Where the method used should be using complex analysis. ∫ c d θ ( p + cos ⁡ θ ) 2 = 2 π p ( p 2 − 1 ) p 2 − 1 ; c : | z | = 1

Question

Where the method used should be using complex analysis.      ∫          c                  d      θ              (      p      +      cos      ⁡      θ              )        2              =            2      π      p              (              p        2            −      1      )                        p          2                −        1              ;  c  :      |    z    |    =  1

Kash Osborn · Accepted Answer

i do my self like this on       |    z      |    =  1  ,  z  =      e          i      θ        ,  d  θ  =            d      z              i      z      using substitution   cos  ⁡  θ  =            z      +              z                  −          1                      2        1    i    ∫                    d        z            z              (      p      +                        z          +                                    z                                      −              1                                      2            )      (      p      +                        z          +                                    z                                      −              1                                      2            )        o  r      1    i    ∫      1          (      2      p      z      +              z        2            +      1      )      (      p      +                        z          +                                    z                                      −              1                                      2            )      and then multiple by       z    z        1    i    ∫      z          (      2      p      z      +              z        2            +      1      )      (      2      p      z      +              z        2            +      1      )      the roots of   (  2  p  z  +      z    2    +  1  ),       z          1        =  −  p  +            p      2        −    1    ;      z          2        =  −  p  −            p      2        −    1  let       z          1        =  a and       z          2        =  bso   a  −  b  =  2            p      2        −    1   and   a  +  b  =  −  2  pand then i use residue theorm, but the result on the right             π      p              2      (              p        2            −      1      )                        p          2                −        1

Christopher Saunders · Accepted Answer

On       |    z      |    =  1, we have   z  =      e          i      θ      , so             d      z              d      θ        =  i      e          i      θ      , and hence   d  θ  =            d      z              i      z      Also on |z|=1, we have   cos  ⁡  θ  =      1    2        (    z    +          1      z        )  Putting this together, you get      1    i        ∫                  |            z              |            =      1            1          z                        (          p          +                      1            2                                (            z            +                          1              z                        )                    )                2                 d  z  .Expand the denominator,                              1          i                          ∫                                    |                        z                          |                        =            1                                    1                      z                                          (                p                +                                  1                  2                                                  (                  z                  +                                      1                    z                                    )                                )                            2                                               d        z                            =                  1          i                          ∫                                    |                        z                          |                        =            1                                    1                      z                          (                              p                2                            +              p              (              z              +              1                              /                            z              )              +                              1                4                            (              z              +              1                              /                            z                              )                2                            )                                                                        =                  1          i                          ∫                                    |                        z                          |                        =            1                                    1                      (                          p              2                        z            +            p                          z              2                        +            p            +                          z              4                        (                          z              2                        +            2            +            1                          /                                      z              2                        )            )                                                            =        ⋯

Where the method used should be using complex analysis. int_c d theta/(p+cos theta)^2=2pi p/((p^2-1)sqrt(p^2-1));c:|z|=1

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