = z -iy$$
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### Division of Complex Numbers
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- Multiply numerator and denominator by the conjugate of the denominator
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<details>
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<summary>
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#### Example
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</summary>
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> \begin{align*}
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z_1 &= 5 + i \\
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z_2 &= 1 -i \\
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\\
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\frac{z_1}{z_2} &= \frac{5+i}{1-i} \cdot \frac{\bar{z_2}}{\bar{z_2}} = \frac{(5+i)(1+i)}{(1-i)(1+i)} \\
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&= \frac{5 + i + 5i -1}{1 + 1} \\
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&= \frac{4 + 6i}{2} = 2 + 3i
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> \end{align*}
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</details>
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### Algebra and Conjugation
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When taking complex conjugate of an algebraic expresion, we can replace $i$ by $-i$ before or after
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doing the algebraic operations:
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\begin{align*}
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\overline{z_1+z_2} &= \bar{z_1}+\bar{z_2} \\
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\overline{z_1z_2} &= \bar{z_1}\bar{z_2} \\
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\overline{z_1/z_2} &= \frac{\bar{z_1}}{\bar{z_2}}
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\end{align*}
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The conjugate of a real number is the same as that number.
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#### Application
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If $z$ is a root of the polynomial equation
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$$0 = az^2 + bz + c$$
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with **real** coefficients $a$, $b$, and $c$, then $\bar{z}$ is also a root because
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\begin{align*}
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0 &= \overline{az^2 + bz + c} \\
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&= \bar{a}\bar{z}^2 + \bar{b}\bar{z} + \bar{c} \\
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&= a\bar{z}^2 + b\bar{z} + c
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\end{align*}
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### The Argand Diagram
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A general complex number $z = x + iy$ has two components so it can can be represented as a point in
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the plane with Cartesion coordinates $(x, y)$.
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\begin{align*}
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4-2i &\leftrightarrow (4, -2) \\
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-i &\leftrightarrow (0, -1) \\
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z &\leftrightarrow (x, y) \\
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\bar z &\leftrightarrow (x, -y)
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\end{align*}
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### Plotting on a Polar Graph
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We can also describe points in the complex plain with polar coordinates $(r, \theta)$:
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\begin{align*}
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z &= r(\cos\theta + i\sin\theta) &\text{ polar form of $z$} \\
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r &= \sqrt{x^2+y^2} &\text{(modulus)}\\
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\theta &= \arg z,\text{ where} \tan \theta = \frac y x &\text{(argument)} \\
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x &= r\cos \theta \\
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y &= r\sin \theta
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\end{align*}
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Be careful when turning $(x, y)$ into $(r, \theta)$ form as $\tan^{-1} \frac y x = \theta$ does not
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always hold true as there are many solutions.
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#### Choosing $\theta$ Correctly
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1. Determine which quadrant the point is in (draw a picture).
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2. Find a value of $\theta$ such that $\tan \theta = \frac y x$ and check that it is consistent.
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If it puts you in the wrong quadrant, add or subtract $\pi$.
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## Exponential Functions
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- The exponential function $f(x) = \exp x$ may be wirtten as an infinite series:
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$$\exp x = e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots $$
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- The function $f(x) = e^{-x}$ is just $\frac 1 {e^x}$
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- Note the important properties:
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\begin{align*}
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e^{a+b} &= e^a e^b \\
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(e^a)^b &= e^{ab}
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\end{align*}
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## Euler's Formula
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$$e^{i\theta} = \cos\theta + i\sin\theta$$
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- Properties of $e^{i\theta}$: For any real angle $\theta$ we have
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$$|e^{i\theta}| = |\cos\theta + i\sin\theta| = \sqrt{\cos^2\theta + \sin^2\theta} = 1$$
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and
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$$ \arg {e^{i\theta}} = \theta $$
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- A complex number in *polar form*, where $r = |z|$, and $\theta = \arg z$, may alternatively be
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written in its *exponential form*:
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$$z = re^{i\theta}$$
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**Note**: $$\bar z = r\cos\theta - ir\sin\theta = re^{-i\theta}$$
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<details>
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<summary>
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### Example 1
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Write $z = -1 + i$ in exponential form
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</summary>
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> $\arg z = \frac {3\pi} 4$
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> $|z| = \sqrt 2$
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>
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> So $z = \sqrt2e^{i\frac{3\pi} 4}$
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</details>
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<details>
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<summary>
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### Example 2
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The equations for a mechanical vibration problem are found to have the following mathematical
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solution:
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$$z(t) = \frac{e^{i\omega t}}{\omega_0^2-\omega^2 + i\gamma}$$
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</summary>
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where $t$ represents time and $\omega$, $\omega_0$ and $\gamma$ are all positive real physical
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constants.
