2 z z 1 3 2 what can you conclude about f justify

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) | ≤ 2 | z | | z - 1 | 3 / 2 . What can you conclude about f ? Justify your assertions. Solution We claim the only possibility is f ( z ) 0. Proof: Clearly the only possible singularity of f ( z ) is at z = 1. Let g ( z ) := ( z - 1) 2 f ( z ). Then | g ( z ) | ≤ (2 | z | ) 3 / 2 | z - 1 | 1 / 2 . Since g ( z ) is bounded near z = 1, it has at most a removable singularity there. Consequently g is an entire function. Because it grows at most like | z | 2 for large z , it must be a quadratic polynomial. But g ( z ) := ( z - 1) 2 f ( z ) so f ( z ) constant. Noticing that g (0) = 0 we conclude that f (0) = 0. Hence f ( z ) 0. B2. Evaluate A = -∞ cos x x 2 + a 2 dx where a > 0. Solution Consider B R := Γ R e iz z 2 + a 2 dz over the semicircle Γ R which is the boundary of the half-disk | z | = R in the upper-half plane y > 0. For R > a the only singularity of the integrand inside Γ R is at z = ia . Thus by the residue theorem B R = 2 πi e - a 2 ia = πe - a a . But also B R = R - R e ix x 2 + a 2 dx + γ R e iz z 2 + a 2 dz = I 1 + I 2 , where γ R is the semi-circle z = Re for 0 θ π . Since | e iz | = | e ix - y | = e - y 1 on γ R , then for R large I 2 = O (1 /R ) 0. Letting R → ∞ we conclude that πe - a a = B R -∞ e ix x 2 + a 2 dx. . Taking the real parts of both sides, we conclude that -∞ cos x x 2 + a 2 dx = πe - a a .
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3 B3. a) Let f ( z ) be holomorphic in | z | ≤ R with | f ( z ) | ≤ M on | z | = R . Show that | f ( z ) - f (0) | ≤ 2 M | z | R (1) Solution To be brief, apply the Schwarz Lemma to g ( z ) := f ( z ) - f (0) and note that | g ( z ) | ≤ 2 M on | z | = R . In greater detail, since g ( z ) /z is holomorphic in | z | ≤ R , by the maximum principle | g ( z ) /z | ≤ 2 M/R .
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  • Fall '09
  • Math, Cauchy Integral Formula, Short Answer Problems, Entire function, Jerry L. Kazdan

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