Let f ( z ) = A 0 + A 1 z + A 2 z 2 + … + A n z n be a complex polynomial of degree n &gt; 0Show that 1 2 π i ∫ | z | = R z n − 1 | f ( z ) | 2 d z = A 0 A n ¯ R 2 n

Question

Let   f  (  z  )  =      A    0    +      A    1    z  +      A    2        z    2    +  …  +      A    n        z    n   be a complex polynomial of degree   n  &amp;gt;  0Show that       1          2      π      i            ∫                  |            z              |            =      R              z          n      −      1            |    f  (  z  )            |        2    d  z  =      A    0                      A        n            ¯            R          2      n

volksgeesr9 · Accepted Answer

Denote by             f      ¯       the polynomial obtained from f by conjugating its coefficients       A    k  . When       |    z            |        2    =  z            z      ¯        =      R    2   then      |    f  (  z  )            |        2    =  f  (  z  )              f      (      z      )        ¯    =  f  (  z  )            f      ¯        (            z      ¯        )  =  f  (  z  )            f      ¯                  (                          R        2            z                  )           .Now      z          n      −      1                  f      ¯                  (                          R        2            z                  )        =                    A        ¯              n                      R                  2          n                    z        +  q  (  z  )     ,where q is a certain polynomial. It follows that            1              2        π        i                  ∫          ∂              D        R                  z          n      −      1            |    f  (  z  )            |        2       d  z  =            1              2        π        i                  ∫          ∂              D        R              f  (  z  )            (                          A        ¯              n                      R                  2          n                    z        +  q  (  z  )            )           d  z  =      A    0                      A        ¯              n        R          2      n           ,because   f  (  z  )  =      A    0    +  z    p  (  z  ) for some polynomial p.

Let f(z)=A_0+A_1z+A_2z^2+…+A_nz^n be a complex polynomial of degree n>0 Show that 1/(2 pi i) int_(|z|=R) z^(n-1) |f(z)|^2 dz=A_0 A_n R^(2n)

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