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Although $z(t)$
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is complex and cannot directly represent a physical solution, it turns out that the real and
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imaginary parts $x(t)$ and $y(t)$ in $z(t) = x(t) + iy(t)$ can. Polar notation can be used to extract
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this physical information efficiently as follows:
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a. Put the denominator in the form
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$$ae^{i\delta}$$
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where you should give explicit expressions for $a$ and $\delta$ in terms of $\gamma$, $\gamma_0$,
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and $\gamma$.
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> \begin{align*}
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a &= \sqrt{\gamma^2 + (\omega_0^2 - \omega^2)^2} \\
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\delta &= \tan^{-1}\frac \delta {\omega_0^2 - \omega^2}
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> \end{align*}
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b. Hence find the constants $b$ and $\varphi$ such that
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$$x(t) = b\cos(\omega t + \varphi)$$
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and write a similar expression for $y(t)$.
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> \begin{align*}
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z &= \frac{e^i\omega t}{ae^{i\delta}} = \frac 1 a e^{i (\omega t - \delta)} \\
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x + iy &= \frac 1 a \cos(\omega t - \delta) + \frac 1 a \sin(\omega t - \delta) \\
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\therefore \Re z &= x = \frac 1 a \cos(\omega t - \delta), \\
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\Im z &= y = \frac 1 a \sin(\omega t - \delta) \\
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\\
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b &= \frac 1 a = \frac 1 {\sqrt{\gamma^2 + (\omega_0^2 - \omega^2)^2}} \\
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\varphi &= -\delta = \tan^{-1}\frac \delta {\omega_0^2 - \omega^2}\\
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\\
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y(t) &= \frac 1 a \sin(\omega t - \delta) \\
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> \end{align*}
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</details>
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## Products of Complex Numbers
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Suppose we have 2 complex numbers:
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$$z_1 = x_1 + iy_1 = r_1e^{i\theta_1}$$
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$$z_2 = x_2 + iy_2 = r_2e^{i\theta_2}$$
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Using $e^a e^b = e^{a+b}$, the product is:
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\begin{align*}
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z_3 = z_1 z_2 &= (r_1e^{i\theta_1})(r_2e^{i\theta_2}) \\
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&= r_1r_2e^{i\theta_1}e^{i\theta_2} \\
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&= r_1r_2e^{i(\theta_1+\theta_2)} \\
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\\
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|z_1z_2| &= |z_1|\times|z_2| \\
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\arg z_1z_2 &= \arg z_1 \times \arg z_2
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\end{align*}
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## de Moivre's Theorem
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Let $z = re^{i\theta}$. Consider $z^n$.
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Since $z = r(\cos\theta + i\sin\theta)$,
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\begin{align*}
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z^n &= r^n(\cos\theta + i\sin\theta)^n &\text{(1)} \\
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\end{align*}
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But also
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\begin{align*}
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z^n &= (re^{i\theta})^n \\
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&= r^n(e^{i\theta})^n \\
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&= r^ne^{in\theta} \\
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&= r^n(\cos{n\theta} + i\sin{n\theta}) &\text{(2)} \\
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\end{align*}
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By equating (1) and (2), we find de Moivre's theorem:
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\begin{align*}
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r^n(\cos\theta +i\sin\theta)^n &= r^n(\cos{n\theta} + i\sin{n\theta}) \\
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(\cos\theta +i\sin\theta)^n &= (\cos{n\theta} + i\sin{n\theta})
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\end{align*}
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<details>
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<summary>
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### Example 1
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Write $1+i$ in polar form and use de Moivre's theorem to calculate $(1+i)^{15}$.
