{ α 3 β 5 = 1 α 7 β 2 = 1
Let α = cos θ 1 + i sin θ 1 and β = cos θ 2 + i sin θ 2 be the complex numbers satisfying the system above, where 0 < θ 1 and θ 2 < 2 π .
If θ 2 θ 1 = b a , where a and b are coprime positive integers , compute a + b .
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Another solution can be obtained by letting 7 θ 1 + 2 θ 2 = 4 π , which is possible, as 0 < 7 θ 1 + 2 θ 2 < 2 9 π . You should edit the problem to remove this solution (you could make θ 1 , θ 2 have the range 0 , 3 π ).
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OK. Will do.
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I see my error, the same one committed by @Guillermo Templado ,
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 then θ 2 = 2 9 2 π and θ 1 = 2 9 1 6 π but this implies that θ 1 > 2 π and this can't happen... And there are no more possibilities... So, sorry, Worranat, your original problem was absolutely right.
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No problem. So back to the same one then.
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.
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By using De Moivres Theorem, we can multiple the angles of the complex numbers and write equations for θ 1 & θ 2 :
3 θ 1 + 5 θ 2 = 2 π
7 θ 1 + 2 θ 2 = 2 π
Solving for both angles, we will get θ 1 = 2 9 3 ( 2 π ) and θ 2 = 2 9 4 ( 2 π ) .
Clearly, θ 2 θ 1 = 4 3 .
Hence, a + b = 7 .