From 841db6edbdd89190d68a937c7ce0aaad4861ccc7 Mon Sep 17 00:00:00 2001 From: Alvie Rahman Date: Tue, 12 Oct 2021 12:30:28 +0100 Subject: [PATCH] Notes on mmme1026 lecture 2 --- mechanical/mmme1026_maths_for_engineering.md | 275 +++++++++++++++++++ 1 file changed, 275 insertions(+) create mode 100755 mechanical/mmme1026_maths_for_engineering.md diff --git a/mechanical/mmme1026_maths_for_engineering.md b/mechanical/mmme1026_maths_for_engineering.md new file mode 100755 index 0000000..5dc6242 --- /dev/null +++ b/mechanical/mmme1026_maths_for_engineering.md @@ -0,0 +1,275 @@ +--- +author: Alvie Rahman +date: \today +title: MMME1026 // Mathematics for Engineering +tags: [ uni, nottingham, mmme1026, maths, complex_numbers ] +--- + +# Lecture 1 // Complex Numbers (2021-10-04) + +## Complex Numbers + +- $i$ is the unit imaginary number, which is defined by: + + $$ i^2 = -1 $$ + +- An arbritary complex number is written in the form + + $$z = x + iy$$ + + Where: + + - $x$ is the real part of $z$ (Re($z$)) + - $y$ is the imaginary part of $z$(Im($z$)) + +- Two complex numbers are equal if both their real and imaginary parts are equal + + e.g. $$(3 + 4i) + (1 -2i) = (3+1) + i(4-2) = 2 + 2i$$ + +### Complex Conjugate + +Given complex number $z$: + +$$z = z + iy$$ + +The complex conjugate of z, $\bar z$ is: + +$$\bar{z} = z -iy$$ + +### Division of Complex Numbers + +- Multiply numerator and denominator by the conjugate of the denominator + +#### Example + +\begin{align*} +z_1 &= 5 + i \\ +z_2 &= 1 -i \\ +\\ +\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)} \\ + &= \frac{5 + i + 5i -1}{1 + 1} \\ + &= \frac{4 + 6i}{2} = 2 + 3i +\end{align*} + +### Algebra and Conjugation + +When taking complex conjugate of an algebraic expresion, we can replace $i$ by $-i$ before or after +doing the algebraic operations: + +\begin{align*} +\overline{z_1+z_2} &= \bar{z_1}+\bar{z_2} \\ +\overline{z_1z_2} &= \bar{z_1}\bar{z_2} \\ +\overline{z_1/z_2} &= \frac{\bar{z_1}}{\bar{z_2}} +\end{align*} + +The conjugate of a real number is the same as that number. + +#### Application + +If $z$ is a root of the polynomial equation + +$$0 = az^2 + bz + c$$ + +with **real** coefficients $a$, $b$, and $c$, then $\bar{z}$ is also a root because + +\begin{align*} +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 $\pi$. + +# Lecture 2 // Complex Numbers (2021-10-11) + +## Exponential Functions + +- The exponential function $f(x) = \exp x$ may be wirtten as an infinite series: + + $$\exp x = e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... $$ + +- 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**: $$\bar z = r\cos\theta - ir\sin\theta = re^{-i\theta}$$ + + Example: + + > Write $z = -1 + i$ in exponential form: + > + > $\arg z = \frac {3\pi} 4$ + > $|z| = \sqrt 2$ + > + > So $z = \sqrt2e^{i\frac{3\pi} 4}$ + +## 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 $z^n$. + +\begin{align*} +\text{Since } z = r(\cos\theta + i\sin\theta) \\ +z^n &= r^n(\cos\theta + i\sin\theta)^n &\text{(1)} \\ +\text{But also} \\ +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)} \\ +\text{By equating (1) and (2), we find:}\\ +(\cos\theta +i\sin\theta)^n &= (\cos{n\theta} + i\sin{n\theta}) +\end{align*} + +### Example 1 + +Write $1+i$ in polar form and use de Moivre's theorem to calculate $(1+i)^{15}$. + +\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*} + +### 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*} + +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*} + +## Complex Roots of Polynomials + +### Example + +Which complex numbers $z$ satisfy + +$$z^3 = 8i$$ + +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 solution