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WACE Stage 3/ATAR past papers archive - the most comprehensive archive of past papers (I think... :D )<br>
<a href="https://atar-wace-archive.github.io/">https://atar-wace-archive.github.io/</a><br>
<br>
Maths page - specialist and methods ramblings<br>
<a href="https://atar-wace-archive.github.io/maths">https://atar-wace-archive.github.io/maths</a><br>
<br>
Programming ramblings - do not expect clean code here!<br>
<a href="https://atar-wace-archive.github.io/code">https://atar-wace-archive.github.io/code</a><br>
<br>
<br>
<br>
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<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>📝 De Moivre's theorem</title>
<meta name="description" content="De Moivre's theorem stuff - year 12 WACE specialist ATAR">
<meta name="viewport" content="width=device-width">
<title>MathJax example</title>
<script>
MathJax = {
tex: { macros: {cis: "\\mathop{\\rm{cis}}\\nolimits"} },
chtml: { displayAlign: 'center', scale: 1.1 }
}
</script>
<script src="https://polyfill.io/v3/polyfill.min.js?features=es6"></script>
<script id="MathJax-script" async
src="https://cdn.jsdelivr.net/npm/mathjax@3/es5/tex-mml-chtml.js">
</script>
<link rel="stylesheet" href="style.css">
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</head>
<body>
<center>
<h1>Polar form and De Moivre's identity</h1>
<h3>(WACE) Mathematics Specialist ATAR</h3>
</center>
<a class="link" style="left:1%; top: 1%;" href="https://npc-strider.github.io/maths">🔗 Back to MATHS home page</a><br>
<a class="link" style="left:1%; top: 1%;" href="https://npc-strider.github.io">🔗 Back to home page</a><br>
Warning: This page requires javascript to render the math.
<hr><br>
<div class="card">
<div class="card-body">
<center>De Moivre's identity</center>
\[\boxed{\left[r\cdot \cis(\theta)\right]^n = r^n\cdot\cis(\theta\cdot n)}\]
where \(\cis(\theta) = \cos(\theta) + i\cdot\sin(\theta)\)<br>
</div>
</div>
<p>
The identity is more obvious when we use Euler's formula, \(\boxed{e^{i\theta} = \cos(\theta) + i\cdot\sin(\theta) = \cis(\theta)}\)<br>
This formula doesn't appear to be taught in the WA curriculum.
\begin{align}
\left[r\cdot e^{i\theta}\right]^n &= r^n\cdot\left[e^{i\theta}\right]^n\\
&= r^n\cdot e^{i(\theta n)} \\
&= r^n\cdot\cis(\theta\cdot n) \\
\end{align}
</p>
<div class="card">
<div class="card-body">
<center>Polar form rules: they're on the formula sheet so <b>don't put them on your notes!</b></center>
\begin{align}
z_1\cdot z_2 &= r_1\cdot r_2 \cdot \cis(\theta_1 + \theta_2) \\
\frac{z_1}{z_2} &= \frac{r_1}{r_2} \cdot \cis(\theta_1 - \theta_2)\\
\cis(\theta_1 + \theta_2) &= \cis(\theta_1)\cdot\cis(\theta_2) \\
\cis(-\theta) &= \frac{1}{\cis(\theta)} \\
\overline{\cis(\theta)} &= \cis(-\theta)
\end{align}
</div>
</div>
<hr>
<center><b>Question 1</b></center>
<hr>
The polar form of a complex number is useful because of its properties.
<center>
Simplify \(\boxed{\frac{(i+1)^{2020}}{(i-1)^{2020}}}\)
</center>
Some defining may be useful right now.
\begin{align}
z_0&=i+1 \\
z_1&=i-1 \\
\end{align}
First steps to convert to polar form is always to obtain the modulus and argument of the complex number
\begin{align}
\lvert z_0 \rvert &= \sqrt{1^2 + 1^2} = \sqrt{2} \\
\lvert z_1 \rvert &= \sqrt{1^2 + 1^2} = \sqrt{2} \\
\arg(z_0) &= \arctan\left(\frac{1}{1}\right) = \frac{\pi}{2} \\
\arg(z_1) &= \arctan\left(\frac{-1}{1}\right) = -\frac{\pi}{2} \\
\therefore z_0 &= \sqrt{2}\cdot\cis\left(\frac{\pi}{2}\right) \\
z_1 &= \sqrt{2}\cdot\cis\left(-\frac{\pi}{2}\right) \\
\end{align}
So our original equation becomes:
\begin{align}
\frac{(i+1)^{2020}}{(i-1)^{2020}} = \frac{z_0^{\phantom{0}2020}}{z_1^{\phantom{0}2020}} &= \frac{\left[\sqrt{2}\cdot\cis(\frac{\pi}{2})\right]^{2020}}{\left[\sqrt{2}\cdot\cis(-\frac{\pi}{2})\right]^{2020}} \\
&= \frac{\sqrt{2}^{2020}}{\sqrt{2}^{2020}}\cdot\frac{\cis(\frac{\pi}{2})^{2020}}{\cis(-\frac{\pi}{2})^{2020}} \\
&= \frac{\cis(\frac{\pi}{2})^{2020}}{\cis(-\frac{\pi}{2})^{2020}} \\
\end{align}
Let's apply De Moivre's theorem.