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</summary>
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> \begin{align*}
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r &= |1+i| = \sqrt2 \\
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\theta &= \arg{1+i} = \frac \pi 4 \\
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\\
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\text{So } 1 + i &= \sqrt2(\cos{\frac pi 4} + i\sin{\frac \pi 4}) = \sqrt2e^{i\frac \pi 4} \text{ and}\\
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(i+i)^{15} &= (\sqrt2)^{15}\left(\cos{\frac \pi 4} + i\sin{\frac \pi 4}\right)^{15} \\
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&= 2^{\frac 15 2} \left(\cos{\frac{15\pi} 4} + i\sin{\frac{15\pi} 4}\right) \\
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&= 2^{\frac 15 2} \left(\frac 1 {\sqrt2} - \frac i {\sqrt2}\right) \\
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&= 2^7 (1 - i) \\
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&= 128 - 128i
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> \end{align*}
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</details>
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<details>
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<summary>
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### Example 2
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Use de Moivre's theorem to show that
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\begin{align*}
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\cos{2\theta} &= \cos^2\theta-\sin^2\theta \\
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\text{and} \\
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\sin{2\theta} &= 2\sin\theta\cos\theta
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\end{align*}
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</summary>
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> Let $n=2$:
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> \begin{align*}
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(\cos\theta+i\sin\theta)^2 &= \cos^2\theta + 2i\sin\theta\cos\theta - \sin^2\theta \\
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\text{Real part: } \cos^2\theta - \sin^2\theta &= \cos{2\theta}\\
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\text{Imaginary part: } 2\sin\theta\cos\theta &= \sin{2\theta}
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> \end{align*}
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</details>
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<details>
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<summary>
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### Example 3
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Given that $n \in \mathbb{N}$ and $\omega = -1 + i$, show that
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$w^n + \bar{w}^n = 2^{\frac n 2 + 1}\cos{\frac{3n\pi} 4}$ with Euler's formula.
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</summary>
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> \begin{align*}
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r &= \sqrt{2} \\
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\arg \omega = \theta &= \frac 3 4 \pi \\
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\\
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\omega^n &= r^n(cos{n\theta} + i\sin{n\theta}) \\
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\bar\omega^n &= r^n(cos{n\theta} - i\sin{n\theta}) \\
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\omega^n + \bar\omega^n &= r^n(2\cos{n\theta}) \\
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&= 2^{\frac n 2 + 1}\cos{\frac {3n\pi} 4}
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> \end{align*}
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</details>
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## Complex Roots of Polynomials
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<details>
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<summary>
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### Example
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Find which complex numbers $z$ satisfy
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$$z^3 = 8i$$
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</summary>
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> 1. Write $8i$ in exponential form,
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>
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> $|8i| = 8$ and $\arg{8i} = \frac \pi 2$
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>
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> $\therefore 8i = 8e^{i\frac \pi 2}$
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>
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>
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> 2. Let the solution be $r = re^{i\theta}$.
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>
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> Then $z^3 = r^3e^{3i\theta}$.
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>
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> 3. $z^3 = r^3e^{3i\theta} = 8e^{i\frac \pi 2}$
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>
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> i. Compare modulus:
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>
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> $r^3 = 8 \rightarrow r = 2$
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>
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> ii. Compare argument:
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>
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> $$3\theta = \frac \pi 2$$
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>
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> is a solution but there are others since
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>
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> $$e^{i\frac \pi 2} = e^{i \frac \pi 2 + 2n\pi}$$
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>
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> so we get a solution whenever
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>
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> $$3\theta = \frac \pi 2 + 2n\pi$$
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>
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> for any integer `n`
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>
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> - $n = 0 \rightarrow z = \sqrt3 + i$
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> - $n = 1 \rightarrow z = -\sqrt3 + i$
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> - $n = 2 \rightarrow z = -2i$
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> - $n = 3 \rightarrow z = \sqrt3 + i$
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> - $n = 4 \rightarrow z = -\sqrt3 + i$
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> - The solutions start repeating as you can see
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>
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> In general, an $n$-th order polynomial has exactly $n$ complex roots.
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> Some of these complex roots may be real numbers.
