Could someone help me through this problem?
Consider Iϵ=∮_(Cϵ) ((z^α)f(z)) dz , α>−1,α real where Cϵ is a circle of radius ϵ centered at the origin and f(z) is analytic inside the circle.
Show that limϵ→0 (Iϵ)=0
Consider Iϵ=∮_(Cϵ) ((z^α)f(z)) dz , α>−1,α real where Cϵ is a circle of radius ϵ centered at the origin and f(z) is analytic inside the circle.
Show that limϵ→0 (Iϵ)=0
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Since f is analytic on the disk, |f| is bounded---and indeed takes its maximum value on the circle (and not in the interior unless it is constant).
Note that using the parametrization z = ε e^(iΘ), you get
∮dz/z = 2πi.
C_ε
Okay, so consider writing z^α = z^(1 + α)/z. Since α > -1, 1 + α > 0. For fixed ε > 0, let
M(ε) = max{|f(z)| | |z| = ε}.
By the maximum modulus theorem, M(ε) ≤ M(δ) whenever ε ≤ δ---that is, M is non-decreasing. Now, observe that (all integrals are over C_ε)
|∮z^α f(z) dz| ≤ ∮|z^(1+α)f(z)| d|z|/|z| ≤ ε^(1 + α)M(ε) 2π.
Since 2πM(ε) is bounded and 1 + α > 0, you can take the limit as ε->0 and obtain the desired result.
Note that using the parametrization z = ε e^(iΘ), you get
∮dz/z = 2πi.
C_ε
Okay, so consider writing z^α = z^(1 + α)/z. Since α > -1, 1 + α > 0. For fixed ε > 0, let
M(ε) = max{|f(z)| | |z| = ε}.
By the maximum modulus theorem, M(ε) ≤ M(δ) whenever ε ≤ δ---that is, M is non-decreasing. Now, observe that (all integrals are over C_ε)
|∮z^α f(z) dz| ≤ ∮|z^(1+α)f(z)| d|z|/|z| ≤ ε^(1 + α)M(ε) 2π.
Since 2πM(ε) is bounded and 1 + α > 0, you can take the limit as ε->0 and obtain the desired result.
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And How can you make without using the maximum modulus theorem?
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