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Complex #34: sneaky use of Rouche’s theorem

June 15, 2011 Leave a comment

Problem #34 is: Prove that there does not exist a polynomial of the form p(z)= z^n+a_{n-1}z^{n-1}+\ldots+a_0 such that |p (z)| < 1 for all z such that |z|=1.

Proof: Suppose that there does exist such a polynomial. Note that |z^n|>|-p(z)| for |z|=1. By Rouche’s theorem, z^n-p(z)=-a_{n-1}z^{n-1}-\ldots-a_0 has the same number of zeros (with multiplicities) as z^n inside the unit disk. But that is impossible, since z^n has n zeros and z^n-p(z) has at most n-1 zeros. QED.

Categories: Complex questions

Complex #27: integrating over a long rectangle

June 15, 2011 Leave a comment

Problem 27: Show that \int_{\gamma} e^{iz}e^{-z^2}dz has the same value on every straight line path \gamma (oriented in the +x direction) parallel to the real axis.

Proof: Consider the rectangle R=[-M,M]\times [0,Y], where Y is some given number and M is a very large number. Since the integrand is entire, the integral over the boundary of R is zero by the Cauchy Integral Theorem, and as M\to\infty, the horizontal pieces correspond to \int_{\gamma} e^{iz}e^{-z^2}dz with \gamma being the line Im (z)=0 or the line Im(z)=Y, with one being negative and the other positive. Thus, the result is proved if we can show that the contributions from the vertical pieces go to zero as M goes to \infty.

Observe that the absolute value of the integrand at z=x+iy is e^{-x^2+y^2-y}, which on each vertical boundary piece is e^{-M^2+y^2-y} for 0\le y\le Y. Integrating this over y will give a value bounded by a constant times e^{-M^2}, which clearly goes to zero as M\to\infty.

QED.

Categories: Complex questions

Complex #26: polynomial properties

June 15, 2011 Leave a comment

Problem #26: If f(z) is an entire function that is not a polynomial, prove that, given arbitrary C > 0, R > 0, and integer m > 0, there exists a z such that  |z|> R and |f(z)| > C|z|^m.

The way this is worded makes my head hurt, so let’s formulate it equivalently as Not B implies Not A instead of A implies B.
Reformulation: Let f(z) be an entire function. Suppose that there exists a C>0R > 0, and an integer m>0  such that |f(z)|\le C|z|^m whenever |z|>R. Prove that f is a polynomial.

Proof: By the given and a limiting process, |f(z)|\le CR^m for |z|=R. By the maximum modulus principle, the same fact is true for |z|\le R. By the Cauchy-Lagrange inequalities if |z|=\frac{R}{2} , we have |f^{(m)}(z)|\le \frac{m! M}{(R/2)^m}, where M is the maximum value of |f(w)| for |w-z|=\frac{R}{2}. Since this circle is completely contained in the disk \{ w:|w|\le R\}, M\le CR^m, and so we have |f^{(m)}(z)|\le (m!) 2^m C for |z|=\frac{R}{2} and thus for |z|\le \frac{R}{2}. By Liouville’s theorem, f^{(m)}(z) is a constant, and thus f is a polynomial.

Categories: Complex questions

Complex #30: sin(z)=z^2 solutions

June 7, 2011 Leave a comment

One of the problems we talked about was the proof of the fact that the equation \sin(z)=z^2 has an infinite number of complex number solutions. I thought of a couple of ways to do this. Here is one way.

Let g(z)=\sin(z)-z^2. In rectangular coordinates, the function is
g(x+iy)=\left[\sin(x)\cosh(y)-x^2+y^2\right]+i \left[\cos(x)\sinh(y)-2xy\right].

For a large positive integer N, consider the rectangle
R=\left\{ z=x+iy: 2\pi \le x \le 2\pi N, 0\le y \le Y\right\}, where Y is a large constant to be chosen conveniently later. We will show that as we go counterclockwise around the boundary of R, the argument of g(z) increases at least by 2\pi (N-2). Then, by the argument principle, the number of zeros inside R is (at least) (N-2) (and there are no poles). So that is all we need for the proof, cause N can be chosen to be arbitrarily large.

Increasing of the argument of g(z):
**On the lower part of the boundary, y=0, and
g(x+i0)=\left[\sin(x)-x^2\right]. For x\ge 2\pi, this is a negative real number, so there is no change in argument as z moves along the lower edge.
**On the left side of the boundary, x=2\pi, and
g(2\pi+iy)=\left[-4\pi^2+y^2\right]+i \left[\sinh(y)-4\pi y\right].
As y moves down from (the large) Y to zero, the real part decreases from positive to negative, and the imaginary part also goes from positive to negative to zero. (Note: you can tell that it is still negative near zero by taking a limit  — use Taylor series.) So this tells us that the argument goes from the first to the third quadrant and ends up at the negative real axis, without looping around too much. Without analyzing further, we have no idea if it g(z) goes above or below the origin, so we don’t know if the argument is increasing or decreasing. Either way, we know the argument increases or decreases by at most \frac{3\pi}{2}.
**On the right side of the boundary, x=2N\pi, and
g(2N\pi+iy)=\left[-4N^2\pi^2+y^2\right]+i \left[\sinh(y)-4N\pi y\right].
As y moves up from  0 to (the large) Y, the real part goes from negative to positive, and the imaginary part starts at zero, gets negative, then eventually goes positive (since \sinh(y) increases exponentially). So in this case, we move from negative real axis to the first quadrant. Again we can conclude that the argument has changed by no more than \frac{3\pi}{2} (plus or minus).
**On the top part of the boundary, y=Y, a large number, and
g(x+iY)=\left[\sin(x)\cosh(Y)-x^2+Y^2\right]+i \left[\cos(x)\sinh(Y)-2xY\right].
As x decreases from 2N\pi to 2\pi, because of the largeness of Y, the graph looks like a small perturbation of the graph of \left[\sin(x)\cosh(Y)\right]+i \left[\cos(x)\sinh(Y)\right], which is an ellipse, so we see that the g(z) goes around the origin counterclockwise (notice \sin and \cos switched from usual roles) (N-1) times.

