BRIEF SKETCHES OF QUIZ SOLUTIONS

Brief Sketches of Quiz Solutions

Week 1
No quiz.

- an ellipse with foci at z=1 and z=-1 and major axis of length 7.
- the unit circle.
Since 1-i has argument -Pi/4 and modulus square_root(2), and -square_root(3)+i has argument 5*Pi/6 and modulus 2, it follows that arg(z) = 5*Pi/6 - Pi/4 = 7*Pi/12 + 2*Pi*k and that |z| = 2*square_root(2). The polar form is z = (2*square_root(2))*e^(7*Pi*i/12).
e^(e^i) = e^(cos(1)+i*sin(1)) = e^(cos(1))*(cos(sin(1))+i*sin(sin(1))) = (e^(cos(1))*cos(sin(1)))+i*(e^(cos(1))*sin(sin(1))).
Since z=0 is not a solution, we can divide by z and get that ((z+1)/z)^5 = 1. This shows that (z+1)/z is a 5-th root of 1. The possible 5-th roots of 1 are of the form r = e^((2*Pi*k*i)/5) (k = 0, 1, 2, 3, 4). If r is one of these roots, then r = (z+1)/z = 1 + (1/z). Solving for z we get z = 1/(r-1). This excludes the possibility that r = 1, so we can only use 5-th roots of 1 that are different from 1. The four solutions of the equation are therefore z = 1/(e^((2*Pi*k*i)/5)-1) for k = 1, 2, 3, 4.

Suppose that U and V are open. We must show that every point of I = (U intersect V) is an interior point of I. If p is any point in I, then p is in U, so there is an epsilon-nhood of p that lies in U. A similar argument shows that there is a delta-nhood of p that lies in V. If gamma = min(delta,epsilon), then the gamma-nhood of p lies in both U and V, so it lies in (U intersect V) = I. This proves that p is an interior point of I.
- (x/(x^2+y^2)) - i(y/(x^2+y^2))
- (e^x*cos(y)) - i(e^x*sin(y))
Yes. It is entire since the composition of entire functions is entire, and e^(e^z) is the composition of e^z with itself.
The derivative is u_x + iv_x = 2y - 2ix.

(Notation: write z^* for the complex conjugate of z.) If z = x+iy, and f(z) = f(x+iy) = u(x,y)+iv(x,y), then f(z^*) = f(x-iy) = u(x,-y) + iv(x,-y). Hence f(z^*)^* = u(x,-y) - iv(x,-y). Thus U(x,y) = u(x,-y) and V(x,y) = -v(x,-y).
Now U_x(x,y) = u_x(x,-y), U_y(x,y) = -u_y(x,-y) (chain rule!), V_x(x,y) = -v_x(x,-y), and V_y(x,y) = -(-v_y(x,-y)) = v_y(x,-y) (chain rule!). If u and v satisfy the CR equations, then U_x(x,y) = u_x(x,-y) = v_y(x,-y) = V_y(x,y) and similarly U_y(x,y) = -u_y(x,-y) = v_x(x,-y) = -V_x(x,y).
v = (3/2)x^2y^2-x^4/4-y^4/4+C. f' = u_x+iv_x = u_x-iu_y = (3x^2y-y^3) + i(3xy^2-x^3).

cosh(z) = cos(iz), so p is a period of cosh(z) if and only if iz is a period of cos(z). Since the periods of cos(z) are the integer multiples of 2*Pi, it follows that the periods of cosh(z) are the integer multiples of 2*Pi*i.
Using the angle addition formula and the identities cos(iz) = cosh(z) and sin(iz) = i*sinh(z), we get sin(1+i) = sin(1)*cos(i) + cos(1)*sin(i) = [sin(1)*cosh(1)] + i*[cos(1)*sinh(1)].
Suppose that F(z) is an entire extension of f(x) = |x^3|. Then F(z) agrees with the entire function G1(z) = z^3 on the positive real line. It follows that F(z) = G1(z) = z^3 in the complex plane. The same argument using G2(z) = -z^3 and the negative real line forces F(z) = -z^3. But we cannot have F(z) = z^3 and also F(z) = -z^3, so there is no entire function F extending f.

