Solution for Final Review Problems 1

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1 Solution for Final Review Problems Final time and location: Dec. Gymnasium, Rows 23, 25 5, 2, Wednesday, 9-2am, Main ) Let fz) be the principal branch of z i. a) Find f + i). b) Show that fz )fz 2 ) λfz z 2 ) for all z, z 2, where λ, e 2π or e 2π. Solution. a) f + i) + i) i expi Log + i)) expiln 2 + πi 4 )) e π/4 cos ln 2 2 ) + i sinln 2 2 )) b) We have Argz ) + Argz 2 ) Argz z 2 ) + 2nπ for some integer n. Since π < Argz ) π, π < Argz 2 ) π and π < Argz z 2 ) π, Therefore, 3π < 2nπ < 3π n Logz ) + Logz 2 ) Logz z 2 ) + 2nπi with n {,, } and fz )fz 2 ) fz z 2 ) 2) Do the following: a) Find sin π 3 + i). expi Log z ) expi Log z 2 ) expi Logz z 2 )) expi Log z + i Log z 2 i Logz z 2 )) exp 2nπ) {e 2π,, e 2π } b) Find the Taylor series of sin z) 2 at z. c) Show that sinz) sinh y ) for all z C, where y Imz). xichen/math3f/fpsol.pdf

2 2 d) Let C R denote the semicircle { z i R, Imz) }. Show that dz R C R z 2 sin z Solution. a) π ) sin 3 + i 2i eiπ/3+i) e iπ/3+i) ) 2i e e πi/3 ee πi/3 ) e cos π 2i 3 + i sin π 3 ) ecos π 3 i sin π ) 3 ) 3 e + ) + i e ) 4 e 4 e b) We have ) e sin z) 2 iz e iz 2 2i 4 e2iz + e 2iz 2) 2i) n z n 2i) n z n + 4 n! 4 n! 2 2 2i) 2n z 2n 2 2n)! 2 ) n 2 2n z 2n 2n)! ) n 2 2n z 2n ) n+ 2 2n z 2n 2n)! 2n)! n c) By triangle inequality, sin z e iz e 2iz 2i 2 n e iz e iz e ix y e ix+y 2 2 e y e y sinh y sinh y ) d) When z C R, Imz). Therefore, sin z sinh). And since for z C R, z z i) + i z i R z 2 sin z R ) 2 sinh)

3 3 for z C R. Therefore, dz C R z 2 sin z dz R ) 2 sinh) C R Since πr R R ) 2 sinh) we conclude R C R dz z 2 sin z πr R ) 2 sinh) 3) For each of the following functions, do the following: find all its singularities in C; write the principal part of the function at each singularity; for each singularity, determine whether it is a pole, a removable singularity, or an essential singularity; compute the residue of the function at each singularity. ) a) fz) z) exp b) fz) z 2 + c) fz) tan z e z d) fz) z 2 z ) z 2 Solution. a) fz) has a singularity at. At z, ) z) exp z)e /z2 z) z 2 n!)z 2n z + So the principal part is n n!)z 2n n!)z 2n n n!)z 2n n n!)z 2n n!)z 2n n n!)z 2n Consequently, fz) has an essential singularity at and Res z fz)!

4 4 b) fz) has two singularities at ±i. We write z 2 + z i)z + i) i 2 z + i ) z i At i, the principal part of fz) is it has a pole of order and i 2 z i Res zi fz) i 2 At i, the principal part of fz) is it has a pole of order and i 2 z + i Res z i fz) i 2 c) fz) has singularities at {cos z } {z kπ + π/2 : k Z}. At z kπ + π/2, we let w z kπ π/2. Then tanz) tan w + kπ + π ) cotw) 2 cos w ) sin w ) n w 2n ) ) n w 2n+ 2n)! 2n + )! ) ) n w 2n ) ) n w 2n w 2n)! 2n + )! ) ) w2 w 2! + w4 4!... w2 3! + w4 5!... w + a n w n z kπ π/2 + a n z kπ π/2) n So the principal part of fz) at kπ + π/2 is z kπ π/2

5 5 fz) has a pole of order at kπ + π/2 and e z z 2 z ) z 2 Res fz) zkπ+π/2 d) fz) has two singularities at and. At z, ) z n ) z n n! z 2 + z! +... ) + z +...) z + 2z) + a 2 n z n z 2 2 z + a n z n So the principal part of fz) at is z 2 2 z it has a pole order 2 at and Res fz) 2 z At z, we let w z and then e z z 2 z ) e w+ + w) 2 w e w ew + w) 2 ) e ) w n ) ) n n + )w n w n! e + w ) w! w +...) e w + a n w n e z + a n z ) n So the principal part of fz) at is e z it has a pole of order and then Res fz) e z 4) Let fz) be an entire function. If fz) z 2 for all z, then fz) az 2 for some constant a C satisfying a.

