MTH 3102 Complex Variables Final Exam May 1, :30pm-5:30pm, Skurla Hall, Room 106

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1 Name (Last name, First name): MTH 02 omplex Variables Final Exam May, 207 :0pm-5:0pm, Skurla Hall, Room 06 Exam Instructions: You have hour & 50 minutes to complete the exam. There are a total of problems. You must show your work. Partial credit may be given even for incomplete problems as long as you show your work.. Find the value of the contour integral of g (z) = z2 in the positive sense. +z around the circle z = 2 Solution By auchy s Residue Theorem, since g is analytic on and inside except for the simple pole at z =, z 2 ( ) z 2 dz = 2πi Res = 2πi z 2 + z z= + z z= = 2πi.

2 2. In each case, determine all the isolated singularites and their type, i.e., either removable singularity, essential singularity, or a pole. If it is a pole then determine the order m of that pole and the corresponding residue B: (a) Log z cosh z ; (b) z z. Solution 2 (a) The function f (z) = Log z z has an isolated singularity only at z = and since Log = 0 and z has a simple pole this implies that f (z) has a removable singularity and in particular Log z lim z z = lim Log z Log z z = d dz z= Log z = z = z= which implies by Riemann s theorem it is a removable singularity. (b) We have cosh z = (2n)! z2n ( z < ) n=0 so that f (z) = cosh z z = = + (0 < z < ) 2! z n=(2n)! z2n implies f (z) = cosh z z has a pole of order m = with residue B = 2. 2

3 . Prove the following statement: If a function f (z) is analytic in a deleted (punctured) neighborhood of z = z 0 and there exists a constant c > 0 such that Re f (z) c for all z in this deleted neighborhood then the function g (z) = e f(z) has a removable singularity at z = z 0. Proof. By hypothesis, the composition g (z) = e f(z) has an isolated singularity at z = z 0 and is bounded in a punctured neighbhood of this point since g (z) = e f(z) = e Re f(z) e c in this punctured neighborhood. Therefore, by Riemann s theorem (Sec. 84) the function g (z) = e f(z) has a removable singularity at z = z 0.

4 4. Using residues, derive the integration formula 0 (x 2 + ) 2 dx = π 4. Solution Let f (z) = (z 2 +) 2 = (z+i) 2 (z i) 2. Then the only singularities are the poles z = ±i each of order 2. The contour = L R + R, R > is a simple closed contour, where L R : z = z (x) = x, R x R, R : z = z (θ) = Re iθ, 0 θ π, and f is analytic on and inside except at the pole z = i. Thus, by auchy s Residue Theorem f (z) dz + f (z) dz = f (z) dz = 2πi Res f (z) = 2πi d L R R z=i dz (z + i) 2 z=i 2 = 2πi (z + i) = 2πi 2 z=i (2i) = 2πi 2 2 ( i) = π 2. And since f (z) ( z 2 ) 2 = (R 2 ) 2 for all z on R then and hence P.V. f (z) dz R πr (R 2 ) 2 0 as R R (x 2 2 dx = lim + ) R + R (x 2 2 dx = lim + ) = π 2 lim f (z) dz = π R + R 2. R + L R f (z) dz Therefore, since f (x) = (x 2 +) 2, x R is an even function, i.e., f ( x) = f (x) for all x R, we know that P.V. (x 2 + ) 2 dx = (x 2 + ) 2 dx = 2 (x 2 + ) 2 dx 0 and so 0 x 2 + dx = π 4. 4

5 5. Using residues and Jordan s lemma, calculate the improper integral where a > 0 is a constant. Solution 4 As f (x) = integral in which case cos (ax) x 2 + dx cos ax x 2 +, x R is an even function then the improper cos ax x 2 + dx exists if and only if the principal value P.V. since g (x) = cos ax x 2 + dx = P.V. [ cos ax x 2 dx = Re P.V. + sin ax x 2 +, x R is odd so that P.V. e iax x 2 + dx cos ax x 2 + dx, and sin ax x 2 + dx = 0 and eiax = cos ax+ i sin ax, x R. Using Jordan s lemma (Sec. 88), the function h (z) = z 2 + is analytic at all points in the upper half plane y 0 that are exterior to the circle z = R 0 = and on the semicircle R : z = z (θ) = Re iθ, 0 θ π (R > R 0 ) we have h (z) R 2 = M R and M R 0 as R so that by Jordan s lemma lim h (z) e iaz dz = 0. R + R By the auchy s Residue Theorem, h (z) e iaz dz+ h (z) e iaz dz = h (z) e iaz dz = 2πi Res h (z) L R R z=i eiaz = 2πi eiaz z + i = πe a, z=i where = L R + R is the simple closed contour with Therefore, P.V. e iax x 2 + dx = L R : z = z (x) = x, R x R. lim R + R + L R h (z) e iaz dz = πe a lim ] R h (z) e iaz dz = πe a and so [ cos (ax) x 2 dx = Re P.V. + e iax x 2 + dx ] = πe a. 5

