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Unformatted text preview: W 1.6 CONTOUR INTEGRATION 71 where y is a circle centred at z = 1. Writing f,(z) = Z4, then _ 11(2)
lewd: _ it (ze 1de . and, sincef1(z) is analytic within and on the circle 9!, it follows from (1.48) that jg f(z)dz = 2mg isfmz) = ﬂj(1222)z=l 2=l so that
Z4
3 dz = 121:}
C (2 e 1)
Evaluate fc(zz + 3x) dz along the following contours I‘ Using the Cauchy integral theorem, evaluate the
C in the complex 2 plane: contour integral
(a) the straight line joining 2 +j0 t0 0 +j2; 22 dz
(b) the straight lines from 2 +j0 to 2 + j2 and then m
to 0 + j2; C
(c) the circle [2] = 2 from 2 +j0 to 0 +j2 in an Where C is anticlockwise direction. I
(a) the elrcle z = 1 Evaluate Msz“ — 23 + 2) dz around the following (13) the circle Izl = 3
closed contours C in the 2 plane: Using the Cauchy integral theorem, evaluate the (a) the CirCle lZI = 1; contour integral (b) the square with vertices at 0 + jO, 1 + jO,
l+jland0+jl; 5zdz (c) the curve consisting of the parabolas :x2 from a 0+;0tol+JIandyzzxfromI+Jt00+Jll L where C is Generalize the result of Example 1.30, and show that _
(a) the Circle z = 3 dz jg“ (n =1) (b) the circle z = 5
C(z—zu)" 0 (incl)
h Using the Cauchy integral theorem, evaluate the
where C is a simple closed contour surrounding following contour integrals:
the point z = 20.
23 +2
. (a) ——. dz
Evaluate the contour integral C (22 + 1)’
dz where C is the unit circle Jz = 1;
Cz—4 4
(b) ﬁl Z 2 dz
where Cis any simple closed curve and z : 4 is C (2  1N2 + 2) (a) outside C (b) inside C where C is the circle lzl = 3. 73 FUNCTIONS OF A COMPLEX VARIABLE so, by the residue theorem, . 1 _21t
[—27C][J_—\§]#\3 Thus Using the residue theorem, evaluate the following contour integrals: Evaluate the integral % 2 dz
2 '3
H +1 (20% (3z+2)dz
where c is c (2—1)<zl+4)'
' ' = l ' v =
(a) the Cerle z[ 2 (b) the Circle H 2 i (i) the circle IE 72‘ = 2
where C15 ” .
Evaluate the integral (11) the Guide lzl = 4
z2 + 3jz — 2
3+9 dz (b) (2222)dz
Z Z 5
( C(z+l)2(22+4) (i) the circle lzl = 3 where C is
(a) the circle 12] 2 1 (b) the circle lzl = 4 where (:15
(ii) the circle Iz+jl = 2 Calculate the residues at all the poles of the function
2 2
+ 2 + 4 ~
ﬂz) = W ( ) 1 dA 1
(z +1)(z +6) (_(z+l)'(z—l}(z2)
Hence calculate the integral
(i) the circle Izl :lE
jgﬂzmz whereas (ii) the circle 12+ 1  :1
C (iii) the rectangle with vertices
where Cis at ij, 3 ij
(a) the circle z = 2 (b) the circle z —jl = 1
(c) the c1rciez= 4 (d) § (z_1)dz
2 4 '
Evaluate the integral 5 (7 r 4“: + 1}
§ dz (1) the circle lzl : g
2 2 2
CZU”) whereas (ii) mecirctez+§ :2
where C is (iii) the triangle with vertices
at —%+i, %—j, 3 +10 (a) the circle 121 = (b) the circle z = 2 —————_ 1.7 ENGINEERING APPLICATION: ANALYSING AC CIRCUITS 79 Using a suitable contour integral, evaluate the °°
(f) 2
xdx
____ “(x2+1)2(x2+2x+2) following real integrals: co °° 23: m
dx dx d5 (L[
b
(a) I_mx2+x+l ()I_M(x2+l)2 (g) In 3~2c056+sin6 (h) I0x4+1
m ﬁx . w dx
c ———__
() I1, (xl+1)(xz+4)2 (I) I_m()cz+4x+5)2 Zn 211 2
c0539 4d9 . c056
0”] 5—4c056d9 mi 5+4sins (“In 3+2cost9d9 1.7 Engineering application: analysing AC circuits 1 ] 1 Z—R+JQJC, Y—Z
Writing 1_1+ijR Z‘ R we clearly have R
Z_ (1.50) Equation (1.50) can be interpreted as a bilinear mapping with Z and C as the two vari ables. We examine what happens to the real axis in the C plane (C varies from 0 to 00
and, of course, is real) under the inverse of the In apping given by (1.50). Rearranging
(1.50), we have
R e Z
C h (1.51)
i.
