diff --git a/mechanical/mmme1026_maths_for_engineering/complex_numbers.md b/mechanical/mmme1026_maths_for_engineering/complex_numbers.md
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+++ b/mechanical/mmme1026_maths_for_engineering/complex_numbers.md
@@ -0,0 +1,390 @@
+---
+author: Alvie Rahman
+date: \today
+title: MMME1026 // Complex Numbers
+tags: [ uni, nottingham, mechanical, engineering, mmme1026, maths, complex_numbers ]
+---
+
+# Complex Numbers
+
+## What is a Complex Number?
+
+- $i$ 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. $$(3 + 4i) + (1 -2i) = (3+1) + i(4-2) = 2 + 2i$$
+
+### The 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
+
+<details>
+<summary>
+
+#### Example
+
+</summary>
+
+> \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*}
+
+</details>
+
+### 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 $\pi$.
+
+## 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>
diff --git a/mechanical/mmme1026_maths_for_engineering.md b/mechanical/mmme1026_maths_for_engineering/systems_and_matrices.md
similarity index 63%
rename from mechanical/mmme1026_maths_for_engineering.md
rename to mechanical/mmme1026_maths_for_engineering/systems_and_matrices.md
index ca00755..5fdb8a5 100755
--- a/mechanical/mmme1026_maths_for_engineering.md
+++ b/mechanical/mmme1026_maths_for_engineering/systems_and_matrices.md
@@ -1,394 +1,9 @@
 ---
 author: Alvie Rahman
 date: \today
-title: MMME1026 // Mathematics for Engineering
-tags: [ uni, nottingham, mechanical, engineering, mmme1026, maths, complex_numbers ]
+title: MMME1026 // Systems of Equations and Matrices
+tags: [ uni, nottingham, mechanical, engineering, mmme1026, maths, systems_of_equations, matrices ]
 ---
-
-# Complex Numbers
-
-## What is a Complex Number?
-
-- $i$ 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. $$(3 + 4i) + (1 -2i) = (3+1) + i(4-2) = 2 + 2i$$
-
-### The 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
-
-<details>
-<summary>
-
-#### Example
-
-</summary>
-
-> \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*}
-
-</details>
-
-### 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 $\pi$.
-
-## 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