Test 2 - Sp 08

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(Take home test.)

  • 1A Show that f(x)=\bar{z} is not analytic.

Just check that the Cauchy Riemann equations don't hold for f(z)= x -iy. Alternately you could take the partial derivatives df / dx = 1, and df / dy = - 1, and since these are not equal, the function is not analytic.

  • 1B Evaluate \int_{|z|=1} \bar{z}^2 dz.

The easiest way is to parameterize z = eit, and get \int_0^{2 \pi} e^{-2 i t} i e^{it} dt=\int_0^{2 \pi} e^{-i t} dt = 0 since the complex exponential is periodic.

  • 2 Prove, or find a counter-example: |\cos(z)| \le 1, for all complex numbers z.

First note if the statement were true, then sinz would be a bounded, entire function (Liouville's Theorem) and hence constant. So we need a counter-example. If z= -i w, for w complex, then  \sin(z) = \frac{e^{iz} - e^{-iz}}{2}= \frac{ e^{w} - e^{-w}}{2} = \sinh{w}. Since the real-valued sinh function is not bounded, neither is the complex sine function. Alternately, plug in something along the imaginary axis and show the value is larger than one:  |\sin{2i}| = |\frac{e^{-2} - e^{2}}{2}|\le (|e^2|-|e^{-2}|)/2 > (4-1)/2 >1.

Keep in mind that just because something is true for he real numbers does not mean it will be true of the complex numbers. There is a theorem that says (roughly) that if analytic functions satisfy an algebraic identity for all real numbers, then this identity will be true for the complex extension as well. So, for example, since sin(2x) = 2sinxcosx, for real numbers x, the same identity is true for complex values.

  • 3 Prove, or find a counter-example: Cauchy's theorem hold seperately for the real and imaginary parts of analytic functions f.

The easiest way to do this is to realize that whenever we have a + bi = c + di then a=c and b=d. In our case we have \int f(z) dz =0, and so both the real and imaginary parts of the integral must be equal to zero.

Alternately, the real and imaginary parts of f are harmonic, and we showed that a version of Cauchy's theorem holds for harmonic functions in chapter 1.

  • 4 For entire functions f, evaluate I(k)=\int_0^{2 \pi} f(z_0+r e^{i \theta} )e^{ik\theta} d\theta,, for  k=0,\pm 1, \pm 2, \cdots.

We saw a problem somewhat like this earlier, in the section on mean values - I think Marla worked the problem in class (though there were others assigned.) The related problem was from parameterizing an integral as follows  \int_{|z-z_0|=r} \frac{f(z)}{z-z_0} dz = \int_0^{2 \pi}  \frac{ f(z_0+r e^{i \theta}) }{r e^{i \theta} } i r e^{i \theta} d \theta = i \int_0^{2 \pi} f(z_0 + r e^{i \theta}) d \theta. So, I(0)= -2 \pi f^{\prime}(z_0), by Cauchy's integral theorem.

Now consider  \int_{|z-z_0|=r} \frac{f(z)}{(z-z_0)^{k+1}} = \frac{2 \pi i f^{(k)}(z_0)}{k!}, by the Cauchy integral theorem. But, parameterizing we see that it is equal to  i/r^k \int f e^{-ik\theta}=i/r^k I(-k).. thus I(-k)= -i r^k \frac{2 \pi i}{k!} f^{(k)}(z_0) = \frac{ 2 \pi r^k}{k!} f^{(k)}(z_0).

Now for the positive values of k. In this case show that if we parameterize \int f(z) (z-z_0)^k dz we get constants x I(k-1), and since the integrand is analytic, the value of I(n), n positive, is zero.

  • Prove | \int_{|z|=1}  \frac{\sin z}{z} dz | \le 2 \pi e.

Some of you applied Cauchy's integral formula to evaluate the integral to icos(0) with absolute value 2 \pi. this is better than the desired result.

If, instead you apply the M-L inequality, you get the exponential: |\frac{\sin z}{z}|=|\sin z| = |\frac{e^{iz} - e^{- i z}}{2} | \le 1/2 (|e^{iz}|+|e^{-iz}|) \le e, where both inequalites are due to the fact that |z|=1.

  • 7a Evaluate  \int \frac{dz}{z^2-1} dz.
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