Could be this : cos n ⁡ ( i ) + sin n ⁡ ( − i ) integer for n&gt;2?

Question

Could be this :      cos    n    ⁡  (  i  )  +      sin    n    ⁡  (  −  i  ) integer for n&amp;gt;2?

pampatsha · Accepted Answer

Clearly n should be even, because   sin  ⁡  (  −  i  )  =  −  sin  ⁡  i is purely imaginary.If n is a multiple of 4, we have  sin  ⁡  (  −  i  )  =  −  sin  ⁡  iIf this is an integer N, then we get  (  1  +      e    2        )    n    +  (  1  −      e    2        )    n    =      2    n        e    n    Nwhich is impossible, because we&#039;d obtain a polynomial of degree 2n with integer coefficients having e as root.So the only possibility is to have   n  ≡  2    (  mod    4  ), when the equality would be            (      1      +              e        2                    )        n                            2        n                    e        n              −            (      1      −              e        2                    )        n                            2        n                    e        n              =  NWith the binomial theorem,      ∑          k      =      0        n                      (                    n      k                      )                  e          2      k        −      ∑          k      =      0        n                      (                    n      k                      )              (  −  1      )    k        e          2      k        =      2    n    N      e    n  so  2      ∑                                                      0              ≤              k              ≤              n                                                          k               odd                                                              (                    n      k                      )                  e          2      k        =      2    n    N      e    n  If n&amp;gt;2, this would again give a polynomial with integer coefficients having e as root.So the only case is indeed n=2.

Could be this : cos n </msup> ⁡ ( i ) + si

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