UMD Analysis Qualifying Exam/Aug08 Complex

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Problem 2[edit]


 \int_0^\infty \frac{1}{x^3+1}dx \!\,

Solution 2[edit]

We will compute the general case:

 \int_0^\infty \frac{1}{x^n+1}dx  \!\,

Find Poles of f(z)[edit]

The poles of  f(z)= \frac{1}{z^n+1}  \!\, are just the zeros of  z^n+1   \!\,, so we can compute them in the following manner:

If z=re^{i \theta}    \!\, is a solution of  z^n+1=0   \!\,,

then  z^n=r^ne^{i\theta n} = -1   \!\,

   \Rightarrow r=1   \!\, and  i n \theta  = i (\pi + 2\pi k )    \!\,

   \Rightarrow \theta = \frac{\pi + 2\pi k}{n}\!\, , k=0,1,2,...,n-1.

Thus, the poles of  f(z)   \!\, are of the form  z=e^{\frac{\pi+2\pi k}{n}}   \!\, with k=0,1,...,n-1    \!\,

Choose Path of Contour Integral[edit]

In order to get obtain the integral of  f(x)    \!\, from 0 to     \infty \!\, , let us consider the path   \gamma  \!\, consisting in a line   A  \!\, going from 0 to   R  \!\,, then the arc  B   \!\, of radius  R   \!\, from the angle 0 to   \frac{2 \pi}{n}  \!\, and then the line  C   \!\, joining the end point of B    \!\, and the initial point of     A\!\,,

Pi n contour.jpg

where   R  \!\, is a fixed positive number such that

the pole  z_0 = e^{i \frac{\pi}{n}}   \!\, is inside the curve     \gamma \!\,. Then , we need to estimate the integral

 \int_{\gamma} f(z) = \underbrace{\int_{0}^R f(z)}_{A} + \underbrace{\int_{S(R)} f(z)}_{B} + \underbrace{\int_{Re^{i\frac{2\pi}{n}}}^0 f(z) }_{C}  \!\,

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


   Res(f,z_0)&= \left. \frac{1}{(z^n+1)' } \right|_{z=z_0}   \\
&= \frac{1}{nz_0^{n-1}} \\
&= \frac{1}{ne^{\frac{i\pi}{n}(n-1)}} \\
&= \frac{e^{\frac{i\pi}{n}}}{ne^{i \pi}} \\
&= \frac{-1}{n}e^{\frac{i\pi}{n}} 



Bound Arc Portion (B) of Integral[edit]

|B|                              &=\left| \int_{S(R)} \frac{dz}{z^n+1} \right|  \\
                                 &\leq \int_{S(R)}\frac{dz}{|z^n+1|} \\
                                 &= \int_{S(R)} \frac{dz}{|R^n+1|} \\
                                 &\leq  \frac{1}{R^n} \int_{S(R)} dz \\
                                 &=     \frac{1}{R^n} \frac{2 \pi}{n} R \\
                                 &= \frac{2 \pi}{nR^{n-1}}

Hence as R \rightarrow \infty \!\,, |B| \rightarrow 0 \!\,

Parametrize (C) in terms of (A)[edit]

Let z=re^{i\frac{2\pi}{n}} \!\, where r \!\, is real number. Then  dz=e^{i\frac{2\pi}{n}}dr \!\,

C    &= \int_{Re^{i\frac{2\pi}{n}}}^0 \frac{dz}{1+z^n}\\
     &= -\int_0^{Re^{i\frac{2\pi}{n}}}\frac{dz}{1+z^n} \\
     &= -\int_0^R \frac{e^{i\frac{2\pi}{n}}}{1+(re^{i\frac{2\pi}{n}})^n}\\
     &= -\int_0^R \frac{e^{i\frac{2\pi}{n}}}{1+r^n}\\
     &= -e^{i\frac{2\pi}{n}} \underbrace{\int_0^R \frac{dr}{1+r^n}}_{A} \\

Apply Cauchy Integral Formula[edit]

From Cauchy Integral Formula, we have,

 A+B+C = 2\pi i \frac{-e^{i\frac{\pi}{n}}}{n } \!\,

As R \rightarrow \infty \!\,, B \rightarrow 0 \!\,. Also C \!\, can be written in terms of A \!\,. Hence

A+B+C &= A+C \\
      &= (1-e^{i\frac{2\pi}{n}}) A 

We then have,

A     &= \frac{2\pi i }{n }\frac{e^{i\frac{\pi}{n}}}{e^{i\frac{2\pi}{n}}-1} \\
      &= \frac{\pi}{n} \frac{2i e^{i\frac{\pi}{n}}}{e^{i\frac{\pi}{n}}(e^{i\frac{\pi}{n}}-e^{-i\frac{\pi}{n}})}\\
      &= \frac{\pi}{n} \frac{1}{\sin(\frac{\pi}{n})}

Problem 4[edit]

Suppose S=\{ z: -\frac{\pi}{2} < \Im(z) < \frac{\pi}{2}\} \!\, and there is an entire function g \!\, with g(S) \subset S \!\,. If g(-1)=0 \!\, and g(0)=1 \!\,, prove that g(z)=z+1 \!\,

Solution 4[edit]

Lemma: Two fixed points imply identity[edit]

Lemma. Let  f  \!\, be analytic on the unit  D \!\,, and assume that |f(z)| < 1 \!\, on the disc. Prove that if there exist two distinct points a\!\, and b\!\, in the disc which are fixed points, that is,                   f(a)=a\!\, and  f(b)=b  \!\,, then    f(z)=z               \!\,.

