MATH 3160: APPLIED COMPLEX VARIABLES FINAL EXAM (VERSION C) 1. Find all solutions z to the equation tanh z = 1 . Solution: There are none. 2. (a) Calculate the Laurent expansion about z = 0 of 1 2 z sin z2 up to 1/z 4 terms. (b) Calculate 2 Res z sin z=0 1 z2 Solution: (a) 2 z sin 1 z2 =1− 1 + ··· 6z 4 (b) 2 Res z sin z=0 3. Calculate the integral Z C 1 z2 =0 cos z dz z (z 2 + 8) where C is the square with corners at the four points ±2 ± 2i. Solution: Only one of the poles is inside the square, so Z cos z cos z π dz = 2πi Res = i. 2 z=0 z (z 2 + 8) 4 C z (z + 8) Date: May 6, 2002. 1 2 MATH 3160: APPLIED COMPLEX VARIABLES FINAL EXAM (VERSION C) 4. Calculate the integral Z |z|=1 cos (1/z) dz . z 2 sin (1/z) Hint: why can’t you just use residues to get it? What happens if you use the coordinate change w = 1/z? Solution: There are infinitely many poles inside the circle, and none outside. Let w = 1/z. Calculate residue at infinity. Z Z cos (1/z) dz cos(w) = dw 2 |w|=1 sin(w) |z|=1 z sin (1/z) = 2πi Res w=0 cos(w) sin(w) = 2πi. 5. Show that the map w= z−1 z+1 1/2 (with the principal branch of the square root) takes the z plane slit along the segment −1 ≤ x ≤ 1 onto the right half plane u > 0 (where z = x+iy and w = u+iv). Hint: it is a composite function. Figure out where the linear fractional transformation Z = (z − 1)/(z + 1) takes the upper and lower half planes, and where it maps the segment −1 ≤ x ≤ 1. Solution: As in figure 5 on the facing page, the map Z = (z − 1)/(z + 1) takes the interval of the real line from z = −1 to z = 1 into the interval of the real line from Z = −∞ to Z = 0. It takes z = ∞ to Z = 1. It takes real numbers to real numbers, and nonreal numbers to nonreal numbers, since we can easily calculate z = −(Z + 1)/(Z − 1). But it takes z = i to Z = (i − 1)/(i + 1) = i, so it must take the upper half plane to the upper half plane, and the lower half plane to the lower half plane. Finally w = Z 1/2 takes the Z plane slit along the negative real axis to the right half plane in the w plane, just dividing angles from the real axis in half. Putting these together, we find the z plane slit along the real axis from −1 to 1 goes to the right half plane. 6. Using conformal mapping, find the harmonic function f which is defined outside of the two circles drawn in figure 6 on page 4 (unfortunately they don’t quite look like circles, because of some MATH 3160: APPLIED COMPLEX VARIABLES FINAL EXAM (VERSION C) 3 z Z w problems with graphics), with f = 0 on the little circle, and f = 1 on the big circle. (The answer looks like figure 6 on page 5.) Show that the isotherms and flow lines are circles, and explain why they are perpendicular to each other. Solution: Using w = 1/z to straighten out the circles, the result is 2 x f= +1 . 3 x2 + y 2 The isotherms are constant lines of f , so (x − c)2 + y 2 = c2 for any constant c while the flow lines are x2 + (y − c)2 = c2 for any constant c. 7. Suppose that f (z) is an entire analytic function, and that the real part of f (z), which we will call u(x, y), has an upper bound: u(x, y) ≤ u0 4 MATH 3160: APPLIED COMPLEX VARIABLES FINAL EXAM (VERSION C) 1 0.5 –1 –0.5 0.5 1 1.5 2 –0.5 –1 for any x, y. Show that f (z) constant. Hint: consider g(z) = ef (z) . Solution: Applying Liouville’s theorem to g(z) we find that |g(z)| = eu is bounded, so g(z) is constant. Taking logarithm of g(z), f (z) must also be constant. MATH 3160: APPLIED COMPLEX VARIABLES FINAL EXAM (VERSION C) 5 1 0.5 0 2 2 1 1 y 0 0 –1 –1 –2 –2 x