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Unformatted text preview: Model Answers Mathematical Methods for Mechanical Sciences ASSIGNMENT 2
Problems 3A 3A1: Express in the form a + z'b:
(1) (2+1)2 + (2 i)2; (ii) Zi—gf— 2:43: [6 ‘lz'aB/(Cr2 +ﬁ2)] Solution: i) Expand and collect real and imaginary parts: (2 +z’)2 + (2 — i)2 (2+i)(2+i)+(2——i)(2—z‘)
4+4i+i2+4—4i+z'2 = 8—1— 1 {because i2 = —1}
6 ii) Crossmultiply: 3A3: Find the modulus and principal value of the argument of: z = ——1, i, 3 + 42', ——i — J5.
[1, 7r;1, ’2'; 5, 0. 927 radians; 2, —— Solution: For 2 = a; + iy, the modulus Izl = V1132 + y2 and the principal value of arg z is
0 = arctan 3% which satisﬁes —7r < 9 5 7r. —1: [2 = «(—152 = 1, 0 = 7r {2 = —1 lies on the —:1; axis.}
2 = i: z = x/l—2= 1, 9 = 7r/2 {z =z' lies on the +3; axis}
2 = 3 + 42': z = v32 + 42 = 5, 0 = arctan4/3 = 0.927 radians z = —z' — J37: lzl = (—ﬁ)? + (—1)? = 2, a = arctanjlg = —57r/6 radians {0 lies in the third
quadrant} 3A5:
Express «5 + 122', \/—5 +12i, ﬂ in the form a + ib. [:i:(3 + 21'), :t(2 + 32'), :l:\%—2(1 + n]. 1 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences Solution: Let a + bi = V5 + 122' then (a + bi)2 = 5 + 122'. Equating real and imaginary parts,
respectively:
a2 — b2 = 5 (1)
2ab = 12. (2) Elimination of a between (1) and (2) yields b4 + 5b2 —— 36 = 0. Solving, we ﬁnd b2 = "5313. We
know b must be real, so take b2 = 4, hence b = :l:2. Substitution into (2) yields a = $65 = :l:3, a+bi = :l:(3+2i). The same procedure applied to m and x/i yields a+ bi i(2 + 32')
and :lzﬁﬂ + 2'), respectively.
3A7:
Find two real numbers a, b such that (1 + i)a + 2(1 — 2i)b — 3 = 0. [a = 2, b = a.
Solution: Expand, collect real and imaginary parts, and obtain respectively:
a + 2b = 3 (1)
a —— 4b = 0 (2) Elimination of a between ( 1) and (2) yields b = 1/2 and substitution into (1) yields a = 2. 3A9: If 21, 22, 23 are complex numbers, show that
(i) '21 + 22'2 + I21 — 222 = 2[212 + 2Zz2
(ii) I221  22 — Z3l2 + I222  Z3  2112 + I223  21 — 22? = 3{l22  23]2 + [Z3 — 21]? + [21 — 22'2} Solution: (i) Since 2 = \/ 22* and 22 = 22* then, 21 + 222 + I21 ~ 222 = (21 + 22)(21 + 22)* + (21 — 22)(21 — 22)*
= (21 + Z2)(Zi + 35) + (21  22)(Zi — 23)
= 2121‘ + 212; + 2122 + 222; + 2121‘ — 212; — 2122 + 222;
= 2212; + 2222; = 22!12 + ZIZQIZ.