Taking Z = x + j y c: R—xiy _ x+iy—R _(x+iyeR)(y+ix) . . — . — 1.52 with +Jy) (MO—ix) wR(x2+y2) ( )
Equating imaginary parts, and remembering that C is real, gives 0=x2+y2—Rx (1.53) which represents a circle, with centre at GR, 0) and of radius éR. Thus the real axis in
the. C plane is mapped onto the circle given by (1.53) in the Z plane. Of course, C is
positive. If C = 0, (1.53) indicates that Z: R. The circuit of Figure 1.34 conﬁrms that the'transformation w = 1/2, w = u +jv, transfonns the circle 3:2 +y2 = 2m: in the
. state the straight line it = l/2a in the w plane.
tang conducting wires of radius a are placed
and parallel to each other, so that their
eetion appears as in Figure 1.41. The
areseparated at O by an insulating gap of ble dimensions, and carry potentials i V0
1"inherited. Find an expression for the potential
"general point (x, y) in the plane of the cross
and sketch the equipotentiais. , the points A(—1, 0), 13(0, 1), C( 3—; g and
— 33(3, 0) in the 2 plane, _ straight liney = 0, the circlex2+y2 =1. te your answer with a diagram showing the
. w planes and shade on the w plane the region
'poridingtoni~y2 <1. “semicircular disc of unit radius, [(x, y): ' s I, y 2: 0], has its straight boundary at
rature 0 “C and its curved boundary at l00 °C.
a that the temperature at the point (x, y) is _ Zﬂtanef 2y ] 1_x2 _y2 iﬁhow that the function
GCIJ) = 2750 — y) . satisﬁes the Laplace equation and construct
itsharmonic conjugate H(x, y) that satisﬁes
'0, 0) =0. Hence obtain, in terms of z, where
fit=3: + jy, the function F such that W: F(z)
here W: G + jH. how that under the mapping w = 1112, the
‘ harmonic function 60:, y) deﬁned in (a) is
mapped into the function Gm, o) = 2e"cos v — e2" sin 21)
Verify that 004, L2) is harmonic. (c) Generalize the result (b) to prove that under
the mapping w =f(z), Wheref’(z) exists, a
harmonic function of (x, y) is transfonned
into a harmonic function of (u, U). Show that ifw : (z + 3)/(z — 3), w = u +jv, z at +9, the circle a2 + v2 = k2 in the w plane
is the image of the circle 1 + k2
1 7 k2 x2+y2+6 x+9=0 (kzil) in the 2 plane. Two long cylindrical wires, each of radius
4 mm, are placed parallel to each other with their
axes 10 mm apart, so that their cross—section
appears as in Figure I .42. The wires carry potentials
1V0 as shown. Show that the potential m, y) at the
point (x, y) is given by =£iln [(x+3)2+y2]h1[(x—3)2+ﬂ} Figure 1.42 Cylindrical wires of Exercise 69. Find the image under the mapping w =j(1ez)
l+z z:x+jy,w=u+jv,of (a) the points AU, 0), 8(0, 1), C(0, —l) in the
2 plane, (b) the straight line y = 0, (c) the circle 3:2 +y2 : I. A circular plate of unit radius, [(x, y): x2 + y2 S 1],
has one half (withy > 0) of its rim, x2 +y2 = l, at temperature 0 °C and the other half (with y < 0) at
temperature 100 °C. Using the above mapping, prove
that the steadystate temperature at the point (x, y) is ...
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 Fall '10
 ChinC.Lee

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