notes/mechanical/mmme1026_maths_for_engineering.md

author	date	title	tags
Alvie Rahman	\today	MMME1026 // Mathematics for Engineering	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 (2021-10-12)

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 1

Write z = -1 + i in exponential form

$\arg z = \frac {3\pi} 4$ |z| = \sqrt 2

So z = \sqrt2e^{i\frac{3\pi} 4}

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}

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\del

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 + \varp

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*}

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*}

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.

\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*}

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 solutions