(ii) Consider the ﬁrst term on the lefthand side: I221 — 22 — 23]2 = [(21  22) + (21 — 23)2, then [(21  22) + (21 — 23)] [(21 — Z2)* + (21  23V] (21 — 22)(21 — Zz)* + (21 — 22)(Zl  Z3)* + (21 — Z3)(21 — 22)* + (21 — Z3)(21 — Z3)"r ll I221  22 — Z3l2 lzl — 22I2 + I21  23? + (21  23)(21 — 22)* + (21 — 22)(21— 23)* 2 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences Now (21  22)(21 — 23)* = (21  Z2)(21 — 22V +(21  22)(Z2 * 2:3)"
= I21 — 222 + (21 — Z2)(22 — Z3)*
2z1 — z2 — 232 = 221 — 222 + In  232 + (Z1 — 23)(21  Z2)* + (21  22)(22 — 23V (1) We can now write down expansions for the second and third terms of the left of (ii) simply permuting the subscripts.
I222 — 23  211i2 = 222  23l2 + I22  21!2 + (22 — 20(22 — 221)" + (22  23)(23 — 21)* (2)
I223 — 2:1 — ZQIZ = 2'23 — A? + '23 — 222 + (2'3  22)(Z3 — 21)* + (Z3  zl)(zl — 22)*. (3) Adding (1),(2),(3): the 3rd term on the right of (1) cancels with the last in (3), the last term on
the right of (1) cancels with the third in (2), and the last term on the right of (2) cancels with the third in (3). This proves (ii). 3A11:
Show that l 0 6 0
+ cos + isin . 1r
_______ = t _ “9"").
l—cosﬁ+isin0 co (2)e 2
Solution: c030 + isin9 = e19 and — c036 + isinO = —e‘w, therefore
1+cos€+isin0 _ 1+ei9
1cos0+z'sin0 _ l—e‘i" II II are“ sine
2(1 — cos 0) Using the half angle formula I—Eiéngj = W = cot (g), this becomes cot (g) (——iei9) cot (g) ei(9—%), {because 1‘ = e—ig} sin0
(1 — cos 0) (—iew) 3 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences which is the desired result. 3A13:
Evaluate the roots: {VT [:lzl, :hi, :t(1 ii)/\/§] Solution: Let w = \e/T = Re“, then R = 11/8 = 1 and = g+ %5 for k = 0,1,2,3,4,5,6,7. '2k_1r 4.21 ~ '1: ‘ZL '33.". ~ in ‘§£ '2: . .
Therefore, w = 1e2 8 = e1 4 = eo', e14, e12, e1 4 , e”, e‘ 4 e1 4 , and e1 4 . Expressmg 1n rectangular coordinates, em: and e"7r = 21:1
'1 6_1r .
e12 and e' 4 = :lzz
i3 i5—" lﬂ
e 4 and e 4 = :l:
\/§
‘3" 1‘11 1—1 3A15:
Evaluate the roots: 22 — (5 + z')z + 8 +i = 0 [z = 3 + 22', 2 — 1'] Solution: Applying the quadratic formula: (5+z'):l: (5+i) —4(8+i) (5+z'):t\/24+10i—32—4i 2 2
_ (5+i)::\/—8+6'i
— 2
5 '21: 1 3'
= (——+—Z)—2(—i——Z) {v—8 + 62' = i(1 + 3i) by the method of 3A5}
= 6+4z and 4;22 3+2z' and 2—21 Problems 3D
3D1: Evaluate: fi21=2 sz: [27rz']. Solution: To evaluate lel=2 £7“ The contour Izl = 2 is a circle of radius 2, centered at the
origin. The integrand has one singularity at z = —z', which is within the circle, but is regular everywhere else. By Cauchy’s Theorem the integration contour can be contracted down to a circle of arbitrarily small radius 6 with center at z = —i: f dz __ f dz
[4:2 2 +i z+ilz6 z + z" 4 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences Let 2 = —i + dew and dz 2 die” d0, then: f“ dz __ f2" (Siewdﬂ
z+i=52+i _ o 68"” II
S
s.
es 
$.
2
: ON:
:3 H
to
:1
S 3D—3: Evaluate: f0 :54};— where C is the square with corners :l:(1 :l: i) [0].