6 6 Proof. Since fz) is entire, fz) for all z. By Cauchy Integral Formula, f n) ) 2πi f n) ) z n n! z R fz)n! z dz n+ for all n and R >. Since fz) z 2, fz)n! z n+ n! n! z n z R R n for z R. Therefore, ) ) fz)n! 2πi z dz n! n+ 2πR) n! 2π R n R. n 2 And since R n!/r n 2 for n > 2, fz)n! dz R 2πi zn+ z R and hence f n) ) for all n > 2. And since R n!/r n 2 for all n < 2, fz)n! dz R 2πi z R zn+ and hence f n) ) for all n < 2. In conclusion, fz) f ) z 2 az 2 2 for some constants a. Finally, by az 2 z 2, we obtain that a. 5) Let z 3 fz) z 2 3z + 2 Find the Laurent series of fz) in each of the following domains: a) < z < 2; b) 2 < z < ; c) < z <.

7 7 Solution. We write fz) as a sum of partial fractions: fz) z z 2 z Then a) For < z < 2, fz) z z 2 z 4 z + 3 z/2) ) z /z) z n z n z z n z + 3 z 2 2 n z n 2 2 n z n n2 z n n z n b) For 2 < z <, fz) z z 2 z z ) ) z 2/z) z /z) z z 2 n z n z z n z z z n+3 z n 2 n+3 )z n 2 n+2 )z n n z n

8 8 c) For < z <, fz) z z 2 z 8 z z ) z z z ) n z 4 7z ) 8 n2 z ) n z 6) Compute the integrals: π a) 2 cos θ) dθ 2 cos2x) b) + x + x dx 2 Solution. a) We parameterize the circle z with z e iθ for π θ π. Then π 2 cos θ) dθ π π π π π π π π π 2i 2i 2 cos θ) 2 dθ 2 e iθ + e iθ )/2) 2 dθ 4 e iθ e iθ ) 2 dθ e 2iθ 4e iθ e 2iθ ) 2 dθ π C e iθ 4e iθ e 2iθ ) 2 deiθ ) z 4z z 2 ) 2 dz

9 9 C and z z dz 2πi Res 4z z 2 ) 2 z2 3 4z z 2 ) 2 z 2πi Res z2 3 z 2 3)) 2 z 2 + 3)) 2 ) z 2πi z 2 + 3)) 2 z2 3 3π 9 i Therefore, π 2 cos θ) dθ 2 3π 2 9 b) Obviously, cos2x) dx Re + x + x2 Re R ) + z + z dz 2 R R + z + z 2 dz We integrate along the closed contour going from R to R and then the semicircle C R { z R, Imz) } counterclockwise. Then R R + z + z dz + e CR 2iz 2 + z + z dz 2 2πi Res z + 3i)/2 + z + z 2 2πi z 3i)/2 2π 3 e 3 cos i sin ) z + 3i)/2 by Cauchy Integral Theorem. For z on C R, e 2y, z 2 + z + z 2 z R 2 R and hence + z + z 2 R 2 R. )

10 Therefore, And since + z + z dz 2 CR we conclude that Thus, and R R R R R πr R 2 R. πr R 2 R, CR dz. + z + z2 2π dz e 3 cos i sin ) + z + z2 3 cos2x) 2π dx e 3 cos. + x + x2 3 7) Compute the following contour integrals. You may apply Cauchy integral theorem and its corollaries wherever possible. a) zdz, b) c) d) L where L is the polygonal path ABC with A, B + i and C i. L z 2 dz where L is the curve in part a). C dz sin 2 z where C is the circle z oriented counter-clockwise. C z 29 z 2 + z + dz where C is the circle z 2 oriented counter-clockwise.

11 L Solution. a) zdz zdz + AB + BC zdz + i)td + i)t) + i) t) + i) + t i)d t) + i) + t i)) i)tdt 2i 2t)i)dt t 2 2i t + i4 ) 2t)2 2i b) Since z 2 is entire, z 2 has a complex anti-derivative z 3 /3 in C and z 2 dz z3 i i L c) /sin z) 2 has singularities at kπ for k Z. Hence z dz sin z) 2 2πi 3 Res zkπ k 3 At z kπ, we let w z kπ and then sin z) 2 sinw + kπ)) 2 sin w) 2 ) 2 ) n w 2n+ 2n + )! sin z) 2 ) 2 ) n+ w 2n w 2 2n + )! n ) n+ w 2n m + ) w 2 2n + )! m n ) m So the Laurent series of /sin w) 2 at w only has terms w n with n even. Therefore, Res zkπ sin z) Res 2 w sin w) 2

12 2 Consequently, z dz sin z) 2 d) We first show that all zeroes of z 2 + z + lie in z < 2. Otherwise, suppose that z 2 + z + for some z 2. Then + z 29 + z 2. But + z 29 + z 2 z 29 z > 2 for z 2. Contradiction. So all zeroes of z 2 + z + lie in z < 2. Therefore, z/z 2 + z + ) is analytic in z > 2. It follows that z 29 z 29 dz 2πi Res C z 2 + z + z z 2 + z + ) z 29 2πi Res z z 2 z 2 + z + 2πi Res z z + z 29 + z 2 ) 2πi