Proof Let h:D \rightarrow D                  \!\, be the automorphism defined as

 h(z)= \frac{a-z}{1-\overline{a}z}  \!\,

Consider now  F(z)= h \circ f \circ h^{-1} (z)    \!\,. Then, F has two fixed points, namely

 F(0)=    h \circ f \circ h^{-1} (0)= h \circ f (a) = h  (a) =0              \!\,

F\left(\frac{a-b}{1-\overline{a}b}\right) =  h \circ f \circ h^{-1} \left(\frac{a-b}{1-\overline{a}b}\right) = h \circ f (b) = h (b) = \frac{a-b}{1-\overline{a}b}        \!\,.

Since  F(0)=0 \!\,,

\frac{a-b}{1-\overline{a}b} \neq 0 \!\, (since  a \!\, is different to  b \!\, ), and

 \left|F\left(\frac{a-b}{1-\overline{a}b}\right)\right|=\left|\frac{a-b}{1-\overline{a}b}\right| \!\,,

by Schwarz Lemma,

F(z)= \alpha z     \!\,.

But, replacing \frac{a-b}{1-\overline{a}b} \!\, into the last formula, we get  \alpha =1 \!\,.


    h \circ f \circ h^{-1} (z)  = z           \!\,,

which implies

         f(z)=z         \!\,

Shift Points to Create Fixed Points[edit]

Let f(z)=g(z)-1 \!\,. Then f(0)=0 \!\, and f(-1)=-1 \!\,.

Notice that S \!\, is an infinite horizontal strip centered around the real axis with height \pi \!\,. Since f(z) \!\, is a unit horizontal shift left, f(S) \subset S \!\,.

Use Riemann Mapping Theorem[edit]

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

Define Composition Function[edit]

Let F=h^{-1}\circ f \circ h \!\,. Then  F\!\, maps D \!\, to D \!\,.

From the lemma, since F(z) \!\, has two fixed points, F(z)=z \!\, which implies  f(z)=z \!\, which implies g=z+1 \!\,.

Problem 6[edit]

Let \mathcal{F} \!\, be the family of functions f \!\, analytic on \{|z|<1\} \!\, so that

\int\int_{|z|<1} |f(x+iy)|^2dx dy \leq 1 \!\,

Prove that \mathcal{F} \!\, is a normal family on \{|z|<1\} \!\,

Solution 6[edit]

Choose any compact set K in D[edit]

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

We want to show that for all f \in \mathcal{F} \!\, and all z \in K \!\, , |f(z)| \!\, is bounded i.e.

|f(z)| < B_K \!\,

where B_K \!\, is some constant dependent on the the choice of K \!\, .

Apply Maximum Modulus Principle to find |f(z0)|[edit]

Choose z_0 \!\, that is the shortest distance from the boundary of the unit disk D \!\, . From the maximum modulus principle, |f(z_0)|=\max_{z \in K} |f(z|  \!\, .

Note that z_0 \!\, is independent of the choice of f \in \mathcal{F}\!\, .

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

We will apply Cauchy's Integral formula to f^2(z_0) \!\, (instead of  f(z_0)\!\, ) to take advantage of the hypothesis.

Choose sufficiently small r_0  > 0\!\, so that  D(z_0,r_0) \!\, \in D

f^2(z_0) &= \frac{1}{2\pi i} \int_{|z-z_0|=r_0} \frac{f(z)}{z-z_0}dz \\
         &= \frac{1}{2\pi i} \int_0^{2\pi} \frac{f^2(z_0+r_0e^{i\theta})ir_0e^{i\theta}}{(z_0+r_0e^{i\theta})-z_0} d\theta\\
         &= \frac{1}{2\pi} \int_0^{2\pi} f^2(z_0+r_0e^{i\theta})d\theta

Integrate with respect to r[edit]

\int_0^{r_0} r f^2(z_0)dr &= \int_0^{r_0} \int_0^{2\pi} f^2(z_0+re^{i\theta})rdr d\theta \\ 
                          &= \frac{1}{2\pi}\int\int_{D(z_0,r)}f^2(x+iy)dxdy   

Integrating the left hand side, we have

 \int_0^{r_0} r f^2(z_0)dr  =   \frac{r_0^2}{2}f^2(z_0)     \!\,


\frac{r_0^2}{2}f^2(z_0)  = \frac{1}{2\pi}\int\int_{D(z_0,r)}f^2(x+iy)dxdy    \!\,

Bound |f(z0)| by using hypothesis[edit]

\left| \frac{r_0^2}{2} f^2(z_0) \right| &= \left| \frac{1}{2 \pi} \int\int_{D(z_0,r_0)} f^2(x+iy)dxdy \right|  \\
&\leq    \frac{1}{2 \pi} \int\int_{D(z_0,r_0)} |f^2(x+iy)|dxdy  \\
 &\leq \frac{1}{2 \pi} \int\int_{D} |f^2(x+iy)|dxdy  \\
 &\leq \frac{1}{2\pi} \\
\mbox{Then     }
\left| \frac{r_0^2}{2} f^2(z_0) \right| &\leq \frac{1}{2 \pi}    \\
\mbox{This implies}\\
|f(z_0)| &\leq  \frac{1}{r_0 \sqrt{\pi}}        

Apply Montel's Theorem[edit]

Then, since any  f \in \mathcal{F} \!\, is uniformly bounded in every compact set, by Montel's Theorem, it follows that              \mathcal{F}\!\, is normal