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>
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> 4. There are three solutions
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</details>
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@ -1,394 +1,9 @@
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---
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author: Alvie Rahman
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date: \today
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title: MMME1026 // Mathematics for Engineering
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tags: [ uni, nottingham, mechanical, engineering, mmme1026, maths, complex_numbers ]
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title: MMME1026 // Systems of Equations and Matrices
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tags: [ uni, nottingham, mechanical, engineering, mmme1026, maths, systems_of_equations, matrices ]
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---
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# Complex Numbers
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## What is a Complex Number?
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- $i$ is the unit imaginary number, which is defined by:
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$$ i^2 = -1 $$
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- An arbritary complex number is written in the form
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$$z = x + iy$$
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Where:
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- $x$ is the real part of $z$ (which you may seen written as $\Re(z) = x$ or Re$(z) = x$)
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- $y$ is the imaginary part of $z$ (which you may seen written as $\Im(z) = y$ or Im$(z) = y$)
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- Two complex numbers are equal if both their real and imaginary parts are equal
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e.g. $$(3 + 4i) + (1 -2i) = (3+1) + i(4-2) = 2 + 2i$$
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### The Complex Conjugate
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Given complex number $z$:
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$$z = z + iy$$
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The complex conjugate of z, $\bar z$ is:
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$$\bar{z} = z -iy$$
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### Division of Complex Numbers
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- Multiply numerator and denominator by the conjugate of the denominator
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<details>
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<summary>
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#### Example
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</summary>
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> \begin{align*}
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z_1 &= 5 + i \\
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z_2 &= 1 -i \\
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\\
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\frac{z_1}{z_2} &= \frac{5+i}{1-i} \cdot \frac{\bar{z_2}}{\bar{z_2}} = \frac{(5+i)(1+i)}{(1-i)(1+i)} \\
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&= \frac{5 + i + 5i -1}{1 + 1} \\
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&= \frac{4 + 6i}{2} = 2 + 3i
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> \end{align*}
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</details>
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### Algebra and Conjugation
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When taking complex conjugate of an algebraic expresion, we can replace $i$ by $-i$ before or after
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doing the algebraic operations:
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\begin{align*}
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\overline{z_1+z_2} &= \bar{z_1}+\bar{z_2} \\
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\overline{z_1z_2} &= \bar{z_1}\bar{z_2} \\
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\overline{z_1/z_2} &= \frac{\bar{z_1}}{\bar{z_2}}
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\end{align*}
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The conjugate of a real number is the same as that number.
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#### Application
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If $z$ is a root of the polynomial equation
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$$0 = az^2 + bz + c$$
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with **real** coefficients $a$, $b$, and $c$, then $\bar{z}$ is also a root because
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\begin{align*}
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0 &= \overline{az^2 + bz + c} \\
|
||||
&= \bar{a}\bar{z}^2 + \bar{b}\bar{z} + \bar{c} \\
|
||||
&= a\bar{z}^2 + b\bar{z} + c
|
||||
\end{align*}
|
||||
|
||||
### The Argand Diagram
|
||||
|
||||
A general complex number $z = x + iy$ has two components so it can can be represented as a point in
|
||||
the plane with Cartesion coordinates $(x, y)$.
|
||||
|
||||
\begin{align*}
|
||||
4-2i &\leftrightarrow (4, -2) \\
|
||||
-i &\leftrightarrow (0, -1) \\
|
||||
z &\leftrightarrow (x, y) \\
|
||||
\bar z &\leftrightarrow (x, -y)
|
||||
\end{align*}
|
||||
|
||||
### Plotting on a Polar Graph
|
||||
|
||||
We can also describe points in the complex plain with polar coordinates $(r, \theta)$:
|
||||
|
||||
\begin{align*}
|
||||
z &= r(\cos\theta + i\sin\theta) &\text{ polar form of $z$} \\
|
||||
r &= \sqrt{x^2+y^2} &\text{(modulus)}\\
|
||||
\theta &= \arg z,\text{ where} \tan \theta = \frac y x &\text{(argument)} \\
|
||||
x &= r\cos \theta \\
|
||||
y &= r\sin \theta
|
||||
\end{align*}
|
||||
|
||||
Be careful when turning $(x, y)$ into $(r, \theta)$ form as $\tan^{-1} \frac y x = \theta$ does not
|
||||
always hold true as there are many solutions.
