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240 lines
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<!DOCTYPE html>
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<html lang="en">
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<head>
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<meta charset="UTF-8">
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<title>📝 De Moivre's theorem</title>
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<meta name="description" content="De Moivre's theorem stuff - year 12 WACE specialist ATAR">
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<title>MathJax example</title>
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<script>
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MathJax = {
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tex: { macros: {cis: "\\mathop{\\rm{cis}}\\nolimits"} },
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chtml: { displayAlign: 'center', scale: 1.1 }
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}
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</script>
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<script src="https://polyfill.io/v3/polyfill.min.js?features=es6"></script>
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<script id="MathJax-script" async
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<link rel="stylesheet" href="https://stackpath.bootstrapcdn.com/bootstrap/4.5.2/css/bootstrap.min.css" integrity="sha384-JcKb8q3iqJ61gNV9KGb8thSsNjpSL0n8PARn9HuZOnIxN0hoP+VmmDGMN5t9UJ0Z" crossorigin="anonymous">
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</head>
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<body>
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<center>
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<h1>Polar form and De Moivre's identity</h1>
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<h3>(WACE) Mathematics Specialist ATAR</h3>
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</center>
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<a class="link" style="left:1%; top: 1%;" href="https://peter-tanner.github.io/maths">🔗 Back to MATHS home page</a><br>
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<a class="link" style="left:1%; top: 1%;" href="https://peter-tanner.github.io">🔗 Back to home page</a><br>
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Warning: This page requires javascript to render the math.
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<hr><br>
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<div class="card">
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<div class="card-body">
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<center>De Moivre's identity</center>
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\[\boxed{\left[r\cdot \cis(\theta)\right]^n = r^n\cdot\cis(\theta\cdot n)}\]
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where \(\cis(\theta) = \cos(\theta) + i\cdot\sin(\theta)\)<br>
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</div>
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</div>
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<p>
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The identity is more obvious when we use Euler's formula, \(\boxed{e^{i\theta} = \cos(\theta) + i\cdot\sin(\theta) = \cis(\theta)}\)<br>
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This formula doesn't appear to be taught in the WA curriculum.
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\begin{align}
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\left[r\cdot e^{i\theta}\right]^n &= r^n\cdot\left[e^{i\theta}\right]^n\\
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&= r^n\cdot e^{i(\theta n)} \\
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&= r^n\cdot\cis(\theta\cdot n) \\
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\end{align}
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</p>
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<div class="card">
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<div class="card-body">
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<center>Polar form rules: they're on the formula sheet so <b>don't put them on your notes!</b></center>
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\begin{align}
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z_1\cdot z_2 &= r_1\cdot r_2 \cdot \cis(\theta_1 + \theta_2) \\
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\frac{z_1}{z_2} &= \frac{r_1}{r_2} \cdot \cis(\theta_1 - \theta_2)\\
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\cis(\theta_1 + \theta_2) &= \cis(\theta_1)\cdot\cis(\theta_2) \\
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\cis(-\theta) &= \frac{1}{\cis(\theta)} \\
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\overline{\cis(\theta)} &= \cis(-\theta)
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\end{align}
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</div>
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</div>
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<hr>
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<center><b>Question 1</b></center>
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<hr>
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The polar form of a complex number is useful because of its properties.
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<center>
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Simplify \(\boxed{\frac{(i+1)^{2020}}{(i-1)^{2020}}}\) (no calculator)
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</center>
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Some defining may be useful right now.
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\begin{align}
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z_0&=i+1 \\
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z_1&=i-1 \\
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\end{align}
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First steps to convert to polar form is always to obtain the modulus and argument of the complex number
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\begin{align}
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\lvert z_0 \rvert &= \sqrt{1^2 + 1^2} = \sqrt{2} \\
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\lvert z_1 \rvert &= \sqrt{1^2 + 1^2} = \sqrt{2} \\
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\arg(z_0) &= \arctan\left(\frac{1}{1}\right) = \frac{\pi}{2} \\
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\arg(z_1) &= \arctan\left(\frac{-1}{1}\right) = -\frac{\pi}{2} \\
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\therefore z_0 &= \sqrt{2}\cdot\cis\left(\frac{\pi}{2}\right) \\
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z_1 &= \sqrt{2}\cdot\cis\left(-\frac{\pi}{2}\right) \\
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\end{align}
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So our original equation becomes:
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\begin{align}
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\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}} \\
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&= \frac{\sqrt{2}^{2020}}{\sqrt{2}^{2020}}\cdot\frac{\cis(\frac{\pi}{2})^{2020}}{\cis(-\frac{\pi}{2})^{2020}} \\
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&= \frac{\cis(\frac{\pi}{2})^{2020}}{\cis(-\frac{\pi}{2})^{2020}} \\
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\end{align}
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Let's apply De Moivre's theorem.
