--- 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 $z$ (which you may seen written as $\Re(z) = x$ or Re$(z) = x$) - $y$ is the imaginary part of $z$ (which you may seen written as $\Im(z) = y$ or Im$(z) = y$) - 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 $$ae^{i\delta}$$ 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*}
## 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*}
### 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 Find 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