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 32 Complex Variables Final Exam May, 27 3:3pm-5:3pm, Skurla Hall, Room 6 Exam Instructions: You have hour & 5 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 () = cos 2 ( 2 +) around the circle = 2 in the positive sense. [Hint: Use Cauchy s Residue Theorem] Solution The function g () = cos 2 ( 2 + ) = cos 2 ( + i) ( i) is analytic on = 2 and inside it except for the isolated singularities at = and = ±i, which are poles of order m = 2 and m =, respectively, with residues (see Theorem in Sec. 8) Res s= g () = [ ] cos ( 2 = + ) = = i = i 2 2 [ cos Res s= i g () = 2 = ( i)] = i ( (sin ) 2 + ) (cos ) 2 ( 2 + ) 2 = =, [ cos Res s=i g () = 2 = ( + i)] =i cos i i 2 2i = i 2 cos i (recall cos = ei + e i ) 2 ) ( (e i2 + e i2 e + ), e cos ( i) i 2 2i cos ( i) = i 2 2i Therefore by Cauchy s Residue Theorem (Sec. 76) we have = i ( e + ). e cos C 2 ( 2 + ) d = 2πi (Res s=g () + Res s=i g () + Res s= i g ()) [ = 2πi i ( e + ) + i ( e + )] e e =.

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) e ( ) ; (b) sin ; (c) cos e2 ; (d). (a) The function f () = e has the convergent Taylor series representation f () = e = n! n ( < ) n= so that the function g () = f ( ) = e ( ) has only one isolated singularity at = with the convergent Laurent series representation ( ) g () = e ( ) = f = n! n+ ( < < ) which has an infinite number of nonero coeffi cients for the terms n, n > implying that = is an essential singular point of g () = e ( ). (b) The function f () = sin has the convergent Taylor series representation n= f () = sin = ( ) n (2n + )! 2n+ ( < ) n= so that the function g () = f() = sin has only at isolated singularity at = with the convergent Laurent series representation g () = sin = f () = ( ) n (2n + )! 2n ( < < ) n= which implies that = is a removable singularity of g () = sin. (c) The function f () = cos has the convergent Taylor series representation f () = cos = (2n)! 2n ( < ) n= so that the function g () = f() = cos has only one isolated singularity at = with the convergent Laurent series representation g () = cos = f () = n=(2n)! 2n ( < < ) = + n=(2n)! 2n which implies that = is a pole of order m = (i.e., a simple pole) with residue B =. 2

3 (d) The function f () = e 2 has the convergent Taylor series representation f () = e 2 = n! (2)n ( < ) = n= 2 n n n! n= so that the function g () = f() = e2 has only one isolated singularity at = with the convergent Laurent series representation g () = e2 = ! = f () = n= ! 2 n n ( < < ) n! + n= 2 n n n! which implies that = is a pole of order m = 3 with residue B = 23 3! = 3. 3

4 3. Prove the following statement: If a function f () is analytic in a deleted (punctured) neighborhood of = and there exists constants c > such that c f () for all in this deleted neighborhood then f() has a removable singularity at =. Proof. By hypothesis the function g () = f() is analytic in a deleted (punctured) neighborhood of = and there exists constants c > such that g () c for all in this deleted neighborhood. Therefore, by Riemann s theorem (Sec. 8) g () has a removable singularity at =. This completes the proof.

5 . Using residues, derive the integration formula x 2 + dx = π 2. Solution 2 Let f () = 2 + = (+i)( i). Then the only singularities are the simple poles = ±i. The contour C = L R +, R > is a simple closed contour, where L R : = (x) = x, R x R, : = (θ) = Re iθ, θ π, and f is analytic on C and inside C except at the pole = i. Thus, by Cauchy s Residue Theorem f () d + f () d = f () d = 2πi Res f () = 2πi L R C =i + i = π. =i And since f () = 2 R 2 for all on R then f () d πr R 2 as R and hence P.V. Therefore, since f (x) = for all x R, we know that R x 2 + dx = lim R x 2 + dx = lim = π lim f () d = π. P.V. x 2 + L R f () d, x R is an even function, i.e., f ( x) = f (x) x 2 + dx = x 2 + dx = 2 x 2 + dx and so x 2 + dx = π 2. 5