4 Solution: The contour encloses singularities of the integrand at z = :l:%. By Cauchy’s Theorem
the integral is equal to the sum of the integrals taken around two small circular contours of arbitrarily small radius 6, centered at z = 3%: f dz _ dz +f dz
1 — 1 1'
C z2 — — Iz+%=5 2'2 — z I %I=5 ”2 ‘ Z 4
For the integration around z = ——§—, set z = —% + 66m and dz = Me” d0, then:
% dz _ 2" (new d6
2+él=6 (2 + %)(z — %) o 6e“’(6e"9 — 1)
_ /2" we
— o (dew  1)
21:"
= —i d9 {letting 6 —+ 0}
0 = —27r1l. For the integration around z = +%, set z = % + 6e” and dz = die“ d0, then: f dz _ 2" Mew d0
Iz—%=6 (z + %)(z — %) — 0 5ei9(1+ 58”)
_ [2" we
— 0 (1 + 6e”)
21:
= 2' d0 {letting 6 —) 0}
0
= 27rz'. Finally, add these two results to obtain the original integral: f 2‘12] =—27ri+27ri=0.
oz —— 4 5 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences 3D5: . . d '
Evaluate. ﬁzgld 31:3; [22]. Solution: To evaluate ﬁz_gl=1 Simdz . The contour z —— g—I = 1 is a circle of radius 1 centered 2
2 22_1r_ 4
at z = g. The integrand has one singularity within the contour at z = %. By Cauchy’s Theorem the contour can be collapsed onto a small circle of arbitrarily small radius 6 enclosing this point.
On this circle set 2 = Z25 + 6e“) and dz = 6iew d6, then: f ’ sinzdz _ [27' sin(§ + 6ei9)6iei0 d0
z——’2£:6 (z + ’2—')(z — g) _ o 6ei9(6ei9 + 7r)
__ ‘/27r sin(§ + 6e“) (10
— z —————.——
o 6619 + 7r
27‘ sin % d0
= ' 1 tt'
1 /0 7r { e mg 6 —) 0}
' 27r
= 1 d0
77 0
= 21'.
3D7:
Evaluate: flzl=glal 2—456. [2W2].
Solution: To evaluate lel=§a 3%. The contour [z] = glal is a circle of radius glal with center at the origin. One singularity of the integrand at z = a is enclosed by the contour. By Cauchy’s Theorem the integration can be taken around a circle of arbitrarily small radius 6 centered on 2 = a. On the circle set 2 = a + 6e“ and dz 2 62'6” (10, then: f4 dz _ /27r 61c“) d0
lzl=6Z—a _ 0 6e“ 21r
2' d0
0
2m'. 3D9:
Evaluate: fz=1Q£2:—§3;ﬁ [2m]. Solution: To evaluate lel=1 ﬂagged—z. The contour z = 1 is a circle of radius 1 with center at the origin. There is one singularity of the integrand inside the contour at z = 0. By Cauchy’s 6 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences Theorem the integration can be performed around a circle of arbitrarily small radius 6 centered on 2 = 0. On this circle set 2: = 6e” and dz = die” d0, then: ?{ (2z — 3)dz _ f2” (2599 — 3)6z'ei9 d0
[4:5 z(z — 3) — 0 6ew(6ew — 3)
2w (26ei0 — 3) d0
/0 6e“ — 3
27r 1' d0 {letting 6 —) 0}
0 27113. Problems 3E
3E—1: Evaluate the residues of: zcosh(3/z). [% at z = 0]. Solution: The argument of the cosh term is singular at z = 0. Now (3)2 (3)4 (2)6
2l+4l+6! cosh(3/z) = 1 + + .. Multiplication by 2 results in (3)2 (§)4 (3)6
2!+4!+6! zcosh(3/z) z 1 + +   z+i+
22 The coefﬁcient of the % term is the residue at z = 0, namely 9
R0 = ‘2‘.
3E—3:
Evaluate the residues of: 24/(22 + 1). [—5 at z = i; g at z = —z'].
