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UMD Analysis Qualifying Exam/Aug08 Complex

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Problem 2

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Compute



Solution 2

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We will compute the general case:



Find Poles of f(z)

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The poles of are just the zeros of , so we can compute them in the following manner:

If is a solution of ,

then

and

, k=0,1,2,...,n-1.

Thus, the poles of are of the form with

Choose Path of Contour Integral

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In order to get obtain the integral of from 0 to , let us consider the path consisting in a line going from 0 to , then the arc of radius from the angle 0 to and then the line joining the end point of and the initial point of ,

where is a fixed positive number such that

the pole is inside the curve . Then , we need to estimate the integral

Compute Residues of f at z0= exp{i\pi /n}

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Bound Arc Portion (B) of Integral

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Hence as ,

Parametrize (C) in terms of (A)

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Let where is real number. Then


Apply Cauchy Integral Formula

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From Cauchy Integral Formula, we have,



As , . Also can be written in terms of . Hence



We then have,


Problem 4

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Suppose and there is an entire function with . If and , prove that

Solution 4

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Lemma: Two fixed points imply identity

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Lemma. Let be analytic on the unit , and assume that on the disc. Prove that if there exist two distinct points and in the disc which are fixed points, that is, and , then .

Proof Let be the automorphism defined as

Consider now . Then, F has two fixed points, namely

.

Since ,

(since is different to ), and

,

by Schwarz Lemma,

.

But, replacing into the last formula, we get .

Therefore,

,

which implies

Shift Points to Create Fixed Points

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Let . Then and .


Notice that is an infinite horizontal strip centered around the real axis with height . Since is a unit horizontal shift left, .

Use Riemann Mapping Theorem

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From the Riemann mapping theorem, there exists a biholomorphic (bijective and holomorphic) mapping , from the open unit disk to .

Define Composition Function

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Let . Then maps to .


From the lemma, since has two fixed points, which implies which implies .

Problem 6

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Let be the family of functions analytic on so that



Prove that is a normal family on

Solution 6

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Choose any compact set K in D

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Choose any compact set in the open unit disk . Since is compact, it is also closed and bounded.


We want to show that for all and all , is bounded i.e.



where is some constant dependent on the choice of .

Apply Maximum Modulus Principle to find |f(z0)|

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Choose that is the shortest distance from the boundary of the unit disk . From the maximum modulus principle, .

Note that is independent of the choice of .

Apply Cauchy's Integral Formula to f^2(z0)

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We will apply Cauchy's Integral formula to (instead of ) to take advantage of the hypothesis.


Choose sufficiently small so that


Integrate with respect to r

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Integrating the left hand side, we have



Hence,


Bound |f(z0)| by using hypothesis

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Apply Montel's Theorem

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Then, since any is uniformly bounded in every compact set, by Montel's Theorem, it follows that is normal