\begin{align}
\frac{(i+1)^{2020}}{(i-1)^{2020}} = \frac{\cis(2020\cdot\frac{\pi}{2})}{\cis(-2020\cdot\frac{\pi}{2})} \\
&= \frac{\cis(1010\cdot\pi)}{\cis(-1010\cdot\pi)}
\end{align}
We have the useful formula, \(\boxed{\frac{\cis(\alpha)}{\cis(\beta)} = \cis(\alpha-\beta)}\). Let's apply it!
\begin{align}
\frac{(i+1)^{2020}}{(i-1)^{2020}} &= \cis\left[1010\cdot\pi-\left(-1010\cdot\pi\right)\right] \\
&= \cis\left[1010\cdot\pi+1010\cdot\pi\right] \\
&= \cis\left[2020\cdot\pi\right] \\
&= \cis\left[1010\cdot2\pi\right] \\
&= \cis\left[2\pi\right] \\
&= 1 \\
\end{align}
All <b>simp</b>lified.<br>
This question forces you to use \(\cis\) due to the large power. If we had, say, \(\frac{(i+1)}{(i-1)}\), we could simply multiply the top and bottom by \((i+1)\), expand and split the fraction.
<hr>
<center><b>Question 2</b></center>
<hr>
Let's try something harder.
<p>
<center>
Express \(\boxed{\prod_{n=1}^{2020}\left[n\cdot\cis\left(\frac{\pi}{2}\right)^n\right]}\) in the form \(\boxed{r\cdot\cis(\alpha)}\) <br>
</center>
What can we do? Firstly, we can expand the capital pi \(\Pi\) to reveal the terms.
\[\prod_{n=1}^{2020}\left[n\cdot\cis\left(\frac{\pi}{2}\right)^n\right] = \cis\left(\frac{\pi}{2}\right) \times 2\cis\left(\frac{\pi}{2}\right)^2 \times 3\cis\left(\frac{\pi}{2}\right)^3 \times [\dots] \times 2020\cis\left(\frac{\pi}{2}\right)^{2020}\]
Secondly, we can recognise that we can bring the coefficients together and use our identity to bring the power \(n\) into the angle
\[\prod_{n=1}^{2020}\left[n\cdot\cis\left(\frac{\pi}{2}\right)^n\right] = \left(1\times 2\times 3\times [\dots]\times 2020 \right)\times \cis\left(\frac{\pi}{2}\right) \times \cis\left(2\cdot\frac{\pi}{2}\right) \times \cis\left(3\cdot\frac{\pi}{2}\right) \times [\dots] \times \cis\left(2020\cdot\frac{\pi}{2}\right)\]
The coefficients together form a factorial! We can express this as:
\[\prod_{n=1}^{2020}\left[n\cdot\cis\left(\frac{\pi}{2}\right)^n\right] = 2020!\times \cis\left(\frac{\pi}{2}\right) \times \cis\left(2\cdot\frac{\pi}{2}\right) \times \cis\left(3\cdot\frac{\pi}{2}\right) \times [\dots] \times \cis\left(2020\cdot\frac{\pi}{2}\right)\]
OK, let's now bring the angles together. \(\cis(\alpha)\times\cis(\beta) = \cis(\alpha + \beta)\)
\begin{align}
\prod_{n=1}^{2020}\left[n\cdot\cis\left(\frac{\pi}{2}\right)^n\right] &= 2020!\times \cis\left(\frac{\pi}{2} + 2\cdot\frac{\pi}{2} + 3\cdot\frac{\pi}{2} + [\dots] + 2020\cdot\frac{\pi}{2}\right) \\
&= 2020!\times \cis\left((1 + 2 + 3 + [\dots] + 2020)\cdot\frac{\pi}{2}\right)
\end{align}
Great! This is looking like the form we need. All that's left is to use our triangular number (sum) formula to calculate the sum of the coefficient of \(\frac{\pi}{2}\) <br>
The formula by the way is: \(\boxed{\sum_{k=1}^n k = \frac{n(n+1)}{2}}\)
\begin{align}
\prod_{n=1}^{2020}\left[n\cdot\cis\left(\frac{\pi}{2}\right)^n\right] &= 2020!\times \cis\left(\frac{2020(2020+1)}{2}\cdot\frac{\pi}{2}\right) \\
&= 2020!\times \cis\left(1020605\pi\right)
\end{align}
By recognising that \(\cis\) is periodic, we can reduce the angle size to simplify further. Because \(1020605=2n+1\), where \(n\) is an integer, it is odd and we can reduce the angle to simply \(\pi\).