Solution: Factorizing the denominator
z4 24 f(z)=z2+1=(7+zT(:7) f(z) has simple poles at z = :Lz', and the residues R33 are calculated from Z4 R: :—
ZZO %(zz+1) 2:20 7 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences i.e.
24 i4 —i
Ri*§;.—z—? a“
2:1
72; = if = (—0.4 = 3
22 —22 2
Z='—1
3E5:
Evaluate the residues of: 9&5. —% at z = 0]. Solution: There is a 3rd order pole at z = 0. {sin z/z4 —) 1/z3 as 2 ——> 0} Now . _ z z z
smz — z—§+g!—ﬁ+
sinz __ 1 z3 25 27
7r‘;z‘a+§‘ﬁ+ _ 1 1 — 23 62
residue — 1 _ 6' 3E—7:
Evaluate the residues of: z2ei. [31 at z = 0].
Solution: There is an essential singularity at z = 0. Expansion of ei yields 1 1 1 1 ez = 1+2+2!22+3!z3+lu'
21 _ 2 1 1 1
26“ ZP+2+52+QE+“'
’ 2 67.
1
'd = —.
1'68] 118 6 3E9: Evaluate the residues of: (cosh 2z)/25. [g at z = 0]. Solution: There is a 5th order pole at z = 0. Now 22:2 224 26
23+<£ (5) cosh2z = 1+ 8 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences cosh2z _ 1 (22:)2 (2z)4 (22)6 25 _ 25[1+ 21+ 4! + 6! +"'
_ 1+ 4 +§+
_ 25 2lz3 4!z residue — §_E
‘ 4!‘3' 3E11: Evaluate by the residue theorem (contours are traversed in the anticlockwise sense): ﬁzn=1 ﬁn [—77r'i]. Solution: By the residue theorem, 26+7 26+7
I: d =f d =2 ' R,
z:122_22 z z=1z(z—2) z m: C where the sum is over the residues of the singularities enclosed by the circle z = 1. There are simple poles at z = 0 and z = 2; only the pole at z = 0 contributes to the integral. The residue at z=0is 26+7
R0 $022 22)
26+?
2z—2
7
_—2. I = 21rz'Ro = 27rz' (_;) = —77ri. z=0 ll z=0 3E13: Evaluate by the residue theorem (contours are traversed in the anticlockwise sense):
ﬁzi=1%’§%dz [m'sin (%)]. Solution: By the residue theorem, sinh z I =
z=1 422 + 1 dz = 27r'i 2 RC,
where the sum is over the residues of the singularities enclosed by the circle z = 1, which are
simple poles at z = 21:5 The residue at z = 20 is given by sinh z
82 sinh 2
gm? + 1) ’RZO = 2:20 2:30 9 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences Ri = = d
a 8e) a“
' hi hi
’R.__i = sm .2 = 3m, 2 {because sinh(—:1:) = —sinha;}
2 8(21) 4’1.
2
Hence,
I = 21rz(’R%+’R,_%)
i
= ' h—
7rs1n 2
— m'sinl because sinh—7’ — eé —e'% —ze% —e_% ' ' 1
— 2' 2 2 ‘ 2i "”1112
3E—15: Evaluate by the residue theorem (contours are traversed in the anticlockwise sense):
ﬁz—%[=1 iii—1‘13 [7r]
Solution: By the residue theorem, z4 dz =f —.——._
z—%=1 22+1  %]=1(Z+z)(z—z) where the sum is taken over the residues of the singularities enclosed by the circle of unit radius 4
dz=27riZRc, with center at z = g. The simple pole at z = 2' is within the contour but the pole at z = ——z' is not. The residue at z = i is z4 z4
Rz‘ = T;— =2—
(Tz(z +1) 2:12 z 221‘
_ i
_ i
_ i
_ 2
I = 27rsz=27ri (:21)=7r 3E17: Evaluate by the residue theorem (contours are traversed in the anticlockwise sense): 1 .
