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+---
+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