Prime Power Roots

Algebra Level 3

{ α 3 β 5 = 1 α 7 β 2 = 1 \begin{cases} \alpha^3 \beta^5 = 1 \\ \alpha^7 \beta^2 = 1 \end{cases}

Let α = cos θ 1 + i sin θ 1 \alpha = \cos \theta_1 + i \sin \theta_1 and β = cos θ 2 + i sin θ 2 \beta = \cos \theta_2 + i \sin \theta_2 be the complex numbers satisfying the system above, where 0 < θ 1 0 < \theta_1 and θ 2 < π 2 \theta_2 < \frac{\pi}{2} .

If θ 1 θ 2 = a b \frac{\theta_1}{\theta_2} = \frac{a}{b} , where a a and b b are coprime positive integers , compute a + b . a+b.


The answer is 7.

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1 solution

By using De Moivres Theorem, we can multiple the angles of the complex numbers and write equations for θ 1 \theta_1 & θ 2 \theta_2 :

3 θ 1 + 5 θ 2 = 2 π 3\theta_1 + 5\theta_2 = 2\pi

7 θ 1 + 2 θ 2 = 2 π 7\theta_1 + 2\theta_2 = 2\pi

Solving for both angles, we will get θ 1 = 3 29 ( 2 π ) \theta_1 = \dfrac{3}{29} (2\pi) and θ 2 = 4 29 ( 2 π ) \theta_2 = \dfrac{4}{29} (2\pi) .

Clearly, θ 1 θ 2 = 3 4 \dfrac{\theta_1}{\theta_2}=\dfrac{3}{4} .

Hence, a + b = 7 a + b = \boxed{7} .

Another solution can be obtained by letting 7 θ 1 + 2 θ 2 = 4 π 7\theta_1 + 2\theta_2 =4\pi , which is possible, as 0 < 7 θ 1 + 2 θ 2 < 9 π 2 0 < 7\theta_1+2\theta_2 < \dfrac{9\pi}{2} . You should edit the problem to remove this solution (you could make θ 1 , θ 2 \theta_1 , \theta_2 have the range 0 , π 3 0, \dfrac{\pi}{3} ).

Alex G - 4 years, 11 months ago

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OK. Will do.

Worranat Pakornrat - 4 years, 11 months ago

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I see my error, the same one committed by @Guillermo Templado ,

Alex G - 4 years, 11 months ago

Firstly, my apologies for @Worranat Pakornrat , I had reported this problem and I was wrong,it wasn't necessary what he has edited it . If we consider the possibility { 3 θ 1 + 5 θ 2 = 2 π 7 θ 1 + 2 θ 2 = 4 π \begin{cases} 3\theta_1 + 5\theta_2 = 2\pi \\ 7\theta_1 + 2\theta_2 =4\pi \end{cases} then θ 2 = 2 π 29 \theta_2 = \frac{2\pi}{29} and θ 1 = 16 π 29 \theta_1 = \frac{16\pi}{29} but this implies that θ 1 > π 2 \theta_1 > \frac{\pi}{2} and this can't happen... And there are no more possibilities... So, sorry, Worranat, your original problem was absolutely right.

Guillermo Templado - 4 years, 11 months ago

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No problem. So back to the same one then.

Worranat Pakornrat - 4 years, 11 months ago

Well, I did not really use the exact values for this. I just equated two after using De Moivres as we are interested in the ratio. (I do not see any problem with this so far.) It worked nicely as 3/4.

Jaeyeon Park - 2 years ago

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