fiz—giﬂ Wdz [67”]
Solution: By the residue theorem, 1 I = ——
lz%=1 23(2  1)2 dz=21riZRC, 10 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences where the sum is over the residues of the singularities enclosed by the circle of unit radius with
center at z = 3. There is a third order pole at z = 0 and a second order pole at z = 1. Only the pole at z = 0 is enclosed by the contour. The residue at z = 0 is
2 _
£502 —— 1) 2
2! 6(—1)4
2 6(z —1)‘4 R0 2!  z: 2: ll 3.
I = 21ri’R0 = 27rz'(3) = 67ri. 3E19: Evaluate by the residue theorem (contours are traversed in the anticlockwise sense):
fizl_ge%/(z— 1)2 dz [0]. 2
Solution: By the residue theorem, 1 ez
[zlzg (Z 1)2 I: dz=27riZRc, where the sum is over the residues of singularities enclosed by the circle of radius % centered at
z = 0. There is an essential singularity at z = 0 and a second order pole at z = 1; both are enclosed by the contour. First, consider the residue at z = 1: 1
= _ : = ; _1 ’2
R1 dze 15:1 e ( )2 2:1
= ei(—1)1'2
= —e. (1) 1 1 1
e%=1+;+——+———+——+. (2) Next (2 — 1)‘2 E (1 — z)‘2, and the binomial theorem gives:
(z—l)‘2 = (1 —z)_2 = 1+2z+3z2+423+525+~ . {valid for z < 1} (3) Multiply (2) and (3) term by term:  1
e%(z—1)_2 1+—+~' 2
+ 2z+2+—+'
2!z 11 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences 3 3
2
+ 32 +3z+2!+—3!z+~
4z 4 4
3 2
+ 42 +42. +——2! +—3!+—4!z+ + etc, and collect the coefﬁcients of the terms in %, 1(1+1+1+1+i+ )
z 2! 3! 4! 1 (61) ’20 = e. (4) residue Hence, from (1) and (4),
I = 2m’(’R1 + R0) = 27m'(—e + e) = 0. Problems 3F
3F1: Evaluate by the residue theorem: f3" ”gt—22,19, a < 1 [ «1L0; . Solution: Let 2 = e“, d0 = 12 and sin6 = (em — e‘i9)/2i = (z — z‘l)/2i, then iz’ I_/27r d0 % dz
_ 0 1+asin0 _ z]=1[1+%(Z—Z_l)]iz _ f dz
_ g 2  _e
Izl=1 22 +22 2 2% dz
a z=1 Z2 + 212  1' 71' The integrand has simple poles at 2i = %(—1 :l: v1 — a2) where [a] < 1. The contour is a circle of radius 1 centered at z = 0, which encloses z... where the residue is 1
’R = ,
%(%22 + ZZ — £21.) z=z+
_ 1
_ az+iz_2+
_ 1
a§(—1+v1—a2)+z
1 II
i—I
I
Q
N Therefore, by the residue theorem 1 271'
I = 27ri’R = 27rz' (———) = .