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\begin{align}
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\frac{(i+1)^{2020}}{(i-1)^{2020}} = \frac{\cis(2020\cdot\frac{\pi}{2})}{\cis(-2020\cdot\frac{\pi}{2})} \\
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&= \frac{\cis(1010\cdot\pi)}{\cis(-1010\cdot\pi)}
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\end{align}
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We have the useful formula, \(\boxed{\frac{\cis(\alpha)}{\cis(\beta)} = \cis(\alpha-\beta)}\). Let's apply it!
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\begin{align}
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\frac{(i+1)^{2020}}{(i-1)^{2020}} &= \cis\left[1010\cdot\pi-\left(-1010\cdot\pi\right)\right] \\
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&= \cis\left[1010\cdot\pi+1010\cdot\pi\right] \\
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&= \cis\left[2020\cdot\pi\right] \\
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&= \cis\left[1010\cdot2\pi\right] \\
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&= \cis\left[2\pi\right] \\
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&= 1 \\
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\end{align}
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All <b>simp</b>lified.<br>
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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.
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<hr>
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<center><b>Question 2</b></center>
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<hr>
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Let's try something harder.
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<p>
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<center>
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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)
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</center>
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What can we do? Firstly, we can expand the capital pi \(\Pi\) to reveal the terms.
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\[\prod_{n=1}^{2020}\left[n\cdot\cis\left(\frac{\pi}{2}\right)^n\right] = \cis\left(\frac{\pi}{2}\right) \times 2\cis\left(\frac{\pi}{2}\right)^2 \times 3\cis\left(\frac{\pi}{2}\right)^3 \times [\dots] \times 2020\cis\left(\frac{\pi}{2}\right)^{2020}\]
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Secondly, we can recognise that we can bring the coefficients together and use our identity to bring the power \(n\) into the angle
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\[\prod_{n=1}^{2020}\left[n\cdot\cis\left(\frac{\pi}{2}\right)^n\right] = \left(1\times 2\times 3\times [\dots]\times 2020 \right)\times \cis\left(\frac{\pi}{2}\right) \times \cis\left(2\cdot\frac{\pi}{2}\right) \times \cis\left(3\cdot\frac{\pi}{2}\right) \times [\dots] \times \cis\left(2020\cdot\frac{\pi}{2}\right)\]
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The coefficients together form a factorial! We can express this as:
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\[\prod_{n=1}^{2020}\left[n\cdot\cis\left(\frac{\pi}{2}\right)^n\right] = 2020!\times \cis\left(\frac{\pi}{2}\right) \times \cis\left(2\cdot\frac{\pi}{2}\right) \times \cis\left(3\cdot\frac{\pi}{2}\right) \times [\dots] \times \cis\left(2020\cdot\frac{\pi}{2}\right)\]
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OK, let's now bring the angles together. \(\cis(\alpha)\times\cis(\beta) = \cis(\alpha + \beta)\)
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\begin{align}
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\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) \\
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&= 2020!\times \cis\left((1 + 2 + 3 + [\dots] + 2020)\cdot\frac{\pi}{2}\right)
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\end{align}
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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>
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The formula by the way is: \(\boxed{\sum_{k=1}^n k = \frac{n(n+1)}{2}}\)
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\begin{align}
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\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) \\
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&= 2020!\times \cis\left(1020605\pi\right)
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\end{align}
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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 \(\pi\).
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\[\prod_{n=1}^{2020}\left[n\cdot\cis\left(\frac{\pi}{2}\right)^n\right] = 2020!\times \cis\left(\pi\right)\]
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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.
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</p>
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<br>
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<hr>
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<center><b>Question 3</b></center>
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<hr>
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Another math puzzle.
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<p>
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<center>
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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)
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</center><br>
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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>
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There are two mindsets to evaluate this expression: one which is 'local' and the other 'global'<br>
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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>
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<b>Method one: Local</b><br>
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A 'local' approach aims to identify how the function or expression behaves at a small or local level.<br>
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Let's define the part within the summation as \(f(x)\)
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\[\text{Define: } f(x) = n\cdot\cis\left(\frac{\pi}{2}\right)^n\]
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Let's see how the function behaves for a *few* values of \(x\).<br>
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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.
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\begin{align}
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f(1) &= 1\cdot\cis\left(\frac{\pi}{2}\right)^1 \\
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f(2) &= 2\cdot\cis\left(\frac{\pi}{2}\right)^2 \\
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f(3) &= 3\cdot\cis\left(\frac{\pi}{2}\right)^3 \\
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f(4) &= 4\cdot\cis\left(\frac{\pi}{2}\right)^4 \\
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f(5) &= 5\cdot\cis\left(\frac{\pi}{2}\right)^5 \\
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f(6) &= 6\cdot\cis\left(\frac{\pi}{2}\right)^6 \\
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f(7) &= 7\cdot\cis\left(\frac{\pi}{2}\right)^7 \\
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f(8) &= 8\cdot\cis\left(\frac{\pi}{2}\right)^8 \\
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\end{align}
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Let's apply De Moivre's theorem.
