Answers and Comments on Test 1

General Remark:

Most of you did a good job on the take-home part. However, some of you were sloppy and careless in your writeup. Next time, I will take off more points for this. Since you have unlimited time, your explanations should be neat, clear, and complete.

In-class part

  1. (a) The given number has polar form 27*exp(i*3*pi/4).

    (b) One root is 3*exp(i*pi/4). The others are 3*exp(i*11*pi/12) and 3*exp(i*19*pi/12).

    (c) One root is in the first quadrant, one in the second, and the third is in the fourth.

  2. (a) 1/3*(exp(3*z)+z3) is one antiderivative.

    (b) 1 + 3*z + 11/2*z2 + 9/2*z3 + ... + (3n/(n!))*zn + ...

    (c) Use the Fundamental Theorem of Calculus and your answer to (a). This gives (1/3)*(exp(3*i) - i - 1) .

    (d) Use the generalized Cauchy Integral Formula. Let f be given by

    f(z) = exp(3*z) + z2.

    Then the integral is (pi*i)*f ''(i) = (pi*i)*(9*exp(3*i) + 2).

  3. (a) No such function exists. If f is an entire function such that Re(f(z)) is positive for all z, then exp(-f(z)) is an entire function that is bounded everywhere by 1. By Liouville's Theorem, this function must be constant, so f must be constant as well. Note: This problem is not well-phrased, since a true response required an example and a false response a proof. For this reason, I was somewhat more lenient with partial credit here than in the other parts.

    (b) This is false. The function f(z) = 1/z satisfies the condition.

    (c) This is true. If f is such a function, then its Taylor series expansion about i is identically zero. Since f is entire, this series represents the function for all z, i.e., f is the identically zero function.

    (d) This is true. Suppose f is such a function. Define g by

    g(z) = f(z) - z2.

    Say g = u + iv is the decomposition of g into real and imaginary parts. We know u is identically zero. So by the Cauchy-Riemann equations, v has zero partial derivatives and hence must be constant. Note that the square function itself is not a counterexample; that function is the case where c = 0.

Takehome

The image of the square is the region bounded between the two parabolas

y2 = 4*(x + 1)       and       y2 = 4*(1 - x)

These equations can be obtained by looking at the images of the boundary lines. For example,

(1 + it)2 = 1 - t2 + 2*t*i

So a parametric representation of the image of the right vertical edge of the square is

x(t) = 1 - t2       and       y(t) = 2*t.

Thus, y = t/2. Substituting this in the equation for x, we have

y2 = 4*(1 - x)

Your sketch should have clearly indicated the maximum y-value of 2 and the maximum x-value of 1.

Lawrence C. Moore < lang@math.duke.edu>

Last modified: March 2, 1999