Thus, the total change in argument \Delta \theta of g(z) as z goes counterclockwise around the boundary of R is bounded by 2\pi(N-1)-3\pi\le \Delta\theta \le 2\pi(N-1)+3\pi, and since
\Delta\theta must be an integral multiple of 2\pi, we must have that the argument of g(z) makes between (N-2) and N revolutions counterclockwise around the origin. By the argument principle, there must be at least (N-2)  zeros inside R.

One minor point: how large do we pick Y? As long as we choose it so that \cosh(Y) is way bigger than (2N\pi)^2+Y^2 and \sinh(Y) is way bigger than 4\pi N Y, we are fine. And we can definitely pick such a Y, since \sinh(Y) and \cosh(Y) increase exponentially in Y.

QED

Categories: Complex questions

Proof of CIFD

June 5, 2011 1 comment

Dear all,

Ken asked me about the proof of the Cauchy Integral Formula for Derivatives that uses the real analysis fact of differentiating under the integral sign.

Statement: If f is holomorphic on a simply connected domain D and \alpha is a counterclockwise oriented Jordan rectifiable curve in D, then for every z inside the interior of \alpha, we have

f^{(n)}(z)=\frac{n!}{2\pi i} \int_{\alpha} \frac{f(w)}{(w-z)^{n+1}} dw.

Proof: For fixed positive integers k and z in the interior of \alpha,  the differential form \frac{f(w)}{(w-z)^{k}} dw has bounded and smooth complex coefficients on \alpha, and likewise all of its derivatives wrt z are smooth and bounded. In particular, the functions mentioned are absolutely integrable over \alpha. Thus, by the theorem on differentiating under the integral sign, any partial derivative of the expression

G(x,y)=\int_{\alpha} \frac{f(w)}{(w-(x+iy))^{k}} dw.

may be computed by differentiating under the integral sign. For instance,

\frac{1}{2}\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right) G(x,y)=\frac{1}{2}\int_{\alpha} \left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)\frac{f(w)}{(w-(x+iy))^{k}} dw.

Therefore, by the Cauchy integral formula

f(z)=\frac{1}{2\pi i} \int_{\alpha} \frac{f(w)}{(w-z)} dw,

f^{(n)}(z)=\frac{1}{2\pi i} \int_{\alpha} \frac{d^n}{dz^n}\frac{f(w)}{(w-z)} dw

=\frac{n!}{2\pi i} \int_{\alpha} \frac{f(w)}{(w-z)^{n+1}} dw.

QED

 

Categories: Complex questions

Complex 3 – Cauchy Ineqs and Schwarz Lemma

May 29, 2011 1 comment

Hi all – Ken U. asked me about Complex #3. I quote it here:

Suppose a and b are positive.

(a) Prove that the only entire functions f for which |f(z)|\le a|z|^{1/2}+b for all z are constant.

(b) What can you prove if |f(z)|\le a|z|^{5/2} +b for all z.

Solution: well, there are many ways to do this. Here’s one way.

Suppose you have such an  f in (a). By the Cauchy Inequalities, |f'(z)|\le \frac{M}{R}, where M is the maximum of f(w) on a disk of radius R around z. If |z|=r\ge 0, choose R=r+1, and the given implies that |f'(z)|\le\frac {a (2r+1)^{1/2}+b}{ r+1}  (since 2r+1 is the biggest possible distance from the origin on that circle of radius r+1 around z). Note that the inequality is true for |z|=r, but by the Maximum Modulus Principle, it is also true for all z such that |z|\le r. The RHS of this inequality is bounded above by a constant for 0\le r<\infty (see powers of r), so Liouville’s Theorem implies that f'(z) is a constant. By taking a limit as r\to\infty, we see that that constant must be zero.  Thus, f'(z)=0, and f is a constant function.

Part (b) is similar. Use the same argument, except use the Cauchy inequality for the third derivative: |f'''(z)|\le \frac{(3!)M}{R^3}. Then you get |f'''(z)|\le\frac {6a (2r+1)^{5/2}+6b}{ (r+1)^3} using the same circles. The same argument shows that f'''(z)=0, and so f is a polynomial of degree two.

Categories: Complex questions