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?

Note by Shaan Bhandarkar
7 years ago

No vote yet
9 votes

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link]( link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}


Sort by:

Top Newest

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 tanπ5+i \tan \frac{\pi}{5} + i . This yields tanπ5+itanπ5i=(tanπ5+i)2tan2π5+1=(tanπ5+isecπ5)2=(sinπ5+icosπ5)2=sin2π5+icos2π5, \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.


There is something very wrong with the above solution. The correct answer should be n=3 n = 3 . Note that tanπ5<tanπ4=1 \tan\frac{\pi}{5} < \tan\frac{\pi}{4} = 1 , so that the argument of the numerator is more than π/4 \pi/4 . So after multiplying by the complex conjugate, the square of the numerator must have an argument greater than π/2 \pi/2 ; i.e., it must have a negative real part. This is not the case with the answer n=2 n = 2 .

I suspect the flaw in my solution arises from not considering more carefully the use of the trigonometric identity tan2θ+1=sec2θ \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.


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 (cosθ+isinθ)n=cosnθ+isinnθ (\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n\theta --I have the sines and cosines switched. Therefore, we should write (sinπ5+icosπ5)2=(cos3π10+isin3π10)2=e6iπ/10=e3iπ/5, \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 ago

Log in to reply

If i were to do this a different way, I would proceed by multiplying numerator and denominator by cosπ5 \cos\frac{\pi}{5} : tanπ5+itanπ5i=sinπ5+icosπ5sinπ5icosπ5=eiπ/5eiπ/5=e2iπ/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 apologies for being so silly. I should have written sinπ5+icosπ5sinπ5icosπ5=cos3π10+isin3π10cos3π10isin3π10=e6iπ/10=e3iπ/5, \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 ago

Log in to reply

Your mistake lies in the expansion of the term (sinπ5+icosπ5)2\large (\sin \frac{\pi}{5}+ i \cos \frac{\pi}{5})^2

The correct expansion turns out to be:

sin2π5cosπ5+isin2π5\large \sin ^2 \frac{\pi}{5} - \cos \frac{\pi}{5} + i \sin \frac{2 \pi}{5}

(cos2π5)+isin2π5\large -(\cos \frac{2 \pi}{5}) + i \sin \frac{2 \pi}{5}

Now note that: cos2π5=cos3π5\cos \frac{2 \pi}{5}= -\cos \frac{3 \pi}{5}

And, sin2π5=sin3π5\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 ago

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.

Michael Tong - 7 years ago

Log in to reply

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 ago

Log in to reply

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

Apply it to this problem, what do you get?

Calvin Lin Staff - 7 years ago

Log in to reply

Consider the argument of that complex number. It is equal to arg(tan(π5)+i)arg(tan(π5)i)\arg(\tan(\frac{\pi}{5})+i)-\arg(\tan(\frac{\pi}{5})-i), which is


To calculate this draw a triangle ABCABC such that the side CB=1,BA=tan(π5)CB=1, BA=\tan(\frac{\pi}{5}) angle ABC=π2ABC=\frac{\pi}{2} and CAB=arctan(1tan(π5))CAB=\arctan(\frac{1}{\tan(\frac{\pi}{5})}). Note that CAB=π2π5=3π10CAB=\frac{\pi}{2}-\frac{\pi}{5}=\frac{3 \pi}{10}, so its argument is 3π5\frac{3 \pi}{5}.

I hope this method helps.

A L - 7 years ago

Log in to reply

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

Shaan Bhandarkar - 7 years ago

Log in to reply

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

hero p. - 7 years ago

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


Problem Loading...

Note Loading...

Set Loading...