HSC 2016 MX2 Complex Numbers Marathon (archive) (3 Viewers)

math man · Apr 14, 2016

Math Man is back and seeing the different marathons thought different topic themed marathons seems like a good idea.
I would love to do mechanics but it's too early for that and I love complex numbers!

Rules:
1. Once a question has been answered you can ask a new question.
2. A new question must be stated by writing NEW at the top

I'll begin with a warm up one:

NEW

Q:

math man · Apr 14, 2016

Re: HSC 2016 Complex Numbers Marathon

Note those are ones on top and bottom if you can't see clearly

Paradoxica · Apr 14, 2016

Re: HSC 2016 Complex Numbers Marathon

$$\noindent Define $ z = \cos{\theta} + i\sin{\theta} \\\\ $Then we have $ \frac{1+i\bar{z}}{1-i\bar{z}} = \frac{(1+i\bar{z})(1+iz)}{(1-i\bar{z})(1+iz)} = \frac{1-1+i(z+\bar{z})}{1+1+i(z-\bar{z})} = \frac{2i\cos{\theta}}{2-2\sin{\theta}}\\=\frac{i\sin{(\frac{\pi}{2}-\theta)}}{1-\cos{(\frac{\pi}{2}-\theta)}}=\frac{2i\sin{(\frac{\pi}{4}-\frac{\theta}{2})}\cos{(\frac{\pi}{4}-\frac{\theta}{2})}}{2\cos^2{(\frac{\pi}{4}-\frac{\theta}{2})}}=i\tan{\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}=i\cot{\left(\frac{\theta}{2}-\frac{\pi}{4}\right)}$

Well one of us is wrong. (I vouch myself)

Paradoxica · Apr 14, 2016

Re: HSC 2016 Complex Numbers Marathon

$$\noindent Sum to infinity the following series: $\\ \cos{\theta}\cos{\theta} + \cos^2{\theta}\cos{2\theta} + \cos^3{\theta}\cos{3\theta} + \cdots + \cos^n{\theta}\cos{n\theta}\\ $State the values of $\theta$ for which the series does not converge.$ \\ $Repeat the above for the following series': $\\ \cos{\theta}\cos{\theta} + 2\cos^2{\theta}\cos{2\theta} + 4\cos^3{\theta}\cos{3\theta} + \cdots + 2^{n-1}\cos^n{\theta}\cos{n\theta} \\ \cos{\theta}\cos{\theta} + 2\cos^2{\theta}\cos{2\theta} + 3\cos^3{\theta}\cos{3\theta} + \cdots + n\cos^n{\theta}\cos{n\theta} \\ \cos{\theta}\cos{\theta} + 4\cos^2{\theta}\cos{2\theta} + 9\cos^3{\theta}\cos{3\theta} + \cdots + n^2\cos^n{\theta}\cos{n\theta}$

Drsoccerball · Apr 14, 2016

Re: HSC 2016 Complex Numbers Marathon

Mods pls close this thread...

leehuan · Apr 14, 2016

Re: HSC 2016 Complex Numbers Marathon

Paradoxica said:
$$\noindent Sum to infinity the following series: $\\ \cos{\theta}\cos{\theta} + \cos^2{\theta}\cos{2\theta} + \cos^3{\theta}\cos{3\theta} + \cdots + \cos^n{\theta}\cos{n\theta}\\ $State the values of $\theta$ for which the series does not converge.$ \\ $Repeat the above for the following series': $\\ \cos{\theta}\cos{\theta} + 2\cos^2{\theta}\cos{2\theta} + 4\cos^3{\theta}\cos{3\theta} + \cdots + 2^{n-1}\cos^n{\theta}\cos{n\theta} \\ \cos{\theta}\cos{\theta} + 2\cos^2{\theta}\cos{2\theta} + 3\cos^3{\theta}\cos{3\theta} + \cdots + n\cos^n{\theta}\cos{n\theta} \\ \cos{\theta}\cos{\theta} + 4\cos^2{\theta}\cos{2\theta} + 9\cos^3{\theta}\cos{3\theta} + \cdots + n^2\cos^n{\theta}\cos{n\theta}$

Looking at this now I feel so dumb for not being able to get it out on the day... so bloody trivial

DatAtarLyfe · Apr 14, 2016

Re: HSC 2016 Complex Numbers Marathon

Drsoccerball said:
Mods pls close this thread...

Why would they close it?...

leehuan · Apr 14, 2016

Re: HSC 2016 Complex Numbers Marathon

DatAtarLyfe said:
Why would they close it?...

