From the fact that 1 5 </mfrac> ( 3 + 4 i ) has infinite order

Waylon Ruiz

Waylon Ruiz

Answered question

2022-05-20

From the fact that 1 5 ( 3 + 4 i ) has infinite order in ( C , ), I'm supposed to infer that 1 π arctan 4 3 is irrational. Proving a arctan 4 3 is irrational.

Answer & Explanation

humanistex3

humanistex3

Beginner2022-05-21Added 9 answers

Let α be your complex number, then
α n = a n 5 n + 1 + b n 5 n + 1 i
and since we have
α n = 1 5 ( 3 + 4 i ) ( a n 1 5 n + i b n 1 5 n ) = 1 5 n + 1 ( 3 a n 1 4 b n 1 + i ( a n 1 + 3 b n 1 ) )
it is clear that a a n = 3 a n 1 4 b n 1 and b n = a n 1 + 3 b n 1 , with a 1 = 15 and b 1 = 20.
Examining everything ( mod 7 ) we find ( a 1 , b 1 ) = ( 1 , 6 ) , ( a 2 , b 2 ) = ( 0 , 5 ) , ( a 3 , b 3 ) = ( 6 , 1 ) and ( a 4 , b 4 ) = ( 0 , 2 ) , ( a 5 , b 5 ) = ( 1 , 6 ) so the cycle continues with no b n = 0.
Hence α n is never real, but if arctan ( 4 / 3 ) = p q π we would have exp ( p q π i 2 q ) = 1 = α 2 q .

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