\[\prod_{n=1}^{2020}\left[n\cdot\cis\left(\frac{\pi}{2}\right)^n\right] = 2020!\times \cis\left(\pi\right)\]
OK, we got it in the form \(r\cdot\cis(\alpha)\) as required. \(2020!\), by the way is a very large number. Vsauce already has a video on \(52!\) which is already very large.
</p>
<br>
<hr>
<center><b>Question 3</b></center>
<hr>
Another math puzzle.
<p>
<center>
Express \(\boxed{\sum_{n=1}^{2020}\left[n\cdot\cis\left(\frac{\pi}{2}\right)^n\right]}\) in the form \(\boxed{\alpha + \beta i}\)
</center><br>
Hold up, haven't we done this already? No. This uses the summation formula. Also we want it in rectangular form - not polar form!<br>
Instantly, let's unpack the summation then, apply De Moivre's theorem.
\begin{align}
\sum_{n=1}^{2020}\left[n\cdot\cis\left(\frac{\pi}{2}\right)^n\right] &= \cis\left(\frac{\pi}{2}\right) + 2\cis\left(\frac{\pi}{2}\right)^2 + 3\cis\left(\frac{\pi}{2}\right)^3 + 4\cis\left(\frac{\pi}{2}\right)^4 + 5\cis\left(\frac{\pi}{2}\right)^5 + 6\cis\left(\frac{\pi}{2}\right)^6 + [\dots] + 2020\cis\left(\frac{\pi}{2}\right)^{2020} \\
&= \cis\left(\frac{\pi}{2}\right) + 2\cis\left(\frac{\pi}{2}\cdot 2\right) + 3\cis\left(\frac{\pi}{2}\cdot 3\right) + 4\cis\left(\frac{\pi}{2}\cdot 4\right) + 5\cis\left(\frac{\pi}{2}\cdot 5\right) + 6\cis\left(\frac{\pi}{2}\cdot 6\right) + [\dots] + 2020\cis\left(\frac{\pi}{2}\cdot 2020\right)
\end{align}
What we have here is essentially a series of rotating vectors. So that means we can expect to simplify quite a few of the cis terms, as they repeat periodically.<br>
For example, \(\cis\left(\frac{\pi}{2}\right)=\cis\left(\frac{5\pi}{2}\right)=\cis\left(\frac{9\pi}{2}\right)=[...]=\cis\left(\frac{\pi}{2}+k\cdot 2\pi\right)\) for integer \(k\)
\[\sum_{n=1}^{2020}\left[n\cdot\cis\left(\frac{\pi}{2}\right)^n\right] = \cis\left(\frac{\pi}{2}\right) + 2\cis\left(\pi\right) + 3\cis\left(\frac{3\pi}{2}\right) + 4\cis\left(2\pi\right) + 5\cis\left(\frac{\pi}{2}\right) + 6\cis\left(\pi\right) + [\dots] + 2020\cis\left(2\pi\right)\]
Alright, let's collect the cis terms according to their (simplified) angle.
