mirror of
https://github.com/peter-tanner/peter-tanner.github.io.git
synced 2024-11-30 20:10:18 +08:00
345 lines
17 KiB
HTML
345 lines
17 KiB
HTML
<!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" />
|
|
<link
|
|
rel="stylesheet"
|
|
href="https://stackpath.bootstrapcdn.com/bootstrap/4.5.2/css/bootstrap.min.css"
|
|
integrity="sha384-JcKb8q3iqJ61gNV9KGb8thSsNjpSL0n8PARn9HuZOnIxN0hoP+VmmDGMN5t9UJ0Z"
|
|
crossorigin="anonymous"
|
|
/>
|
|
<meta name="robots" content="noindex, nofollow" />
|
|
</head>
|
|
|
|
<body>
|
|
<b
|
|
>PLEASE DO NOT USE ANY CONTENT FROM HERE! ALL UNMAINTAINED AND THERE'S
|
|
PROBABLY NOTHING USEFUL HERE! CONTENT IS NOW UN-INDEXED TO PREVENT
|
|
CONFUSION, SO YOU CAN ONLY ACCESS THIS PART OF THE SITE THROUGH LINKS.
|
|
CONTENT IS ONLY KEPT HERE FOR HISTORY OF THE SITE -2023</b
|
|
>
|
|
<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://peter-tanner.github.io/legacy_site/maths"
|
|
>🔗 Back to MATHS home page</a
|
|
><br />
|
|
<a
|
|
class="link"
|
|
style="left: 1%; top: 1%"
|
|
href="https://peter-tanner.github.io/legacy_site"
|
|
>🔗 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
|
|
\(\boxed{\frac{(i+1)^{2020}}{(i-1)^{2020}}}\) (no
|
|
calculator)
|
|
</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 \(\boxed{r\cdot\cis(\alpha)}\) (no
|
|
calculator)
|
|
</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 \(\boxed{\alpha + \beta i}\) (3
|
|
marks, no calculator)
|
|
</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 />
|
|
There are two mindsets to evaluate this expression: one which is 'local'
|
|
and the other 'global'<br />
|
|
To be honest, I just skipped this question entirely. It was 3 marks and I
|
|
was unprepared for this sort of question (I was close to a pattern, but
|
|
the coefficients were not periodic and threw me off.)<br />
|
|
<b>Method one: Local</b><br />
|
|
A 'local' approach aims to identify how the function or expression behaves
|
|
at a small or local level.<br />
|
|
Let's define the part within the summation as \(f(x)\) \[\text{Define: }
|
|
f(x) = n\cdot\cis\left(\frac{\pi}{2}\right)^n\] Let's see how the function
|
|
behaves for a *few* values of \(x\).<br />
|
|
We're trying to see how the function within the summation of the original
|
|
expression behaves at a local level, so we don't test all 2020 terms. The
|
|
goal is to find a pattern that applies for the remaining terms.
|
|
\begin{align} f(1) &= 1\cdot\cis\left(\frac{\pi}{2}\right)^1 \\ f(2) &=
|
|
2\cdot\cis\left(\frac{\pi}{2}\right)^2 \\ f(3) &=
|
|
3\cdot\cis\left(\frac{\pi}{2}\right)^3 \\ f(4) &=
|
|
4\cdot\cis\left(\frac{\pi}{2}\right)^4 \\ f(5) &=
|
|
5\cdot\cis\left(\frac{\pi}{2}\right)^5 \\ f(6) &=
|
|
6\cdot\cis\left(\frac{\pi}{2}\right)^6 \\ f(7) &=
|
|
7\cdot\cis\left(\frac{\pi}{2}\right)^7 \\ f(8) &=
|
|
8\cdot\cis\left(\frac{\pi}{2}\right)^8 \\ \end{align} Let's apply De
|
|
Moivre's theorem. \begin{alignat}{2} f(1) &=
|
|
1\cdot\cis\left(\frac{\pi}{2}\right) &&= \cis\left(\frac{\pi}{2}\right) \\
|
|
f(2) &= 2\cdot\cis\left(\frac{2\pi}{2}\right) &&=
|
|
2\cdot\cis\left(\pi\right) \\ f(3) &=
|
|
3\cdot\cis\left(\frac{3\pi}{2}\right) &&=
|
|
3\cdot\cis\left(\frac{3\pi}{2}\right)\\ f(4) &=
|
|
4\cdot\cis\left(\frac{4\pi}{2}\right) &&= 4\cdot\cis\left(2\pi\right) \\
|
|
f(5) &= 5\cdot\cis\left(\frac{5\pi}{2}\right) &&=
|
|
5\cdot\cis\left(\frac{\pi}{2}\right) \\ f(6) &=
|
|
6\cdot\cis\left(\frac{6\pi}{2}\right) &&= 6\cdot\cis\left(\pi\right) \\
