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)

Jazmin Clark

Jazmin Clark

Answered question

2022-08-13

Let f ( z ) = A 0 + A 1 z + A 2 z 2 + + A n z n be a complex polynomial of degree n > 0
Show that 1 2 π i | z | = R z n 1 | f ( z ) | 2 d z = A 0 A n ¯ R 2 n \

Answer & Explanation

volksgeesr9

volksgeesr9

Beginner2022-08-14Added 15 answers

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.

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