Mandelbrot 2002-2003 Round 2 #6

The quantity $\frac{\tan\frac{\pi}{5}+i}{\tan\frac{\pi}{5}-i}$ is the tenth root of unity. In other words, it is equal to $\cos\frac{2n\pi}{10}+i \sin\frac{2n\pi}{10}$ for some integer $n$ between 0 and 9 inclusive. Which value of n?

Does anyone know a way to proceed in this question?

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Comments

It's obvious that one of the steps is to clear the denominator of any complex valued expression by multiplying numerator and denominator by its complex conjugate $\tan \frac{\pi}{5} + i$ . This yields $\begin{aligned} \frac{\tan\frac{\pi}{5} + i}{\tan\frac{\pi}{5} - i} &= \frac{(\tan \frac{\pi}{5} + i)^2}{\tan^2 \frac{\pi}{5} + 1} \\ &= \left( \frac{\tan \frac{\pi}{5} + i}{\sec \frac{\pi}{5}}\right)^2 \\ &= \left( \sin \frac{\pi}{5} + i \cos \frac{\pi}{5} \right)^2 \\ &= \sin \frac{2\pi}{5} + i \cos \frac{2\pi}{5}, \end{aligned}$ where we used DeMoivre's formula in the last step.

EDIT:

There is something very wrong with the above solution. The correct answer should be $n = 3$ . Note that $\tan\frac{\pi}{5} < \tan\frac{\pi}{4} = 1$ , so that the argument of the numerator is more than $\pi/4$ . So after multiplying by the complex conjugate, the square of the numerator must have an argument greater than $\pi/2$ ; i.e., it must have a negative real part. This is not the case with the answer $n = 2$ .

I suspect the flaw in my solution arises from not considering more carefully the use of the trigonometric identity $\tan^2 \theta + 1 = \sec^2 \theta$ , but I have not figured out exactly what is going on yet. I will look into the matter further.

EDIT 2:

Oh for crying out loud... after I looked at my other solution I immediately realized the elementary mistake I made. De Moivre's formula is $(\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n\theta$ --I have the sines and cosines switched. Therefore, we should write $\left(\sin\frac{\pi}{5} + i \cos \frac{\pi}{5}\right)^2 = \left( \cos\frac{3\pi}{10} + i \sin \frac{3\pi}{10} \right)^2 = e^{6i\pi/10} = e^{3i\pi/5},$ and that's the correct result.

hero p. - 7 years, 11 months ago

If i were to do this a different way, I would proceed by multiplying numerator and denominator by $\cos\frac{\pi}{5}$ : $\frac{\tan\frac{\pi}{5} + i}{\tan\frac{\pi}{5} - i} = \frac{\sin \frac{\pi}{5} + i \cos \frac{\pi}{5}}{\sin \frac{\pi}{5} - i \cos\frac{\pi}{5}} = \frac{e^{i\pi/5}}{e^{-i\pi/5}} = e^{2i \pi/5}.$ This still gives the wrong answer, yet I am at a loss as to explain why.

Wow, I shouldn't be doing math so late at night...my apologies for being so silly. I should have written $\frac{\sin \frac{\pi}{5} + i \cos \frac{\pi}{5}}{\sin \frac{\pi}{5} - i \cos\frac{\pi}{5}} = \frac{\cos \frac{3\pi}{10} + i \sin \frac{3\pi}{10}}{\cos\frac{3\pi}{10} - i \sin\frac{3\pi}{10}} = e^{6i\pi/10} = e^{3i\pi/5},$ which gives the correct answer after all. Now I have to figure out what I did wrong in my original solution.

hero p. - 7 years, 11 months ago

Your mistake lies in the expansion of the term $\large (\sin \frac{\pi}{5}+ i \cos \frac{\pi}{5})^2$

The correct expansion turns out to be:

$\large \sin ^2 \frac{\pi}{5} - \cos \frac{\pi}{5} + i \sin \frac{2 \pi}{5}$

$\large -(\cos \frac{2 \pi}{5}) + i \sin \frac{2 \pi}{5}$

Now note that: $\cos \frac{2 \pi}{5}= -\cos \frac{3 \pi}{5}$

And, $\sin \frac{2 \pi}{5}= \sin \frac{3 \pi}{5}$

Hence, our results occur. Correct me if I did anything wrong here.

Forgive me. I didn't notice that you have already edited your original solution and found your mistake.(Faulty Internet connection).

Aditya Parson - 7 years, 11 months ago

The problem with your first solution is you have sic, not cis. By simple confunctions, you can turn sin 2pi / 5 + i cos 2pi / 5 into cos 3pi / 5 + i sin 3pi / 5. Sorry for not using latex, I'm at work and only have a small time before I need to get back.

Michael Tong - 7 years, 11 months ago

Take i common in both the numerator and denominator in, (tanπ/5+i)/(tanπ/5−i) to make it, (-itanπ/5+1)/(-itanπ/5−1) which again becomes, (-iSinπ/5+Cosπ/5)/(-iSinπ/5−Cosπ/5) or, (iSinπ/5-Cosπ/5)/(iSinπ/5+Cosπ/5) Using De moiver's theorum in this( In the numerator the argument is 4π/5) We get n=3.

Nishant Rai - 7 years, 11 months ago

A useful way to deal with fractions of the form $\frac{a+b}{a-b} = \frac{c}{d}$ is to use "componendo and dividendo" to conclude that $\frac{ a } { b} = \frac{ c+d}{c-d}$ .

Apply it to this problem, what do you get?

Calvin Lin Staff - 7 years, 11 months ago

Consider the argument of that complex number. It is equal to $\arg(\tan(\frac{\pi}{5})+i)-\arg(\tan(\frac{\pi}{5})-i)$ , which is

$\arctan(\frac{1}{\tan(\frac{\pi}{5})})-\arctan(\frac{-1}{\tan(\frac{\pi}{5})})=2\arctan(\frac{1}{\tan(\frac{\pi}{5})})$

To calculate this draw a triangle $ABC$ such that the side $CB=1, BA=\tan(\frac{\pi}{5})$ angle $ABC=\frac{\pi}{2}$ and $CAB=\arctan(\frac{1}{\tan(\frac{\pi}{5})})$ . Note that $CAB=\frac{\pi}{2}-\frac{\pi}{5}=\frac{3 \pi}{10}$ , so its argument is $\frac{3 \pi}{5}$ .

I hope this method helps.

A L - 7 years, 11 months ago

The correct answer is n=3. Now, i understand. Thank you for the neat explanation!

Shaan Bhandarkar - 7 years, 11 months ago

Your answer is actually correct; it is mine that is wrong.

hero p. - 7 years, 11 months ago

No calculation has to be done!! It is clear that any value of n belongs to N should satisfy the equation!!!!

Subhrodipto Basu Choudhury - 7 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$