iv1—a2 \/1—0.2 12 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences 3F3:
 . 27f cos 26 d9 27f 2
Evaluate by the re31due theorem. f0 W’ [pl < 1 [1:57;]. Solution: Let z— — e19, d0— — 0030: (e w+e‘i0)/2 = (z+z“1)/2, and cos 20 = (ei2a+e"i20)/2 = (22 +2 )/2, then 17’ [27' cos 20 d0
0 1  2pcosl9+p2 f (z + 2 2)dz
IZI=1 [1 — 21164?) +p2] iz __L]{ (z2 + 2’2)dz
Zip Izllz2—(p+11—,)z+1 _L?{ (22 + 52W”
Zip z=1 (2 P)(Z *1/10). For Ipl < 1 there is a simple pole at z = p and a double pole at z = 0 within the contour, with respective residues RP and R0 II
A
A
N

“6
V
A
N
l
C
'6
v
v Therefore, by the residue theorem I<%)2m(’Rp+Ro)— p (pw_1)+ p 3F5:
Evaluate by the residue theorem: far ﬁ—{l‘gfg, a > 1, [7r (go — a3 + (a,2 — 1)§)]. Solution: Let z— — e9 ,—d6 — E, sin0= (em —e'w)/2i = (z —z‘1)/22', and c030 = (em +e‘w)/2 =
(z + z‘l)/2, then Sin4 0 d0 21f Sin4 0 do
 t 0 =
I: /o a + cos 0 2 —1/ a + cos —’_9 {by Symmetry abou 7r} 13 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences 1% ‘72—VL £5
2z=1[ _ 1 (22 — 1)4dz
‘ 2 Izl=1 24(22 + 2oz + l)8i
—i (1 — 22)4dz 16 z=1 24(22 + 2oz + 1) There is a 4th order pole at z = 0 and two simple poles at z = a = —a + Va2 — 1 and
z = ,6 = —a —— Vaz — 1. Now aﬂ : 1 and a > 1, therefore the pole at z = a is within the contour [2 2 1 and that at z = ﬂ is not. Thus —2' (1 — Z2)4dz I = '13 le=1 z4(z  a)(2 —ﬁ)’ At 2 = a, the residue is (1 — 22V
24(2  ﬂ)
(1 — 012)4
04(0  £3)
(a  1/ 01)4 (a — 3)
(a — m3 {because 1/01 = ,6} 8(a2 — 1)3/2. 2:0 II II At 2 = O, the residue R0 is found by expanding in powers of 2: (1—22)4 1 2 2 22 23
M ;(1—42 +"')[1—(2GZ+Z)+(2GZ+Z) —(2az+z) +"‘] 1
23(1— 422 +   ) [1 — 2oz + 22(4a2 — 1) + 4oz3 —— 80,323 + .  ] ll 1
coefﬁcient — = 8a + 4a — 8a3 = 12a — 8:13. R0 =
Z
Hence, by the residue theorem
I = (F) (2711') [Ra + R0]
_ :1 ~ _ 3 2 _ 3/2
_ (16) (2711) [12a 8a + 8(a 1) ]
= 71' [—413 + 33a + (a2 — Uzi/2] 14 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences 3F7: Evaluate by the residue theorem: f6” W [%]. Solution: Consider dz dz
I = ya: (1 +22)(4+z2) = ‘74; (z+i)(z—i)(z+2)(z —2) along the contour 0 consisting of the real axis from ——R to R, and the semicircle 7 above the real
axis having this line as diameter, traversed in the anticlockwise sense. The simple poles at z = i and z = 22' lie within 0, with residues 1 1 1
57(24 + 522 +4) m. _ 423 +10z Fi 6i 1 _ 1 4(2i)3 +10(2i) ‘E' 731' and 7221' Therefore, by the residue theorem . 1 1 7r
I“(a‘ﬁa’ RM”
i.e.