6 5. Using residues and Jordan s lemma, derive the integration formula cos x x 2 + dx = π 2 e. Solution 3 As f (x) = cos x x 2 +, x R is an even function then the improper cos x cos x integral x 2 +dx exists if and only if the improper integral x 2 +dx exists if cos x and only if the principal value P.V. dx, and in which case cos x x 2 + dx = 2 x 2 + cos x x 2 + dx = 2 P.V. = [P.V. 2 Re e ix x 2 + dx since g (x) = sin x x 2 +, x R is odd so that P.V. ] sin x cos x x 2 + dx x 2 + dx = and eix = cos x + i sin x, x R. Using Jordan s lemma (Sec. 88), the function h () = 2 + is analytic at all points in the upper half plane y that are exterior to the circle = R = and on the semicircle : = (θ) = Re iθ, θ π (R > R ) we have h () R 2 = M R and M R as R so that by Jordan s lemma lim h () e i d =. By the Cauchy s Residue Theorem, h () e i d+ L R h () e i d = C h () e i d = 2πi Res =i h () ei = 2πi where C = L R + is the simple closed contour with Therefore, P.V. and so e ix x 2 + dx = L R : = (x) = x, R x R. lim L R h () e i d = πe [ cos x x 2 + dx = 2 Re P.V. e ix x 2 + dx lim ] e i + i = πe, =i h () e i d = πe = π 2 e. 6

7 6. Using residues and an indented path, derive the integration formula for the Dirichlet s integral sin x x dx = π 2. Solution Notice that for g () = ei, C\ {} it has a simple pole at = [see problem 2 (b) above] with residue Res g () = B = = and we have Re g (x) = f (x) = sin x, x R\ {}. x By Jordan s lemma we have where lim By the theorem in Sec. 89, we have where e i d =, : = (θ) = Re iθ, θ π (R > ). lim e i C ρ d = B πi = πi, C ρ : = (θ) = ρe iθ, θ π ( < ρ < R). By Cauchy-Goursat theorem, with the simple closed contour C = L + + L 2 + C ρ we have where Therefore, L e i d + L 2 e i d + e i d + C ρ e i d = C πi = L : = (x) = x, ρ x R, L 2 : = (x) = x, R x ρ. lim ρ, = lim ρ, = 2i lim ρ, e i ( e i L ( R d e i C ρ d d + ) e i L 2 d ρ sin x x dx ) = 2i sin x x dx e i d =, 7

8 implying sin x x dx = π 2. 8

9 7. Let C denote the unit circle = with positive orientation. Determine the winding number 2π C arg f () when and compute the contour integral f () = C (2 )7 3 f () f () d. Solution 5 The meromorphic function f () is analytic in C (including on C) except at the pole = of order m = 3. And has a ero of order 3 at = 2. The number of eros Z and poles P, counting multiplicities, of f () inside C is Z = 7, P = 3. Therefore by the argument principle (Sec. 93) we have 2πi C f () f () d = 2π C arg f () = Z P = so that C f () d = 8πi. f () 9

10 8. Using Rouché s Theorem, determine the number of eros, counting multiplicities, of the polynomial inside the circle = Solution 6 Let f () = 6 5 and g () = 3 2. Then f and g are analytic on and inside the circle C : = with f () 5 6 = > + 2 = = g () and so by Rouché s Theorem f () = 6 5 and f ()+g () = have the same number of eros, counting multiplicities, inside the circle =. And since f () = 6 5 = ( 2 5 ) has only the ero = of multiplicity inside the circle = then f () + g () = has eros inside the circle =, counting multiplicities.

11 9. Find all points C where the derivative f () exists for the function f () = 2. Solution 7 As the Cauchy-Riemann equations are not satisfied at any points except (x, y) = (, ) since f () = 2 = x 2 + y 2 + i, u (x, y) = x 2 + y 2, v (x, y) =, u x = 2x, u y = 2y, v x =, v y =, u x = v y and u y = x x if and only if (x, y) = (, ) this implies by the Theorem in Sec. 23 that the derivative f () exists for the function f () = 2 nowhere except possibly at (x, y) = (, ). 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 (, ); (b) those partial derivatives are continuous at (, ) and satisfy the Cauchy-Riemann equations at (, ). Thus, by the Theorem in Sec. 23 the derivative of f () exists and f () = u x (, ) + iv x (, ) =. Therefore, the derivative f () of the function f () = 2 doesn t exist anywhere in C except at =.

12 . Evaluate the contour integral C sin ( 2 + 9) 5 d where C denotes any closed contour lying in the open disk < 2. Solution 8 The function f () = sin is analytic on and inside any closed ( 2 +9) 5 contour C lying in the open disk < 2 so that by the Cauchy-Goursat theorem (Sec. 5) sin C ( 2 + 9) 5 d = f () d =. C 2

13 . Find all roots of the equation + 6 = that lie in the left-half of the complex plane. Solution 9 All the roots of the polynomial f () = + 6 are k = ( 6) / = = 6e i( π + 2kπ ) = 2e i( π + 2kπ ( = 2 cos ) (6e i(π+2kπ)) / ( π + 2kπ ) ( π + i sin + 2kπ )), k =,, 2, 3 which form the vertices of a square centered at the origin. Hence those that lie in the left-half of the complex plane are ( ( ) ( )) 3π 3π = 2 cos + i sin, ( ( ) ( )) 5π 5π 2 = 2 cos + i sin. 3

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

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