= 2πi Res. z=0 z (1 z) z 5. z=0. = 2πi 4 5z

Transcription

1 MTH30 Spring 07 HW Assignment 7: From [B4]: hap. 6: Sec. 77, #3, 7; Sec. 79, #, (a); Sec. 8, #, 3, 5, Sec. 83, #5,,. The due date for this assignment is 04/5/7. Sec. 77, #3. In the example in Sec. 76, two residues were used to evaluate the integral 4 5 ( ) d, where is the positively oriented circle. Evaluate this integral once again by using the theorem in Sec. 77 and finding only one residue. Solution. We have f () 4 5 ( ) 5 + which analytic everywhere in except for the isolated singular points 0, which are interior to the positively oriented simple close contour : (θ) e iθ. Therefore, by the theorem in Sec. 77 we have [ ( )] 4 5 d πi ( ) 0 f [ ( 4 ) ] 5 πi ( 0 ) (( ) ) [ ] 4 5 πi 0 ( ) πi 4 5 8πi. 0 Sec. 77, #7. Let the degrees of the polynomials P () a a n n (a n 0), Q () b b m m (b m 0) be such that m n +. Use the theorem in Sec. 77 to show that if all of the eros of Q () are interior to a simple closed contour, then [ompare with Exercise 4(b).] P () d 0. Q ()

2 Proof. Using the theorem in Sec. 77 we have [ ( P () P ) ] d πi Q () 0 Q ( ) [ ( a ) n ] n + + a0 πi ( 0 b ) m m + + b0 [ a n n ] + + a 0 πi 0 b m m + + b 0 [ an n +m + + a 0 m ] πi 0 b m + + b 0 m since 0 a n n +m + + a 0 m b m + + b 0 m is analytic at 0 because by hypothesis m n+ and b m 0. This completes the proof. Sec. 79, #. In each case, determine all the isolated singularites and their type, i.e., either removable singularity, essential singularity, or a pole: (a) e ( ) ; (b) + ; (c) sin cos ; (d) ; (e) ( ) 3. Solution. (a) The function f () e has the convergent Taylor series representation f () e n! n ( < ) n0 so that the function g () f ( ) e ( ) has only one isolated singularity at 0 with the convergent Laurent series representation ( ) g () e ( ) f n! n+ (0 < < ) which has an infinite number of nonero coeffi cients for the terms n, n > 0 implying that 0 is an essential singular point of g () e ( ). (b) The function f () + has the convergent Taylor series representation in powers of + : f () n0 ( + ) + + ( + ) ( + ) ( ) + ( + ) (0 < + ) +

3 implying it has a simple pole at. (c) The function f () sin has the convergent Taylor series representation f () sin ( ) n (n + )! n+ ( < ) n0 so that the function g () f() sin has only at isolated singularity at 0 with the convergent Laurent series representation g () sin f () ( ) n (n + )! n (0 < < ) n0 which implies that 0 is a removable singularity of g () sin. (d) The function f () cos has the convergent Taylor series representation f () cos (n)! n ( < ) n0 so that the function g () f() cos has only one isolated singularity at 0 with the convergent Laurent series representation g () cos f () n0(n)! n (0 < < ) + n(n)! n which implies that 0 is a pole of order m (i.e., a simple pole). (e) The function f () is already in Laurent series form so we see ( ) 3 ( ) 3 that it has a pole at of order m 3. Sec. 79, #(a). Show that the singular point of each of the following function f () cosh 3 is a pole and then determine the order m of that pole and the residue B. Solution. We have cosh (n)! n ( < ) n0 so that f () cosh 3 (n)! n 3 + (0 < < ) n implying the order of the pole is m and residue B. 3

4 Sec. 8, #. Show that (a) (b) i (a) i /4 + + i ( > 0, 0 < arg < π) ; Log ( + ) π + i ; 8 / ( + ) i 8 ( > 0, 0 < arg < π). Proof. (a) We have /4 e 4 log ( > 0, 0 < arg < π) is analytic at e iπ with ( ) /4 e 4 log( ) e i π + i 4 so that ( > 0, π < arg < π) is analytic at i with deriva- (b) We have Log (+i) tive so that d d (c) We have / /4 + /4 + i. Log ( + i) i i e log (+i) (+i) ( + i) ( + i) Log ( + i) 4 i (i) (i) Log i (i) 4 4i 4i Log i 6 4i 4i π 6 Log ( + ) d d π + i 8 Log ( + i) π + i. 8 i i ( > 0, 0 < arg < π) is analytic at i with 4