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\begin{alignat}{2}
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f(1) &= 1\cdot\cis\left(\frac{\pi}{2}\right) &&= \cis\left(\frac{\pi}{2}\right) \\
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f(2) &= 2\cdot\cis\left(\frac{2\pi}{2}\right) &&= 2\cdot\cis\left(\pi\right) \\
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f(3) &= 3\cdot\cis\left(\frac{3\pi}{2}\right) &&= 3\cdot\cis\left(\frac{3\pi}{2}\right)\\
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f(4) &= 4\cdot\cis\left(\frac{4\pi}{2}\right) &&= 4\cdot\cis\left(2\pi\right) \\
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f(5) &= 5\cdot\cis\left(\frac{5\pi}{2}\right) &&= 5\cdot\cis\left(\frac{\pi}{2}\right) \\
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f(6) &= 6\cdot\cis\left(\frac{6\pi}{2}\right) &&= 6\cdot\cis\left(\pi\right) \\
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f(7) &= 7\cdot\cis\left(\frac{7\pi}{2}\right) &&= 7\cdot\cis\left(\frac{3\pi}{2}\right) \\
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f(8) &= 8\cdot\cis\left(\frac{8\pi}{2}\right) &&= 8\cdot\cis\left(2\pi\right) \\
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\end{alignat}
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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>
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If we try summing by the pattern that exists (the \(\cis\)), we may see a pattern emerge
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\begin{align}
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\sum_{n=1}^{4}\left[f(n)\right] &= f(1)+f(2)+f(3)+f(4) \\
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&= i -2 -3i +4 \\
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&= 2 - 2i \\
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\sum_{n=5}^{8}\left[f(n)\right] &= f(5)+f(6)+f(7)+f(8) \\
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&= 5i -6 -7i +8 \\
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&= 2 - 2i
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\end{align}
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In fact we do see a pattern - this sort of local behavior applies for the whole 2020 terms.<br>
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\begin{align}
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\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] \\
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&= (2-2i) + (2-2i) + [\dots] + (2-2i) \\
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&= \frac{2020}{4}\cdot(2-2i) \\
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&= 1010 - 1010i
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\end{align}
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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>
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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>
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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.
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<br>
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<b>Method two: Global</b><br>
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The 'global' approach observes how the function or expression behaves as a whole.<br>
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Instantly, let's unpack the summation then, apply De Moivre's theorem.
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\begin{align}
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\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} \\
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&= \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)
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\end{align}
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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>
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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 \(k\)
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\[\sum_{n=1}^{2020}\left[n\cdot\cis\left(\frac{\pi}{2}\right)^n\right] = \cis\left(\frac{\pi}{2}\right) + 2\cis\left(\pi\right) + 3\cis\left(\frac{3\pi}{2}\right) + 4\cis\left(2\pi\right) + 5\cis\left(\frac{\pi}{2}\right) + 6\cis\left(\pi\right) + [\dots] + 2020\cis\left(2\pi\right)\]
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Alright, let's collect the cis terms according to their (simplified) angle.
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\[\sum_{n=1}^{2020}\left[n\cdot\cis\left(\frac{\pi}{2}\right)^n\right] = (1+5+9+13+[\dots])\times\cis\left(\frac{\pi}{2}\right) + (2+6+10+14+[\dots])\times\cis\left(\pi\right) + (3+7+11+15+[\dots])\times\cis\left(\frac{3\pi}{2}\right) + (4+8+16+20+[\dots])\times\cis\left(2\pi\right)\]
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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, \(a_0\)
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\[\boxed{\sum_{k=1}^{n-1}\left[a_0+k\cdot d\right] = \frac{n}{2}\left[(n-1)\cdot d+2a_0\right]}\]
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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>
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Let's apply the series formula to bring these terms together.
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\begin{align}
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\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) \\
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&= 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) \\
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&= 509545\times[0+i] + 510050\times[-1+0i] + 510555\times[0-i] + 511060\times[1+0i] \\
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&= 511060 - 510050 + (509545 - 510555)i\\
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\sum_{n=1}^{2020}\left[n\cdot\cis\left(\frac{\pi}{2}\right)^n\right] &= 1010 - 1010i
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\end{align}
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And there we have it!<br><br>
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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>
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For this particular question, there is also the downside of requiring previous year content (the AP formula), which I certainly did not remember.<br>
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<br>
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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.
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</p>
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<a class="link" style="left:1%; bottom: 1%;" href="https://peter-tanner.github.io/maths">🔗 Back to MATHS home page</a><br>
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<a class="link" style="left:1%; bottom: 1%;" href="https://peter-tanner.github.io">🔗 Back to home page</a><br>
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</body>
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</html> |