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Solution Manual for Fundamentals of Complex Analysis with Applications to Engineering, Science, and Mathematics, 3rd Edition

Solution Manual for Fundamentals of Complex Analysis with Applications to Engineering, Science, and Mathematics, 3rd Edition simplifies even the toughest textbook questions with step-by-step solutions and easy explanations.

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    Solutions Manual S. Garrett R. Campbell-Wright D. Levinson for the book Fundamentals of Complex Analysis, 3" ed. by E. B. Saff and A. D. Snider Prentice-Hall 2003 + StudyXxy

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    Downloaded from StudyXY.com ® + StudyXY Sd Ye. o> \ | iF ’ pr E \ 3 S Stu dy Anything This ContentHas been Posted On StudyXY.com as supplementary learning material. StudyXY does not endrose any university, college or publisher. Allmaterials posted are under the liability of the contributors. wv 8) www.studyxy.com

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    CHAPTER 1: Complex Numbers EXERCISES 1.1: The Algebra of Complex Numbers 1. —i=a+bi=>a=0andb=-1= (=i)? = (a® = ¥) + (2ab)i = —¥* = —1 2. The Commutative and Associative laws for addition follow directly from the real counterparts. Commutative law for multiplication: - (a+ bi)(c+di) = (ac—bd)+ (be+ ad)i = (ca— db) + (da + cb)i = (c+di)(a+ bi) Associative law for multiplication: [(a + bi)(c + di)l(e + fi) = [(ac — bd) + (bc + ad)i)(e + f7) = [(ac— bd)e — (bc + ad) f] + [(bc + ad)e + (ac — bd) fli = [a(ce — df) — b(de + cf)] + [b(ce — df) + a(de + cf )}i = (a+bi)l(ce — df) + (de + cf)i] = (a+b){c+di)(e+ fi) Distributive law: (a+ bi)[(c+ di) + (e+ fi)] = (a+ bi)[(c + €) + (d+ f)i] = [a(c+e)—b(d+ + bc +e) +a(d+ fli = [(ac — bd) + (bc + ad)i] + (ae — bf) + (be + af)i] = (a+bi)(c+di)+ (a+ bi)(e+ fi) 3 a zm=z—z etfiz=(c—a)+(d-bi=(c+di)— (a+b) += e=c—aand f=d-b <= eta=cand f+b=d += (e+ fi)+(a+b)=c+di <= »

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    b. (e+ fi)(c+di) =a+bi <= ce—fd=aand fc+ed=b <= a+b ac+bd bc—ad. . o& - are tera ctrl _ (ec=fd)c+ (fe + ed)d - +d? 4 (fe+ edjc— (ec — fd)d, 2 +d? = e+ fi 4. Suppose z; # 0. Then JRL J) z zn 3). -3. 5. a 0+ (-3)i=5 b.3+0:=3 c. 04 (-2)i=-2 6. a 0+ (-2)i=-2 b. 6+ (=3)i =6—3i c. 4+ mm 7. a 8+1i=8+: b.1+li=1+: [ 0+ (3)i= id . 3 BE) 33 19. 8 55550 61 107. % 185 185° 253 204 . 10 —1595 ~ 1325" 11. 240: =2 12. —9+ (=T)i

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    13. 6+ 51 14. z =a + bi. Re(iz) = Re(ai — b) = —b= —Im =z 15. i = (ff =1F=1 Fe LT PAE tk 2 1 (—1) = —1 PHS = EB =] (mi) = —i 16. a. —t b. —1 c. —1 d. —i 17. 320H3 4 63° + 8i~5(4) 4 1043 = 3(~1) + 6(—1) + 8(1) + (—i) =8 — 10: 18. (=1+4d)2+2(-1+i)+2=-2i+(-2+2)+2=0 19. The real equations are : Re(z® + 52%) = Re(z+3i) Im(z* +52?) = Im(z+ 3). If z = a + bi these can be rewritten as a® —3ab? + 52> — 50 —a=0 3a’ — 5° + 10ab— 5-3 =0. 20. a= =2 bso l= _2 5 "PTR TE TW 1g c. z=0, 1+ d. z=24:

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    21. (=i)[(1 = i)z1 + 325] + (1 = )fiza + (1 + 24)z) = —i(2-3i) + (1-9)(1) = z= 2-3 1=>z1=1+4+1 TTT TT eT 22. 0=z'-16= (z= 2)(z + 2)(z — 2i)(2 + A) => z=2,-2,2, -2 23. Suppose z = a + bi. 1 a—1b a Re (3) =r (55%) CEA whenever a > 0. 24. Suppose z = a + bi. I G) 1 LE —b SCY PERN ph b = ay < 0 whenever b> 0. 25. Let z; = a+ bi and z, = c+di. The hypotheses specify that a +c < 0, b+d=0,ac—bd <0, and ad + bc = 0. b=0=d=0=> 2 and z, are real. b#0=>d=—band ad + bc = a(~b) + bd = ~b(a — c) = 0 = a = c, a contradiction of the fact that zz, < 0. 26. By induction: The case when n = 1 is obvious. Assume Re(TT, z) = YT; Re(z;) for all positive integers m < n n n—1 Re(3"z | = Re(Y z+z i=1 j=1 n—1 =) Re(z) + Re(z,) = = 2 Re(z;) = The corresponding result for the imaginary parts follows by replacing “Re” by “Im” in the above proof. — StudyXY 1-4

