BRIEF ANSWERS TO TEST 1

• z = -(1/2), z = -(1/2) - (1/2)i, z = -(1/2) + (1/2)i.
• • False, the inequality is backwards.
  • False. Need to change v_y to v_x.
  • True. If z = r*exp(i*theta), then exp(2*log(z)) = exp(2*ln(r) + 2*i*{theta + 2*Pi}) = exp(2*ln(r))*exp(2*i*{theta + 2*Pi}) = r^2*exp(2*i*theta), and z*z yields the same thing.
  • False. The left and right sides are multiple-valued functions, and for z = 1 the number 2*Pi*i is a value assumed by the left side but not the right side.
• v = -(1/4)x^4 + (3/2)x^2y^2 -(1/4)y^4. [u+iv]' = (3x^2y-y^3) + i(3xy^2-x^3).
• sinh^(-1)(z) = z + (z^2+1)^(1/2). [sinh(z)]' = 1/(z^2+1)^(1/2).
• • The upper half plane.
  • The lower half plane.

BRIEF ANSWERS TO TEST 2

• 2*Pi*i*r^2*z_0.
• Use Gauss's mean value theorem.
• • 0.
  • -2*Pi*i/9
• Assume |z| = 1. Then |1/z| = 1, so |P(1/z)| is less or equal to 1. Thus |Q(z)| = |z|^n*|P(1/z)| is less or equal to 1. But Q(0) = 1. Thus |Q(z)| assumes a max on the interior of the unit circle, which implies that Q(z) is constant. Hence Q(z) = 1, implying all a_i = 0, implying that P(z) = z^n.
• • f(z) is analytic everywhere except possibly z=0 since it is a ratio of analytic functions everywhere, and the denominator is nonzero except at z = 0. At z=0 we have: lim_(z goes to 0) f(z) = derivative of sin(z) at 0 = cos(0) = 1 = f(0), so f is continuous.
  • By Cauchy-Goursat, the integral around any simple closed contour that doesn't loop around z=0 is zero. By the Cauchy Integral Theorem the integral around any simple closed contour that does loop around z=0 is sin(0) = 0. By Morera's Thm, f is entire. (We used part (a) because the hypothesis of Morera's Thm requires that f be continuous.)