Mandelbrot 2002-2003 Round 2 #6

The quantity tanπ5+itanπ5i\frac{\tan\frac{\pi}{5}+i}{\tan\frac{\pi}{5}-i} is the tenth root of unity. In other words, it is equal to cos2nπ10+isin2nπ10\cos\frac{2n\pi}{10}+i \sin\frac{2n\pi}{10} for some integer nn between 0 and 9 inclusive. Which value of n?

Does anyone know a way to proceed in this question?

Note by Shaan Bhandarkar
6 years, 3 months ago

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9 votes

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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.

EDIT:

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.

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 (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. - 6 years, 3 months ago

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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 night...my 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. - 6 years, 3 months ago

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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 - 6 years, 3 months ago

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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 - 6 years, 3 months ago

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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 - 6 years, 3 months ago

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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 - 6 years, 3 months ago

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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

arctan(1tan(π5))arctan(1tan(π5))=2arctan(1tan(π5))\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 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 - 6 years, 3 months ago

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

Shaan Bhandarkar - 6 years, 3 months ago

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

hero p. - 6 years, 3 months ago

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

Subhrodipto Basu Choudhury - 6 years, 3 months ago

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