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    n n Disprove: Re | [] z; | = [] Re(zj) and J=1 i=1 n n Im | I] 2; | = IJ Im(z;)- j=1 3=1 Re[(a + bi)(c + di)] = ac— bd Re(a + bi) Re(c + di) = ac These are not equal whenever bd # 0. Im[(a + bi)(c + di)] = ad + bc Im(a + b) Im(c + di) = bd These are not equal whenever ad + bc # bd. (For example, consider the pair 2 and i.) 27. By induction: The case when n = 1 is obvious. Assume (at+z)" = 4+ (F)era +e + WEE +2F for all positive integers m < n. Recall that, for positive integers r and swithr >s, r r r+1 T r (+ (5) = (3) = ©)-0) = (21+ 22)"= (21 + 2)" a1 + 2) -1 =z Nz +2) + (" 1 Jaret + 23) —1 Foot (" k Ea +2) +t (n+ 2) -1 = +2 n+ (" 1 Jaa + 277223) -1 n— tot (PTE AEE gn 5 I~ StudyXY

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    n n—1 n—1 ne a+ 0 )+( 1 IE 12 n—1 n—1 ne [03+ (5) n—1 n—1 nk _k =f) (13) Feet zy n NY ae ny a =z + (1): lz 4 (3): 222 + 4 (arta vm 5 28. 2° + 0) 24(—1) + () (=i)? + (3) 22(—i)® + (3)et-ar + (—1)° =32—-80i — 80 +40; + 10 — i = —38 — 414 29. Suppose z = 2 where p and q are relatively prime integers, and that q 2 =2. 2 (2) =2 =p’ = 2¢* = p? = 4k for some integer k and ¢° = 2k, a contradiction (If p? is an even integer so is p.). 30. By contradiction. Suppose there is a nonempty subset P of the complex numbers satisfying (i), (ii), and (iii) and suppose i is in P. Then, by (iii), ¢ = —1 and (=1)i = —i. This violates (i). Similarly (i) is violated by assuming —i belongs to P. 31. Purpose: to add, subtract, multiply and divide z; = a + bi and zy = c+ di. Input a,b,c,d Set sum= (a + ¢,b+d) Print “21 + 22 =”; sum Set diff=(a — ¢,b— d) Print “21 — 22 = 7; diff Set prod=(a*c—bxd,bxc+a*d) I-¢ lea)

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    Print “z1 # z2 ="; prod Set denom = ¢*2+d"2 If denom = 0, print “there is no quotient” Else Set quot=((a * c + b* d)/(denom), (b*c—a* d)/(denom)) Print “z1/22 =; quot Endif Stop 32. prod=(a*xc—bxd, (a+b) *(c+d)—arc—bxd) EXERCISES 1.2: Point Representation of Complex Numbers; Absolute Value and Complex Conjugates 1. The real and imaginary parts of natn _ nt nty T ~~ 2 TV 2 give the familiar algebra formula for the midpoint of the line segment joining two points in R% Alternatively, one could establish that (z; + 2;)/2 is a point on the line through z; and 2 and that |z — (2 + 2)[2] = lz — (21 + 22) /2]- 9 5= 21 +4) +(-31) +3012) +56) _ 25 7. EE 271+3+5 =n 3. -3 a? «7 13 2; - ras = + (=m) * z v1 Vz EE EW - - Z PR z

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    5. The three side lengths are equal: 1, V3, 1 V3 (rg) =h- (5-2) 1, V3! 1 V3 |e) - (2 6. The Pythagorean theorem is satisfied: 10+10=|(3+1) — 62 +|(3 +i) — (4 +4) = 16 — (4440 = 20 7. a. All points on the horizontal line through z = —2i b. All points on the circle of radius 3 with center at 1 —1 - 1. c. All points on the circle of radius 2 with center at d. The points must be equidistant from 1 and —i, thus lie on the perpendicular bisector of the line through 1 and —i. e. The equation can be written as z = w — 1. The points lie on this parabola. f. The points z have the property that their distance from 1 added to their distance from —1 is always 7, so the points lie on an ellipse with foci £1, with z intercepts + and y intercepts +315 9 g. All points on the circle of radius : with center at 3 h. All points in the half plane z > 4 i. All points inside the circle of radius 2 centered at ¢ j. All points outside the circle of radius 6 centered at the origin | 1-8 — StudyXY

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    8. {a+ bi)—1|=yfla=1P +B = la 17+ (0 =la+t—1 z—1 «Zz z-1 Zz 9. |rz] = r(a + bi)| = Ira + rbil = y/(ra)? + (rb)? = ria? +¥) = ra +B =rlz| 10. [Rez] = ja] = V&? < V@ +P = 4] mz| =p = VF < VEZ + BE =z 1. a=le+bij=vVa@ +P =>a20and b=0 zn (e2E- (aa; + bibs) + (ashy = aba) 12. a. (2)- ar + bai BN (COELU RACE) (a2 + bby) + (—azb + arbad Bh +8 _ a-bhi_ F T abi #m a EE 1