\[\sum_{n=1}^{2020}\left[n\cdot\cis\left(\frac{\pi}{2}\right)^n\right] = (1+5+9+13+[...])\times\cis\left(\frac{\pi}{2}\right) + (2+6+10+14+[...])\times\cis\left(\pi\right) + (3+7+11+15+[...])\times\cis\left(\frac{3\pi}{2}\right) + (4+8+16+20+[...])\times\cis\left(2\pi\right)\]
We need to use the arithmetic progression sum formula, because we have a common difference \(d\) of 4 between each number, not 1 like last time. Also for our terms, we have different starting values, \(a_0\)
\[\boxed{\sum_{k=1}^{n-1}\left[a_0+k\cdot d\right] = \frac{n}{2}\left[(n-1)\cdot d+2a_0\right]}\]
Now, we have \(\frac{2020}{4}\) terms for each unique angle for \(\cis\) because 2020 is divisible by 4 (The number of \(\cis\) with unique angles) evenly. In other scenarios this may not be the case and \(n\) may vary for each sum.<br>
Let's apply the series formula to bring these terms together.
\begin{align}
\sum_{n=1}^{2020}\left[n\cdot\cis\left(\frac{\pi}{2}\right)^n\right] &= \frac{505}{2}\left[(505-1)\cdot 4+2\cdot1\right]\times\cis\left(\frac{\pi}{2}\right) + \frac{505}{2}\left[(505-1)\cdot 4+2\cdot 2\right]\times\cis\left(\pi\right) + \frac{505}{2}\left[(505-1)\cdot 4+2\cdot 3\right]\times\cis\left(\frac{3\pi}{2}\right) + \frac{505}{2}\left[(505-1)\cdot 4+2\cdot 4\right]\times\cis\left(2\pi\right) \\
&= 509545\times\cis\left(\frac{\pi}{2}\right) + 510050\times\cis\left(\pi\right) + 510555\times\cis\left(\frac{3\pi}{2}\right) + 511060\times\cis\left(2\pi\right) \\
&= 509545\times[0+i] + 510050\times[-1+0i] + 510555\times[0-i] + 511060\times[1+0i] \\
&= 511060 - 510050 + (509545 - 510555)i\\
\sum_{n=1}^{2020}\left[n\cdot\cis\left(\frac{\pi}{2}\right)^n\right] &= 1010 - 1010i
\end{align}
And there we have it!
</p>
<a class="link" style="left:1%; bottom: 1%;" href="https://npc-strider.github.io/maths">🔗 Back to MATHS home page</a><br>
<a class="link" style="left:1%; bottom: 1%;" href="https://npc-strider.github.io">🔗 Back to home page</a><br>
</body>
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<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>📝 Maths!</title>
<meta name="description" content="year 12 WACE specialist ATAR stuff">
<meta name="viewport" content="width=device-width">
<title>MathJax example</title>
<script>
MathJax = {
tex: { macros: {cis: "\\mathop{\\rm{cis}}\\nolimits"} },
chtml: { displayAlign: 'center', scale: 1.1 }
}
</script>
<script src="https://polyfill.io/v3/polyfill.min.js?features=es6"></script>
<script id="MathJax-script" async
src="https://cdn.jsdelivr.net/npm/mathjax@3/es5/tex-mml-chtml.js">
</script>
<link rel="stylesheet" href="style.css">
<link rel="stylesheet" href="https://stackpath.bootstrapcdn.com/bootstrap/4.5.2/css/bootstrap.min.css" integrity="sha384-JcKb8q3iqJ61gNV9KGb8thSsNjpSL0n8PARn9HuZOnIxN0hoP+VmmDGMN5t9UJ0Z" crossorigin="anonymous">
</head>
<body>
<h1>Hello!</h1>
Welcome to my math page.<br>
The content here will apply throughout the universe, because as far as I know, logic is the same everywhere. \(1+1\) does equal \(2\) whether you are in WA or on Mars. <br>
However, do take note that these ramblings are influenced by WACE curriculum. For instance, we don't get taught Euler's formula \(e^{i\theta}=\cos(\theta)+i \sin(\theta)\). Instead, we get taught the rarer \(\cis(\theta)=\cos(\theta)+i \sin(\theta)\)<br>
Also note that I'm writing these because I <b>want</b> to, not because I have to. So these pages will probably never cover the whole curriculum and they will probably be incomplete, only showcasing the most interesting or unique problems. Therefore, they are <b>not a guide or textbook</b>.<br><br>
<div class="card">
<div class="card-body">
Here is a list of pages:<br>
<a href="https://npc-strider.github.io/math/de moivre theorem.html">Polar form of complex number, De Moivre's theorem</a>
</div>
</div>
<a class="link" style="left:1%; bottom: 1%; position: absolute;" href="https://npc-strider.github.io">🔗 Back to home page</a><br>
</body>

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body {
max-width:100%;
min-width:300px;
margin:0 auto;
padding:20px 10px;
font-size:14px;
font-size:1.4em;
}