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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 \(\cis(\theta) = \cos(\theta) + i\cdot\sin(\theta)\)
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}
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}
Question 1
The polar form of a complex number is useful because of its properties.
Simplify
\(\boxed{\frac{(i+1)^{2020}}{(i-1)^{2020}}}\) (no
calculator)
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.
Question 2
Let's try something harder.
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)
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.
Question 3
Another math puzzle.
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)
Hold up, haven't we done this already? No. This uses the summation
formula. Also we want it in rectangular form - not polar form!
There are two mindsets to evaluate this expression: one which is 'local'
and the other 'global'
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.)
Method one: Local
A 'local' approach aims to identify how the function or expression behaves
at a small or local level.
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 \(x\).
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.
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.
\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.
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)
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.
Method two: Global
The 'global' approach observes how the function or expression behaves as a
whole.
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,
\(\cis\left(\frac{\pi}{2}\right)=\cis\left(\frac{5\pi}{2}\right)=\cis\left(\frac{9\pi}{2}\right)=[\dots]=\cis\left(\frac{\pi}{2}+k\cdot
2\pi\right)\) 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.
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!
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.
For this particular question, there is also the downside of requiring
previous year content (the AP formula), which I certainly did not
remember.
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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