Do we really need this thread when we can use the actual 4u marathon lol

Kinda like my SOS thread that got closed (and all "subsidiaries")

DatAtarLyfe · Apr 14, 2016

HSC 2016 Complex Numbers Marathon

leehuan said:
Do we really need this thread when we can use the actual 4u marathon lol

Kinda like my SOS thread that got closed (and all "subsidiaries")

You can say the same thing about the integration marathon and the combinatorics marathon as well. By closing this one, you would have to close the other two to keep it fair

leehuan · Apr 14, 2016

Re: HSC 2016 Complex Numbers Marathon

DatAtarLyfe said:
You can say the same thing about the integration marathon and the combinatorics marathon as well. By closing this one, you would have to close the other two to keep it fair

Those two gained popularity mainly cause integration and combinatorics have 20x the amount of questions as anything else in the scope of the syllabus I suppose.

DatAtarLyfe · Apr 14, 2016

HSC 2016 Complex Numbers Marathon

Lol popularity, i can probs count on one hand the amount of times the combinatorics marathon has been active.
And iirc, in the "favourite 4u topic", complex had more votes than harder 3u.
What im saying is, if the mods left the integration and combinatorics marathon, then it would be hypocritical of them to shut this one down

InteGrand · Apr 14, 2016

Re: HSC 2016 Complex Numbers Marathon

DatAtarLyfe said:
Lol popularity, i can probs count on one hand the amount of times the combinatorics marathon has been active.
And iirc, in the "favourite 4u topic", complex had more votes than harder 3u.
What im saying is, if the mods left the integration and combinatorics marathon, then it would be hypocritical of them to shut this one down

I think the main reason the Combinatorics Marathon is rarely active is that most people hate/dislike that topic.

RealiseNothing · Apr 14, 2016

Re: HSC 2016 Complex Numbers Marathon

InteGrand said:
I think the main reason the Combinatorics Marathon is rarely active is that most people hate/dislike that topic.

Still beyond me how anyone can hate the GOAT topic.

Drsoccerball · Apr 14, 2016

Re: HSC 2016 Complex Numbers Marathon

What's the purpose of the 4U marathon(Which were made by the mods) ?

seanieg89 · Apr 14, 2016

Re: HSC 2016 Complex Numbers Marathon

DatAtarLyfe said:
Lol popularity, i can probs count on one hand the amount of times the combinatorics marathon has been active.

Thanks to all your extensive counting practice from the combinatorics marathon!

InteGrand · Apr 14, 2016

Re: HSC 2016 Complex Numbers Marathon

Drsoccerball said:
What's the purpose of the 4U marathon(Which were made by the mods) ?

I think it's a sort of BOS tradition (the old marathons were more popular I think, maybe more high-schoolers posted maths things on BOS then).

KingOfActing · Apr 14, 2016

Re: HSC 2016 Complex Numbers Marathon

Paradoxica said:
$$\noindent Sum to infinity the following series: $\\ \cos{\theta}\cos{\theta} + \cos^2{\theta}\cos{2\theta} + \cos^3{\theta}\cos{3\theta} + \cdots + \cos^n{\theta}\cos{n\theta}\\ $State the values of $\theta$ for which the series does not converge.$ \\ $Repeat the above for the following series': $\\ \cos{\theta}\cos{\theta} + 2\cos^2{\theta}\cos{2\theta} + 4\cos^3{\theta}\cos{3\theta} + \cdots + 2^{n-1}\cos^n{\theta}\cos{n\theta} \\ \cos{\theta}\cos{\theta} + 2\cos^2{\theta}\cos{2\theta} + 3\cos^3{\theta}\cos{3\theta} + \cdots + n\cos^n{\theta}\cos{n\theta} \\ \cos{\theta}\cos{\theta} + 4\cos^2{\theta}\cos{2\theta} + 9\cos^3{\theta}\cos{3\theta} + \cdots + n^2\cos^n{\theta}\cos{n\theta}$

$\text{Write }\cos{\theta} = \frac{z+z^{-1}}{2}\\$

$\begin{align*}\sum_{n=1}^{\infty}\cos^n\theta \cos{n\theta} &= \sum_{n=1}^{\infty} \left(\frac{z+z^{-1}}{2}\right)^n\left(\frac{z^n+z^{-n}}{2} \right ) \\&=\frac{1}{2} \sum_{n=1}^{\infty} \left(\frac{z(z+z^{-1})}{2}\right)^n+\left(\frac{z^{-1}(z+z^{-1})}{2} \right )^n\end{align*}$

$\text{Using the GP formula, and simplifying we obtain:}\\ \sum_{n=1}^{\infty}\cos^n\theta \cos{n\theta} = 0,\,|\cos{\theta}| < 1$