5 (i / e log i e i π 4 +i ) derivative so that d / d ( + i) / ( + i) ( + i) / i ( + i) 4 ( ) i / (i) (i) i / (i) 4 i ( +i ) (i) (i) (i) 4 ( ) ( ) +i 4i +i ( ) 6 ( ) i +i 4i 6 ( ) +i ( i) 4i ( + i) 6 ( + i) + ( i + ( i )) 8 ( + i) + ( i + ) i 8 8 / ( + ) d / d ( + i) i 8. i i Sec. 8, #3. In each case, find the order m of the pole and the corresponding residue B at the singularity 0: Solution. (a) We have (a) sinh 4 ; (b) (e ). sinh 4 4 n0(n + )! n+ (n + )! n 3 n (0 < < ) 3! 5

6 so it has a pole of order m 3 and a residue B 3! 6. (b) We have (e ) n! n n (n+)! n n0 (! + ) n! n n + (0 < < ) so it has a pole of order m and a residue B. Sec. 8, #5. Find the value of the integral 3 ( + 4) d taken counterclockwise around (a) ; (b) + 3. Proof. The function f () 3 (+4) has poles at 0 and 4 of orders 3 and with residues that these poles 0 3 ( + 4) d! d ( + 4) 0 d ( + 4) d 4 ( + 4) , 3 ( + 4) 3 4 ( 4) 3 64, respectively. Hence by auchy s idue Theorem we have (b) (a) : : ( + 4) 3 ( + 4) d πi 0 d πi 0 ( πi ( + 4) πi 64 πi 3, 3 ( + 4) ) 0. + πi 4 3 ( + 4) Sec. 83, #5. Let denote the positively oriented circle and evaluate the integrals (a) tan d; (b) sinh() d. 6

7 Solution. (a) The function f () tan sin cos is analytic on and inside except for the simple poles at ± π with residue deduced from Theorem in Sec. 83 [since sin ( ) ( ) ± π ±, cos ± π 0, d d cos ± π sin ( ) ± π (±)] ) ± π tan sin ( ± π d d cos ± π Thus, by auchy s idue Theorem we have (a). tan d πi tan + πi tan 4πi. π π (b) The function g () sinh() is analytic on and inside except for the simple poles at the eros of sinh () which occur at 0, ±i π since 0 sinh () e e ( e e 4 ) if and only if e 4 if and only if inπ log (n 0, ±, ±,...). 4 To calculate its residues at these points we use Theorem in Sec. 83 since (b) d πi sinh () sinh () 0, for 0, ±i π, d sinh () cosh (), d cosh (0), [ ( cosh ±i π )] cosh [iπ] e iπ + e iπ Thus, by auchy s idue Theorem we have [ ] 0 sinh () + i π sinh () + i π sinh () [ ] πi cosh (0) + cosh ( ) i π + cosh ( ) i π [ πi + + ] πi. Sec. 83, #. Recall (Sec. ) that a point 0 is an accumulation point of a set S if each deleted neighborhood of 0 contains at least one point of S. One form of the Bolano-Weierstrass theorem can be stated as follows: an infinite set of points lying in a closed and bounded region R has at least one accumulation 7

8 point in R. Use that theorem and Theorem in Sec. 8 to show that if a function f is analytic in the region R consisting of all points inside and on a simple closed contour, except possibly for poles inside, and if all the eros of f in R are interior to and are of finite order, then those eros must be finite in number. Proof. The region R described by the hypothesis must be closed and bounded (see Jordan curve theorem, the interior to is an open set and its boundary is hence R is a closed set which is also bounded). First, there must be only a finite number of eros of f inside for otherwise there would be an accumulation point 0 of these eros in R by the Bolano-Weierstrass theorem. This point 0 cannot be a pole of f by Theorem 4 in Sec. 84 and hence must be a point of analyticity of f and hence a ero of f. alculating all orders of derivatives of f along the sequences of eros of f that converge to 0, we find that all the derivatives of f are ero at 0 and so using the convergent power series representation of f in a neighborhood of 0 (since f is analytic at 0 ) we find that f is ero in a neighborhood of 0 implying that it is ero on all of R. This completes the proof. Sec. 83, #. Let R denote the region consisting of all points inside and on a simple closed contour. Use the Bolano-Weierstrass theorem and the fact that poles are isolated singular points to show that if f is analytic in the region R except for poles interior to, then those poles must be finite in number. Proof. The region R described by the hypothesis must be closed and bounded (see Jordan curve theorem, the interior to is an open set and its boundary is hence R is a closed set which is also bounded). Now there must be only a finite number of poles of f inside for otherwise there would be an accumulation point 0 of these poles in R by the Bolano-Weierstrass theorem, but then 0 cannot be an isolated singularity of f, a contradiction. 8