|
||||
|
||||
#### Choosing $\theta$ Correctly
|
||||
|
||||
1. Determine which quadrant the point is in (draw a picture).
|
||||
2. Find a value of $\theta$ such that $\tan \theta = \frac y x$ and check that it is consistent.
|
||||
If it puts you in the wrong quadrant, add or subtract $\pi$.
|
||||
|
||||
## Exponential Functions
|
||||
|
||||
- The exponential function $f(x) = \exp x$ may be wirtten as an infinite series:
|
||||
|
||||
$$\exp x = e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots $$
|
||||
|
||||
- The function $f(x) = e^{-x}$ is just $\frac 1 {e^x}$
|
||||
- Note the important properties:
|
||||
|
||||
\begin{align*}
|
||||
e^{a+b} &= e^a e^b \\
|
||||
(e^a)^b &= e^{ab}
|
||||
\end{align*}
|
||||
|
||||
## Euler's Formula
|
||||
|
||||
$$e^{i\theta} = \cos\theta + i\sin\theta$$
|
||||
|
||||
- Properties of $e^{i\theta}$: For any real angle $\theta$ we have
|
||||
|
||||
$$|e^{i\theta}| = |\cos\theta + i\sin\theta| = \sqrt{\cos^2\theta + \sin^2\theta} = 1$$
|
||||
|
||||
and
|
||||
|
||||
$$ \arg {e^{i\theta}} = \theta $$
|
||||
|
||||
- A complex number in *polar form*, where $r = |z|$, and $\theta = \arg z$, may alternatively be
|
||||
written in its *exponential form*:
|
||||
|
||||
$$z = re^{i\theta}$$
|
||||
|
||||
**Note**: $$\bar z = r\cos\theta - ir\sin\theta = re^{-i\theta}$$
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
### Example 1
|
||||
|
||||
Write $z = -1 + i$ in exponential form
|
||||
|
||||
</summary>
|
||||
|
||||
> $\arg z = \frac {3\pi} 4$
|
||||
> $|z| = \sqrt 2$
|
||||
>
|
||||
> So $z = \sqrt2e^{i\frac{3\pi} 4}$
|
||||
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
### Example 2
|
||||
|
||||
The equations for a mechanical vibration problem are found to have the following mathematical
|
||||
solution:
|
||||
|
||||
$$z(t) = \frac{e^{i\omega t}}{\omega_0^2-\omega^2 + i\gamma}$$
|
||||
|
||||
</summary>
|
||||
|
||||
where $t$ represents time and $\omega$, $\omega_0$ and $\gamma$ are all positive real physical
|
||||
constants.
|
||||
Although $z(t)$
|
||||
is complex and cannot directly represent a physical solution, it turns out that the real and
|
||||
imaginary parts $x(t)$ and $y(t)$ in $z(t) = x(t) + iy(t)$ can. Polar notation can be used to extract
|
||||
this physical information efficiently as follows:
|
||||
|
||||
a. Put the denominator in the form
|
||||
|
||||
$$ae^{i\delta}$$
|
||||
|
||||
where you should give explicit expressions for $a$ and $\delta$ in terms of $\gamma$, $\gamma_0$,
|
||||
and $\gamma$.
|
||||
|
||||
|
||||
> \begin{align*}
|
||||
a &= \sqrt{\gamma^2 + (\omega_0^2 - \omega^2)^2} \\
|
||||
\delta &= \tan^{-1}\frac \delta {\omega_0^2 - \omega^2}
|
||||
> \end{align*}
|
||||
|
||||
b. Hence find the constants $b$ and $\varphi$ such that
|
||||
|
||||
$$x(t) = b\cos(\omega t + \varphi)$$
|
||||
|
||||
and write a similar expression for $y(t)$.