KingOfActing · Apr 14, 2016

Re: HSC 2016 Complex Numbers Marathon

$\begin{align*} \sum_{n=1}^{\infty}2^{n-1}\cos^n{\theta}\cos{n\theta} &= \frac{1}{2}\sum_{n=1}^{\infty}2^{n}\left(\frac{z+z^{-1}}{2}\right)^n\left( \frac{z^n+z^{-n}}{2}\right ) \\&=\frac{1}{4}\sum_{n=1}^{\infty}\left(z(z+z^{-1})\right )^n + \left(z^{-1}(z+z^{-1}) \right )^n \\&= \frac{1}{4} \sum_{n=1}^{\infty}(z^2+1)^n + (1+z^{-2})^n \\ &=\frac{1}{4}\left( \frac{z^2+1}{1-(z^2+1)} + \frac{1+z^{-2}}{1-(1+z^{-2})} \right )\\&=\frac{1}{4}\left(\frac{2+z^2+z^{-2}}{-1} \right ) \\&=\frac{-1}{2}(1+\cos{2\theta}) = -\cos^2{\theta}\\\\&|2\cos{\theta}| < 1 \implies |\cos{\theta}| < \frac12 \end{align*}$

KingOfActing · Apr 14, 2016

Re: HSC 2016 Complex Numbers Marathon

I like these (but I don't like typing them...

)

$\begin{align*}\text{Lemma: } &\sum_{n=1}^{\infty} nr^n = \frac{r}{(1-r)^2}, |r| < 1\\\text{Proof: }\sum_{n=1}^{\infty}r^n &= \frac{r}{1-r} \\ \frac{d}{dr} \sum_{n=1}^{\infty}r^n &= \frac{d}{dr}\frac{r}{1-r}\\ \sum_{n=1}^{\infty}nr^{n-1} &= \frac{1}{(1-r)^2}\\ \sum_{n=1}^{\infty}nr^{n} &= \frac{r}{(1-r)^2}\end{align*}$

$\begin{align*} \sum_{n=1}^{\infty}n\cos^n{\theta}\cos{n\theta} &= \frac{1}{2}\sum_{n=1}^{\infty}n\left(\frac{z^2+1}{2} \right )^n + n\left( \frac{1+z^{-2}}{2}\right)^n \\&= \frac{1}{2}\left(\frac{\frac{z^2+1}{2}}{\left(1-\left(\frac{z^2+1}{2} \right )\right)^2} + \frac{\frac{1+z^{-2}}{2}}{\left(1-\left(\frac{1+z^{-2}}{2} \right )\right)^2} \right )\\&= \dots\\&=\left( \frac{z^2+1}{z^2-1} \right )^2\\&=(-i\cot{\theta})^2 = -\cot^2{\theta}, |\cos{\theta}| < 1 \end{align*}$

KingOfActing · Apr 14, 2016

Re: HSC 2016 Complex Numbers Marathon

$\begin{align*}\text{Lemma: } &\sum_{n=1}^{\infty} n^2r^n = \frac{-r(r+1)}{(1-r)^3}, |r| < 1\\\text{Proof: }\sum_{n=1}^{\infty}nr^{n} &= \frac{r}{(1-r)^2} \\ \frac{d}{dr} \sum_{n=1}^{\infty}nr^n &= \frac{d}{dr}\frac{r}{(1-r)^2}\\\sum_{n=1}^{\infty}n^2r^{n-1} &= \frac{-(r+1)}{(1-r)^3}\\\sum_{n=1}^{\infty} n^2r^n &= \frac{-r(r+1)}{(1-r)^3}\end{align*}$

$\begin{align*} \sum_{n=1}^{\infty}n^2\cos^n{\theta}\cos{n\theta} &= \frac{1}{2}\sum_{n=1}^{\infty}n^2\left(\frac{z^2+1}{2} \right )^n + n^2\left( \frac{1+z^{-2}}{2}\right)^n \\&= \frac{1}{2}\left( \frac{-\left( \frac{z^2+1}{2}\right )\left(\frac{z^2+1}{2} + 1\right)}{\left(1-\frac{z^2+1}{2} \right )^3} + \frac{-\left( \frac{z^{-2}+1}{2}\right )\left(\frac{z^{-2}+1}{2} + 1\right)}{\left(1-\frac{z^{-2}+1}{2} \right )^3} \right)\\&= \frac{-(z^2+1)(z^2+3)}{(1-z^2)^3} + \frac{-(z^{-2}+1)(z^{-2}+3)}{(1-z^{-2})^3}\\&=\frac{-(z^2+1)(z^2+3)+z^2(z^2+1)(3z^2+1)}{(1-z^2)^3}\\&=\dots\\&=-3\left(\frac{z^2+1}{z^2-1}\right )^2\\&=3\cot^2{\theta}, \, |\cos{\theta}| < 1\end{align*}$

HSC 2016 MX2 Complex Numbers Marathon (archive) (3 Viewers)

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