6 6. Let denote the unit circle z = 2 with positive orientation. Determine the winding number 2π arg f (z) when and compute the contour integral f (z) = cos z z f (z) f (z) dz. Solution 5 The meromorphic function f (z) is analytic in (including on ) except at the pole z = 0 of order m =. And has two zeros each of order at z = ± π 2 inside, but none on. The number of zeros Z and poles P, counting multiplicities, of f (z) inside is Z = 2, P =. Therefore by the argument principle (Sec. 9) we have 2πi f (z) f (z) dz = 2π arg f (z) = Z P = so that f (z) dz = 2πi. f (z) 6

7 7. Using Rouché s Theorem, determine the number of zeros, counting multiplicities, of the polynomial inside the circle z = 2. z 4 2z + 9z 2 + z Solution 6 Let f (z) = 9z 2 and g (z) = z 4 2z + z. Then f and g are analytic on and inside the circle : z = 2 with f (z) = 9 z 2 = 6 > 5 = = z 4 + 2z + z + g (z) and so by Rouché s Theorem f (z) = 9z 2 and f (z)+g (z) = z 4 2z +9z 2 +z have the same number of zeros, counting multiplicities, inside the circle z = 2. And since f (z) = 9z 2 has only the zero z = 0 of multiplicity 2 inside the circle z = 2 then (z) + g (z) = z 4 2z + 9z 2 + z has 2 zeros inside the circle z = 2, counting multiplicities. 7

8 8. Find all points z where the derivative f (z) exists for the function f (z) = x + i ( y) and calculate this derivative at those point(s). Solution 7 onsider the function f (z) = x + i ( y) on. Here Re f (z) = u (x, y) = x, Im f (z) = v (x, y) = ( y). As the auchy-riemann equations are not satisfied at any points except (x, y) = (0, ) since u x = x 2, u y = 0, v x = 0, v y = ( y) 2, u x = v y and u y = v x if and only if x 2 = ( y) 2 if and only if (x, y) = (0, ). This implies by the Theorem in Sec. 2 that the derivative f (z) exists for the function f (z) = x + i ( y) nowhere except possibly at (x, y) = (0, ). Now (a) the first-order partial derivatives of the functions u and v with respect to x and y exist everywhere in the neighborhood of (0, ); (b) those partial derivatives are continuous at (0, ) and satisfy the auchy-riemann equations at (0, ). Thus, by the Theorem in Sec. 2 the derivative of f (i) exists and f (i) = u x (0, ) + iv x (0, ) = 0. Therefore, the derivative f (z) of the function f (z) = x + i ( y) doesn t exist anywhere in except at z = i whose derivative there exists and its value is f (i) = 0. 8

9 9. Evaluate the contour integral tan zdz where denotes any closed contour lying in the open disk z < π 2. Solution 8 The function f (z) = tan z = sin z cos z is analytic on and inside any closed contour lying in the open disk z < π 2 (as cos z has zeros π 2 + 2nπ all with π 2 + 2nπ π 2 for n = 0, ±, ±2,...) so that by the auchy-goursat theorem (Sec. 50) tan zdz = f (z) dz = 0. 9

10 0. Find all roots of the equation z + 8 = 0 that lie in the right-half of the complex plane. Solution 9 All the roots of the polynomial f (z) = z + 8 are z k = ( 8) / = = 8e i( π + 2kπ ) = 2e i( π + 2kπ ( = 2 cos ) (8e i(π+2kπ)) / ( π + 2kπ ) ( π + i sin + 2kπ )), k = 0,, 2, which form the vertices of a triangle centered at the origin. Those vertices that lie in the right-half of the complex plane are ( ( π ) ( π )) z 0 = 2 cos + i sin, ( ( ) ( )) 5π 5π z 2 = 2 cos + i sin ( ( π ) ( π )) = z 0 = 2 cos i sin. 0

11 . Prove that the function u (x, y) = e y sin x is a harmonic function on R 2. Proof. First proof: The entire function f (z) = e iz = e y e ix has real part Re f (z) = e y sin x = u (x, y) which implies u (x, y) is a harmonic function on R 2. Second proof: The function u (x, y) = e y sin x on R 2 has continuous partial derivatives of the first and second order and satisfies Laplace s equation, i.e., u xx + u yy = ( e y sin x ) xx + ( e y sin x ) yy = e y (cos x) x + ( e y) y sin x = e y sin x + e y sin x = 0. This proves u (x, y) is a harmonic function on R 2.