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# 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
4 years, 1 month ago

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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 $$\tan \frac{\pi}{5} + i$$. 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. · 4 years, 1 month 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. · 4 years, 1 month 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). · 4 years, 1 month 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. · 4 years, 1 month 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. · 4 years, 1 month ago

@A L The correct answer is n=3. Now, i understand. Thank you for the neat explanation! · 4 years, 1 month ago

@A L Your answer is actually correct; it is mine that is wrong. · 4 years, 1 month 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. · 4 years, 1 month 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? Staff · 4 years, 1 month ago