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< title > 📝 De Moivre's theorem< / title >
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< h1 > Polar form and De Moivre's identity< / h1 >
< h3 > (WACE) Mathematics Specialist ATAR< / h3 >
< / center >
2021-06-14 16:25:47 +08:00
< a class = "link" style = "left:1%; top: 1%;" href = "https://peter-tanner.github.io/maths" > 🔗 Back to MATHS home page< / a > < br >
< a class = "link" style = "left:1%; top: 1%;" href = "https://peter-tanner.github.io" > 🔗 Back to home page< / a > < br >
2020-11-21 16:46:57 +08:00
Warning: This page requires javascript to render the math.
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< 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 >
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< 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}
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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 >
\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}
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< hr >
< center > < b > Question 1< / b > < / center >
< hr >
The polar form of a complex number is useful because of its properties.
< center >
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Simplify \(\boxed{\frac{(i+1)^{2020}}{(i-1)^{2020}}}\) (no calculator)
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< / 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 >
2020-12-08 15:07:12 +08:00
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 >
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 >
2020-12-08 15:07:12 +08:00
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 >
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 >
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 >
2020-11-21 16:46:57 +08:00
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 >
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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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\[\sum_{n=1}^{2020}\left[n\cdot\cis\left(\frac{\pi}{2}\right)^n\right] = \cis\left(\frac{\pi}{2}\right) + 2\cis\left(\pi\right) + 3\cis\left(\frac{3\pi}{2}\right) + 4\cis\left(2\pi\right) + 5\cis\left(\frac{\pi}{2}\right) + 6\cis\left(\pi\right) + [\dots] + 2020\cis\left(2\pi\right)\]
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, \(a_0\)
\[\boxed{\sum_{k=1}^{n-1}\left[a_0+k\cdot d\right] = \frac{n}{2}\left[(n-1)\cdot d+2a_0\right]}\]
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}
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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.
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2021-06-14 16:25:47 +08:00
< a class = "link" style = "left:1%; bottom: 1%;" href = "https://peter-tanner.github.io/maths" > 🔗 Back to MATHS home page< / a > < br >
< a class = "link" style = "left:1%; bottom: 1%;" href = "https://peter-tanner.github.io" > 🔗 Back to home page< / a > < br >
2020-11-21 16:46:57 +08:00
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