No quiz.

No quiz.

The first segment can be parameterized by z(t) = z_0 + a*t, t\in [0,1]. The second segment can be parameterized by z(t) = z_0 + a + i*b*(t-1), t\in [1,2]. The other two segments are similar.
The answer is 2*i*a*b.
One possible parameterization is z(t) = cos(t) + i*sin(t). (Another is z(t) = e^(it).) The answer is Pi*i/2.

Since [sinh(z)]' = cosh(z), the integral is sinh(2*Pi + 2*Pi*i) - sinh(0) = sinh(2*Pi + 2*Pi*i). Since sinh(z) is periodic of period 2*Pi*i, this is just sinh(2*Pi), which is real.
For any circle C in the plane not containing 0, there is a branch of z^(1/2) that is analytic on and inside C.
Int_C f(z) dz = Int_C (u(x(t),y(t)) + i*v(x(t),y(t)))*(x'(t)+i*y'(t)) dt = Int_C (u dx - v dy) + i* Int_C (v dx + u dy). When f(z) = conjugate(z), then u = x and v = -y, so this integral reduces to Int_C (x dx + y dy) + i* Int_C (-y dx + x dy). By Green's Theorem this is just (Int Int_R 0 dA) + i* (Int Int_R 2 dA) = 2*i*Area(R). Thus, Area(R) = (1/(2*i)) Int_C conjugate(z) dz.

No quiz.

No quiz.

tan(z) = sin(z)/cos(z) = (z - (z^3/6) + (z^5/120) - ...)/(1 - (z^2/2) + (z^4/24) - ...) = (z + (z/3) + (2z/15) + ...). The radius of convergence is the distance from zero (the center) to the nearest point where tan(z) fails to be analytic, which is z=Pi/2. Thus the radius of convergence is Pi/2.
If we multiply the integral by 1/(2*Pi*i), then we have the integral that calculates the (n-1)-Taylor coefficient, which in this problem is 1/(n-1)^(n-1). Therefore the integral is 2*Pi*i/(n-1)^(n-1).

... + (3/z^4) + (2/z^3) + (1/z^2) - (1/2) - (z/4) - (z^2/8) - (z^3/16) - ...
Suppose that f(z) is the function defined by the first series and g(z) is the function defined by the second series. Suppose that the first series has a radius of convergence equal to R and the second has a radius of convergence equal to S where S is greater than R. From the form of the two series, we have f'(z)=g(z) on the disk of radius R centered at zero. But since g is analytic on a disk of radius S centered at zero, and disks are simply connected, g has an antiderivative G on this larger disk. We may assume that G(0) = f(0). Since f is also an antiderivative for g on the smaller disk, and f(0) = G(0), it must be that f(z)=G(z) on the smaller disk. G has a Taylor series centered at zero that converges on the disk of radius S (since G is analytic n this larger disk), but this Taylor series is computed by taking derivatives at zero, so it will be the same as the Taylor series for f. Thus, it follows that the series for f IS the Taylor series for G on the larger disk, so the series for f must converge on the larger disk. This contradicts the assumption that the series for f has radius of convergence R.

See Example 2 in Section 6.1.
This is only slightly different from Example 2 in Section 6.3.
Extra Credit: The answer is (1/2)*Integral[0,2*Pi] cos(2*cos(t)) dt = (1/2)*Integral_{|z|=1} (cos(z+(1/z))/(i*z)) dz = Pi*(Res(cos(z+(1/z))/z,0) = Pi*(the constant term of the Laurent series for cos(z+(1/z)) at 0) = Pi*(the constant term of 1 - (z+(1/z))^2/2! + (z+(1/z))^4/4! - (z+(1/z))^6/6! + ...) = Pi*(1 - 1/(2!)^2 + 1/(3!)^2 - 1/(4!)^2 + ...)

Last modified on Aug 23, 2002