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Polar form and De Moivre's identity

+

(WACE) Mathematics Specialist ATAR

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De Moivre's identity
+ \[\boxed{\left[r\cdot \cis(\theta)\right]^n = r^n\cdot\cis(\theta\cdot n)}\] + where \(\cis(\theta) = \cos(\theta) + i\cdot\sin(\theta)\)
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+ The identity is more obvious when we use Euler's formula, \(\boxed{e^{i\theta} = \cos(\theta) + i\cdot\sin(\theta) = \cis(\theta)}\)
+ 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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Polar form rules: they're on the formula sheet so don't put them on your notes!
+ \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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Question 1
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+ The polar form of a complex number is useful because of its properties. +
+ Simplify \(\boxed{\frac{(i+1)^{2020}}{(i-1)^{2020}}}\) +
+ 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 simplified.
+ 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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Question 2
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+ Let's try something harder. +

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+ Express \(\boxed{\prod_{n=1}^{2020}\left[n\cdot\cis\left(\frac{\pi}{2}\right)^n\right]}\) in the form \(\boxed{r\cdot\cis(\alpha)}\)
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+ 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 \(\frac{\pi}{2}\)
+ 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. +

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Question 3
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+ Another math puzzle. +

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+ Express \(\boxed{\sum_{n=1}^{2020}\left[n\cdot\cis\left(\frac{\pi}{2}\right)^n\right]}\) in the form \(\boxed{\alpha + \beta i}\) +

+ Hold up, haven't we done this already? No. This uses the summation formula. Also we want it in rectangular form - not polar form!
+ 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.
+ 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.
+ 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! +

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Hello!

+ Welcome to my math page.
+ 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.
+ 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 \(\cis(\theta)=\cos(\theta)+i \sin(\theta)\)
+ Also note that I'm writing these because I want 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 not a guide or textbook.

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+ Polar form of complex number, De Moivre's theorem +
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