|
|
f(7) &= 7\cdot\cis\left(\frac{7\pi}{2}\right) &&=
|
|
7\cdot\cis\left(\frac{3\pi}{2}\right) \\ f(8) &=
|
|
8\cdot\cis\left(\frac{8\pi}{2}\right) &&= 8\cdot\cis\left(2\pi\right) \\
|
|
\end{alignat} At this point we seem to be close to a pattern (periodic
|
|
nature of \(\cis\)), but it doesn't quite seem like one due to the
|
|
coefficients which are not repeating.<br />
|
|
If we try summing by the pattern that exists (the \(\cis\)), we may see a
|
|
pattern emerge \begin{align} \sum_{n=1}^{4}\left[f(n)\right] &=
|
|
f(1)+f(2)+f(3)+f(4) \\ &= i -2 -3i +4 \\ &= 2 - 2i \\
|
|
\sum_{n=5}^{8}\left[f(n)\right] &= f(5)+f(6)+f(7)+f(8) \\ &= 5i -6 -7i +8
|
|
\\ &= 2 - 2i \end{align} In fact we do see a pattern - this sort of local
|
|
behavior applies for the whole 2020 terms.<br />
|
|
|
|
\begin{align}
|
|
\sum_{n=1}^{2020}\left[n\cdot\cis\left(\frac{\pi}{2}\right)^n\right] &=
|
|
\sum_{n=1}^{4}\left[f(n)\right] + \sum_{n=5}^{8}\left[f(n)\right] +
|
|
[\dots] + \sum_{n=2015}^{2020}\left[f(n)\right] \\ &= (2-2i) + (2-2i) +
|
|
[\dots] + (2-2i) \\ &= \frac{2020}{4}\cdot(2-2i) \\ &= 1010 - 1010i
|
|
\end{align} Hopefully this explains what I mean by 'local' : a bit of an
|
|
odd term but it differentiates this line of thinking from the next one I
|
|
will show.<br /><br />
|
|
The main downside to this mindset is that you may end up wasting time by
|
|
testing to find a pattern. For a 3 mark question, I didn't even consider
|
|
using this method (although this was the intended method, from what I can
|
|
see)<br />
|
|
Compared to the global method it has some benefits. In this case it avoids
|
|
the use of a complicated summation formula. More generally, it also
|
|
requires smaller calculations as we're not looking at the large behavior -
|
|
this can minimize mistakes.
|
|
|
|
<br />
|
|
<b>Method two: Global</b><br />
|
|
The 'global' approach observes how the function or expression behaves as a
|
|
whole.<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,
|
|
\(\cis\left(\frac{\pi}{2}\right)=\cis\left(\frac{5\pi}{2}\right)=\cis\left(\frac{9\pi}{2}\right)=[\dots]=\cis\left(\frac{\pi}{2}+k\cdot
|
|
2\pi\right)\) 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.
|
|
\[\sum_{n=1}^{2020}\left[n\cdot\cis\left(\frac{\pi}{2}\right)^n\right] =
|
|
(1+5+9+13+[\dots])\times\cis\left(\frac{\pi}{2}\right) +
|
|
(2+6+10+14+[\dots])\times\cis\left(\pi\right) +
|
|
(3+7+11+15+[\dots])\times\cis\left(\frac{3\pi}{2}\right) +
|
|
(4+8+16+20+[\dots])\times\cis\left(2\pi\right)\] 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!<br /><br />
|
|
The downside to this method is that you end up needing to handle large
|
|
numbers - which also wastes time and increases chances of making a
|
|
mistake.<br />
|
|
For this particular question, there is also the downside of requiring
|
|
previous year content (the AP formula), which I certainly did not
|
|
remember.<br />
|
|
|
|
<br />
|
|
To be completely honest, this question should be worth more than just
|
|
three marks. In the exam, I decided that this was not worth my time.
|
|
</p>
|
|
|
|
<a
|
|
class="link"
|
|
style="left: 1%; bottom: 1%"
|
|
href="https://peter-tanner.github.io/legacy_site/maths"
|
|
>🔗 Back to MATHS home page</a
|
|
><br />
|
|
<a
|
|
class="link"
|
|
style="left: 1%; bottom: 1%"
|
|
href="https://peter-tanner.github.io/legacy_site"
|
|
>🔗 Back to home page</a
|
|
><br />
|
|
</body>
|
|
</html>
|