|
||||
|
||||
> \begin{align*}
|
||||
z &= \frac{e^i\omega t}{ae^{i\delta}} = \frac 1 a e^{i (\omega t - \delta)} \\
|
||||
x + iy &= \frac 1 a \cos(\omega t - \delta) + \frac 1 a \sin(\omega t - \delta) \\
|
||||
\therefore \Re z &= x = \frac 1 a \cos(\omega t - \delta), \\
|
||||
\Im z &= y = \frac 1 a \sin(\omega t - \delta) \\
|
||||
\\
|
||||
b &= \frac 1 a = \frac 1 {\sqrt{\gamma^2 + (\omega_0^2 - \omega^2)^2}} \\
|
||||
\varphi &= -\delta = \tan^{-1}\frac \delta {\omega_0^2 - \omega^2}\\
|
||||
\\
|
||||
y(t) &= \frac 1 a \sin(\omega t - \delta) \\
|
||||
> \end{align*}
|
||||
|
||||
</details>
|
||||
|
||||
## Products of Complex Numbers
|
||||
|
||||
Suppose we have 2 complex numbers:
|
||||
|
||||
$$z_1 = x_1 + iy_1 = r_1e^{i\theta_1}$$
|
||||
$$z_2 = x_2 + iy_2 = r_2e^{i\theta_2}$$
|
||||
|
||||
Using $e^a e^b = e^{a+b}$, the product is:
|
||||
|
||||
\begin{align*}
|
||||
z_3 = z_1 z_2 &= (r_1e^{i\theta_1})(r_2e^{i\theta_2}) \\
|
||||
&= r_1r_2e^{i\theta_1}e^{i\theta_2} \\
|
||||
&= r_1r_2e^{i(\theta_1+\theta_2)} \\
|
||||
\\
|
||||
|z_1z_2| &= |z_1|\times|z_2| \\
|
||||
\arg z_1z_2 &= \arg z_1 \times \arg z_2
|
||||
\end{align*}
|
||||
|
||||
## de Moivre's Theorem
|
||||
|
||||
Let $z = re^{i\theta}$. Consider $z^n$.
|
||||
|
||||
Since $z = r(\cos\theta + i\sin\theta)$,
|
||||
\begin{align*}
|
||||
z^n &= r^n(\cos\theta + i\sin\theta)^n &\text{(1)} \\
|
||||
\end{align*}
|
||||
|
||||
But also
|
||||
|
||||
\begin{align*}
|
||||
z^n &= (re^{i\theta})^n \\
|
||||
&= r^n(e^{i\theta})^n \\
|
||||
&= r^ne^{in\theta} \\
|
||||
&= r^n(\cos{n\theta} + i\sin{n\theta}) &\text{(2)} \\
|
||||
\end{align*}
|
||||
|
||||
By equating (1) and (2), we find de Moivre's theorem:
|
||||
|
||||
\begin{align*}
|
||||
r^n(\cos\theta +i\sin\theta)^n &= r^n(\cos{n\theta} + i\sin{n\theta}) \\
|
||||
(\cos\theta +i\sin\theta)^n &= (\cos{n\theta} + i\sin{n\theta})
|
||||
\end{align*}
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
### Example 1
|
||||
|
||||
Write $1+i$ in polar form and use de Moivre's theorem to calculate $(1+i)^{15}$.
|
||||
|
||||
</summary>
|
||||
|
||||
> \begin{align*}
|
||||
r &= |1+i| = \sqrt2 \\
|
||||
\theta &= \arg{1+i} = \frac \pi 4 \\
|
||||
\\
|
||||
\text{So } 1 + i &= \sqrt2(\cos{\frac pi 4} + i\sin{\frac \pi 4}) = \sqrt2e^{i\frac \pi 4} \text{ and}\\
|
||||
(i+i)^{15} &= (\sqrt2)^{15}\left(\cos{\frac \pi 4} + i\sin{\frac \pi 4}\right)^{15} \\
|
||||
&= 2^{\frac 15 2} \left(\cos{\frac{15\pi} 4} + i\sin{\frac{15\pi} 4}\right) \\
|
||||
&= 2^{\frac 15 2} \left(\frac 1 {\sqrt2} - \frac i {\sqrt2}\right) \\
|
||||
&= 2^7 (1 - i) \\
|
||||
&= 128 - 128i
|
||||
> \end{align*}
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
### Example 2
|
||||
|
||||
Use de Moivre's theorem to show that
|
||||
|
||||
\begin{align*}
|
||||
\cos{2\theta} &= \cos^2\theta-\sin^2\theta \\
|
||||
\text{and} \\
|
||||
\sin{2\theta} &= 2\sin\theta\cos\theta
|
||||
\end{align*}
|
||||
|
||||
</summary>
|
||||
|
||||
> Let $n=2$:
|
||||
|
||||
> \begin{align*}
|
||||
(\cos\theta+i\sin\theta)^2 &= \cos^2\theta + 2i\sin\theta\cos\theta - \sin^2\theta \\
|
||||
\text{Real part: } \cos^2\theta - \sin^2\theta &= \cos{2\theta}\\
|
||||
\text{Imaginary part: } 2\sin\theta\cos\theta &= \sin{2\theta}
|
||||
> \end{align*}
|
||||
|
||||
</details>
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
### Example 3
|
||||
|
||||
Given that $n \in \mathbb{N}$ and $\omega = -1 + i$, show that
|
||||
$w^n + \bar{w}^n = 2^{\frac n 2 + 1}\cos{\frac{3n\pi} 4}$ with Euler's formula.
