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?

No vote yet

9 votes

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestIt'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 \[ \begin{align*} \frac{\tan\frac{\pi}{5} + i}{\tan\frac{\pi}{5} - i} &= \frac{(\tan \frac{\pi}{5} + i)^2}{\tan^2 \frac{\pi}{5} + 1} \\ &= \left( \frac{\tan \frac{\pi}{5} + i}{\sec \frac{\pi}{5}}\right)^2 \\ &= \left( \sin \frac{\pi}{5} + i \cos \frac{\pi}{5} \right)^2 \\ &= \sin \frac{2\pi}{5} + i \cos \frac{2\pi}{5}, \end{align*} \] 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.

Log in to reply

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.

Log in to reply

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).

Log in to reply

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.

Log in to reply

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.

Log in to reply

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

Log in to reply

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

Log in to reply

Take i common in both the numerator and denominator in, (tanπ/5+i)/(tanπ/5−i) to make it, (-i

tanπ/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.Log in to reply

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?

Log in to reply

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

Log in to reply