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    13. 2? -22=0=(z-2)(z+2)=0= either: Z—z = 0 = 2:lmz = 0 = z is real, or Z+4z=0=2Rez = 0 => z is pure imaginary. 14. a2) = (22) (7%) = (27) (227) = |af'|=f 15. By induction: The case when k = 0 is obvious. Assume (Z)™ = (z™) for all positive integers m < k. (2) = (3) = (FE =F =F Also, 1 1 1 — hes = (=) =2F EO =r == (3)=7 16. Let z = a + bi. Since |z|2 =a? +8? = 1, 1 1 (1—a)+biy 1 Re(;=) =R 1) _pe(Uzatty 1 fp (===) Re 2-2 ) 3 17.5%" + a5" +--+ anaZo ton =z tmz + tan1z0+ 8, =0=0 —ay £4/a? — 4a. 18. The roots of z% + ¢,z + a = 0 are z = a a? — 4a; > 0=> Both roots are real = Each root is its own conjugate a? — 4a; < 0== £/a} — 4a, = £i\/4a, — af == The roots are complex conjugates. | 1-10 4

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    19. The line ax+by=c can be represented in the complex plane as z=rcos6 + irsin + c/a where 6= tan" (-a/b) and —oo < r < co, By working with triangles you can obtain cos8 = -b/(a” + b?) and sin = a/N(a® + b%). To get to point z write the equation from point c/a down the line and make a turn on the perpendicular as z=x+y = rcosB + irsin@ + c/a — ssin6 +iscos® with —o <'s < co. Equating the real and imaginary parts X —c/a=rcosf — ssinb; y = rsin® + scosd Solve for s as s = ( — sinB(x-c/a) + ycosB) = (-ax + c- by)(a® +b?). The distance from the point z to the line ax + by =c is s. Denote the reflected point by z.. The reflected point lies s units on the other side of the line. ze =2-2s(-a~ ib))N(a® +b?) = x + iy — 2{(-ax + c- by)\ra® +b?) }(-a — ib)A(a® + b?) ={[(b — a*)x —2aby + 2ac] + i[(a’-b?)y ~2abx + 2bc]}A(a® + b) = [2ic + (b-ai)(x-iy))/(b+ai) 20. (a) Suppose u'Au=0 forall n by 1 column vectors with complex entries. Let u=[00... 1..0]" with the i" entry being the only nonzero entry. Then u'Au= (ai) = 0 for i=1 to n. Let u be all zeros except for ¥ + iV3/2 On the ith row and ¥2 -iV3/2 on the jthrow. Now ul Au = (ag)(Vs -iV3/2 2 + (a) (te + V3/2)* = -(1/2 - iV3/2)(ay) - (Va + iV3/2)(aj) = 0. Setting the real and imaginary parts equal to zero yields a;=0and a; =0 for all i,j= 1 to n. Consequently A = 0. (b) Let A=[01;-10]. Now u'Au=0 forall 2 by 1 real column vectors . 21. The matrix A is Hermitian A’ = A. Observe (Au) =u’ A" =u’ A. (2) (u Au)’ is the conjugate transpose of the matrix u’ Au which is a one by one matrix, so (u' Aw)" =u Afu =u" Aubecause A is Hermitian. The conjugate is equal to the number only when the number is real. (b) (B'B)' = B'B and therefore is Hermitian. (© @'B'Bu)" = (Bu)'(u'B")" = u'BBu a real number. Ly

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    EXERCISES 1.3: Vectors and Polar Forms 1. az+z=3 | 3 b.ozn—z=1-2 1-2 c. 22 —32z,=1-5¢ - 1-5¢

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    2. |z12223) = |(2122) 23] = |2122]|23] = |21]]22]) 25] 3. Ja + 2 +a — 2 = 2a + 2] 4. By induction: The case when k = 0 is obvious. Assume |2™| = |z|™ for all positive integers m < k. 124) = 17 2] = 2572) = Jee] = 2 Also, " 1 1 1 —k EEE E FTTH TRF Il 5. al b. 5v26 EG) To? d. 1 6. a. b. 3 3n Tds| 4 f 1 EE EE ae a a a a Sr 3 - . - Pp .(—=m i ) ci 6 1-13 + StudyXY

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    c. d. (27m ) ®) 3 a 2 ) cis == 4) | | ERC ICE aE x= = = -1 7. (Only the value of Arg z is given for each of the following.) 1. EL a. Fis7 b. 32s 7) c. weis (-3) a ses (-F) e. 2v2s (13) £ 4cis (3) 2s 3) h Vit, (=) & Bn gr) 12 8. Suppose |z;| = r. Then 21 + 2 lies on the circle in the figure and |21 + za] is greatest when arg 2 = arg z; are 9. Itis a vector of length |z| and angle of inclination arg 2+¢; it is obtained by rotating z by angle ¢ in the counterclockwise direction. mt

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