|
||||
|
||||
</summary>
|
||||
|
||||
> \begin{align*}
|
||||
r &= \sqrt{2} \\
|
||||
\arg \omega = \theta &= \frac 3 4 \pi \\
|
||||
\\
|
||||
\omega^n &= r^n(cos{n\theta} + i\sin{n\theta}) \\
|
||||
\bar\omega^n &= r^n(cos{n\theta} - i\sin{n\theta}) \\
|
||||
\omega^n + \bar\omega^n &= r^n(2\cos{n\theta}) \\
|
||||
&= 2^{\frac n 2 + 1}\cos{\frac {3n\pi} 4}
|
||||
> \end{align*}
|
||||
|
||||
</details>
|
||||
|
||||
## Complex Roots of Polynomials
|
||||
|
||||
<details>
|
||||
<summary>
|
||||
|
||||
### Example
|
||||
|
||||
Find which complex numbers $z$ satisfy
|
||||
|
||||
$$z^3 = 8i$$
|
||||
|
||||
</summary>
|
||||
|
||||
> 1. Write $8i$ in exponential form,
|
||||
>
|
||||
> $|8i| = 8$ and $\arg{8i} = \frac \pi 2$
|
||||
>
|
||||
> $\therefore 8i = 8e^{i\frac \pi 2}$
|
||||
>
|
||||
>
|
||||
> 2. Let the solution be $r = re^{i\theta}$.
|
||||
>
|
||||
> Then $z^3 = r^3e^{3i\theta}$.
|
||||
>
|
||||
> 3. $z^3 = r^3e^{3i\theta} = 8e^{i\frac \pi 2}$
|
||||
>
|
||||
> i. Compare modulus:
|
||||
>
|
||||
> $r^3 = 8 \rightarrow r = 2$
|
||||
>
|
||||
> ii. Compare argument:
|
||||
>
|
||||
> $$3\theta = \frac \pi 2$$
|
||||
>
|
||||
> is a solution but there are others since
|
||||
>
|
||||
> $$e^{i\frac \pi 2} = e^{i \frac \pi 2 + 2n\pi}$$
|
||||
>
|
||||
> so we get a solution whenever
|
||||
>
|
||||
> $$3\theta = \frac \pi 2 + 2n\pi$$
|
||||
>
|
||||
> for any integer `n`
|
||||
>
|
||||
> - $n = 0 \rightarrow z = \sqrt3 + i$
|
||||
> - $n = 1 \rightarrow z = -\sqrt3 + i$
|
||||
> - $n = 2 \rightarrow z = -2i$
|
||||
> - $n = 3 \rightarrow z = \sqrt3 + i$
|
||||
> - $n = 4 \rightarrow z = -\sqrt3 + i$
|
||||
> - The solutions start repeating as you can see
|
||||
>
|
||||
> In general, an $n$-th order polynomial has exactly $n$ complex roots.
|
||||
> Some of these complex roots may be real numbers.
|
||||
>
|
||||
> 4. There are three solutions
|
||||
|
||||
</details>
|
||||
|
||||
# Systems of Equations (Simultaneous Equations)
|
||||
|
||||
## Gaussian Elimination
|
||||
Loading…
Reference in New Issue
Block a user