lim R —————dx + / ————————dz — 5
R—mo —R (1+ar2)(4+:1:2) 7 (1 +22)(4+22) _ 6' Now the integral over 7 vanishes like 1 / R3 as R —> oo,
/°° d2: __ 1/“ d9: _ _7_r_
0 (1 +m2)(4+a:2) ‘ 2 00 (1 +5122)(4+£II2)_12' 3F11: Evaluate by the residue theorem: f0°° ﬁg? [ﬂ . Solution: Consider
I _ % dz _ f dz
‘ C (1 +22)2 ” C (z+z')2(z —i)2
along the contour 0 consisting of the real axis from —R to R, and the semicircle '7 above the real axis having this line as diameter, traversed in the anticlockwise sense. The double pole at z = 2'
lies within C, with residue 1 = —2(2 + i)"3 z=i z=i 4i d ._2
R — Zz‘CZi'z) 15 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences Therefore, by the residue theorem 1
I: [—Ijz(1+—1:‘2)2 2+/——(1+2=27riR=27ri (4—1,): The integral over 7 vanishes like 1 / R3 as R —+ 00, [”LJ/md—fci
0 (1+m2)2—2 _oo(1+m2)2‘4' g, R—+oo. 3F15: Evaluate by the residue theorem: f0°° %# [5733]. I % 2e?” dz _j£‘ ze2iZ dz
(1+22) (z+i)(z—i) along the contour C consisting of the real axis from —R to R, and the semicircle 7 above the real Solution: Consider axis having this line as diameter, traversed in the anticlockwise sense. The simple pole at z = 1' lies within 0, with residue R = d ze2iz : 2821.2 = _:L2_
E(1 + 22) z=i 22 z=i 2e
Therefore, by the residue theorem
R 11:62ialc da: zem dz 1 7rz'
— —2 R— — 2 = —, R —> ,
/—R 1+m2 + 7 1+z2 7” 7”(2e2) e2 00 i.e. R wcos 2x da: , R xsin2a: dac ze2iz dz «2'
/———————+2/ —————+ =—, R—>00
—R 1+2;2 —R 1+3? 7 1 + 22 e2
The integral with the cosine vanishes because the integrand is an odd function, and Jordan’s lemma ensures that the integral over 7 goes to zero as R ——) oo, /°° xsin2a: da:_1/: xsin2x d1: L
2
0 1+1;2 1+$2 =2e2' 3F17: Evaluate by the residue theorem: 303:0 9———°s If?“ d" ,(a, k real, k > 0) [—1r sin ka]. Solution: oo oo ika:
I— ][ cos___lc__xdx=Re[ e dx]=Re[m'R], oo (II—a —oo$"_a 16 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences Where ’R is the residue of e17” / (z — a) at z = a, namely ’R = cum. I Re [wieika] = Re [m’(cos 1m + isin 19(1)] —7r sin ka. 3F19: Evaluate by the residue theorem: ff?» Eff—m7, (a > 0) [7:5]. Solution:
m
I = f —d—$—.— = —7riR,
_00 (11(2) — at) where R is the residue of the simple pole of 1/z(z — cm") at z = 0: 1
R = —,
—az I = —m'7a=I a 17 AM 400 Engineering Mathematics Model Answers Mathematical Methods for Mechanical Sciences Essay: Write a short essay (not more than two sides of paper) in which you give the deﬁnition of
a regular function f (z) of the complex variable 2, deduce the CauchyRiemann equations, and state and prove Cauchy’s theorem fa f(z)dz = 0 for a simple closed contour C’. Solution: A function f (z) of z = (I: + iy is diﬁ’erentiable at z with derivative f’(z) if f’(z) = lim W (1) 62—)0 (52 , exists when 2 + 62 ——) z along any path. f (z) is said to be regular in a domain D of the complex plane if f’ (2) exists at all points of D. The condition that the limit should exist independently of the path by which 62 —) 0 imposes
a severe restriction on a regular function. Let f (z) = u(m,y) + iv($,y), where u and v are
respectively the real and imaginary parts of f (2:) In equation (1) take dz = 6a: and then take
62 = My. Because of path independence it then follows that f’(z)—@+ia—v—i§[email protected]
_6x 8m. 6y By. By equating real and imaginary parts we deduce the CauchyRiemann equations 6u_6v 8u_ 612 53—5313 6—y_—8—x' (2) These equations can be used to prove Cauchy’s theorem that, when f(z) is regular within and
on a simple closed curve C, then f0 f (z)dz = 0. If ds = dz > 0 is the element of arc length on C, and n is the outward normal, then nds = (dy, —d:r). Therefore £f(z)dz 74C(— Many), —u(m,y)) . (...
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