Caushys Theorem
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Transcript of Caushys Theorem
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Homework #7 Solutions
Math 128, Fall 2013Instructor: Dr. Doreen De Leon
1 HW #7(a)
1.1 p. 135: 1, 3
Use parametric representation of C to evaluate C f (z) dz1. f (z) = (z2 )z and C is
(a) the semicircle z = 2 ei (0 )Solution:
C f (z) dz =
0f [z()]z () d
=
0(2e
i+ 2)
2e i 2ie i d
= 2 i
0(ei + 1) d
= 2 i1i
e i + |0
= 2 i1i
e i + 1i
ei (0) + 0
= 2 i 1i +
1i
= 2 i 2i + = 4 + 2 i .
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(b) the semicircle z = 2 ei ( 2 )Solution:
C f (z) dz = 2
f [z()]z () d
= 2
(2ei + 2)
2ei 2ie i d
= 2 i 2
(e i + 1) d
= 2 i1i
ei + |2= 2 i
1i
ei2 + 2 1i
ei ( ) +
= 2 i1i + 2
1i +
= 2 i 2i +
= 4 + 2 i .
(c) the circle z = 2e i (0 2 )Solution: Let C 1 be the contour from part (a) and C 2 be the contour from part (b).Then C = C 1 + C 2, so
C f (z) dz = C 1 f (z ) dz + C 2 f (z) dz= (
4 + 2 i ) + (4 + 2 i )
= 4 i .
3. f (z) = exp(z ) and C is the boundary of the square with vertices at 0 , 1, 1 + i, and i, theorientation of C being counterclockwise.Solution:
1
11 + iC 3
C 2
C 1
C 4
So, C = C 1 + C 2 + C 3 + C 4, and a parameterization for the curves is
C 1 : z(t ) = t, 0 t 1 = z (t ) = 1
C 2 : z(t ) = 1 + ( t 1)i, 1 t 2 = z (t) = i
C 3 : z(t ) = (3 t ) + i, 2 t 3 = z (t ) = 1
C 4 : z(t ) = (4 t )i, 3 t 4 = z (t ) = i
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Integrate f (z) along each contour.
C 1 f (z) dz = 1
0f [z(t )]z (t ) dt
= 1
0 et
1 dt= et 10= e 1.
C 2 f (z) dz = 2
1f [z(t )]z (t ) dt
= 2
1 exp( (1 (t 1)i)) i dt
= 2
1ie e (t 1) i dt
= e 21 ie ( t 1) i dt= e e ( t
1) i 2
1
= e 1 ei
= 2 e .
C 3 f (z) dz = 3
2f [z(t )]z (t ) dt
=
3
2
exp( ((3
t )
i))(
1) dt
= 3
2e i e (3 t ) dt
= e i e (3 t )3
2
= 1 (1 e )= e 1.
C 4 f (z) dz = 4
3f [z(t )]z (t ) dt
=
4
3 exp(
(4
t )i)(
i) dt
= 4
3 ie ( t 4) i dt
= e ( t 4) i 4
3
= 1 ei= 2.
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So,
C f (z) dz = C 1 f (z) dz + C 2 f (z) dz + C 3 f (z) dz + C 4 f (z) dz= e
1 + 2 e + ( e
1)
2
= 4 e 4 = 4 ( e 1) .
1.2 Problem 2
Evaluate C 1(z 1) dz , where C is the circle of radius 2 centered at 1, traversed counterclockwise.Solution: Parameterization of C : z() = 1 + 2 e i , 0 2 . Then, z () = 2 ie i . Therefore,
C
f (z) dz =
2
0
f [z()]z () d
= 2
0
1(1 + 2 ei ) 1
2ie i d
= 2
0i d
= i|20= 2 i .
1.3 Problem 3
Evaluate C z2 dz along two paths joining 0 to 1 + i as follows:
(a) C is the straight line segment joining 0 to 1 + i.
(b) C is the union of the line segment joining 0 to 1, then joining 1 to 1 + i.
What do you observe?Solution: Let f (z) = z2.
(a)
1
11 + i
A parameterization of C is
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z (t ) = (1 + i)t, 0 t 1= z
(t ) = 1 + i.
Then,
C f (z) dz = 1
0f [z(t )]z (t ) dt
= 1
0[(1 + i)t]2(1 + i) dt
= (1 + i) 1
0(1 i)2t 2 dt
= 2i(1 + i) 1
0t 2 dt
= 2i(1 + i) t331
0
= 2i(1 + i)13
= 23
23
i .
(b)
C 1
C 21
C 1 : z(t) = t, 0 t 1 = z (t ) = 1C 2 : z(t) = 1 + ( t 1)i, 1 t 2 = z (t) = i Then,
C 1 f (z) dz = 1
0[t]2 1 dt
= 1
0t 2 dt
= t3
3
1
0
= 13
.
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C 2 f (z) dz = 2
1[(1 + ( t 1)i)]2 i dt
= i 2
1(1 + (1 t )i)2 dt
= i 2
1(1 + 2(1 t )i (1 t )2) dt
= i t (1 t )2i + 13
(1 t )32
1
= i 2 i 13 1
= i23 i
= 1 + 23
i.
Then,
C f (z) dz = C 1 f (z) dz + C 2 f (z) dz=
13
+ 1 + 23
i
= 43
+ 23
i .
The integrals in (a) and (b) yield different results. Therefore, the integral is not path-independent.
2 HW #7(b)
2.1 Problem 1
Find a number M such that
C dzz 2 + 2 M,where C is the upper half of the unit circle.Solution:
1z2 + 2
1
|z2 + 2 |
1
|z2 (2)|
1
||z|2 | 2|| =
1
||1|2 2| = 1 .
Therefore,
C dz
z2 + 2 1(length of C ) = = M.
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2.2 p. 141: 4
Let C R denote the upper half of the circle |z| = R (R > 2), taken in the counterclockwise direction.Show that
C R2z2
1
z4 + 5 z2 + 4 dz R (2R 2 + 1)(R 2 1)(R 2 4) .
Then, show that the value of the integral tends to 0 as R .Solution:
Find an upper bound on 2 z2 1.
|2z2 1| 2|z |2 + 1= 2 R 2 + 1 .
Find a lower bound on z4
+ 5z2
+ 4.
|z4 + 5 z2 + 4 | = |(z2 1)(z2 4)|= |z2 1||z2 4| ||z|2 1|||z|2 4| = ( R 2 1)(R 2 4).
Therefore,2z2 1
z4 + 5 z2 + 4 2R 2 + 1
(R 2 1)(R 2 4).
So,
C R 2z2 1
z4 + 5 z 2 + 4 dz
2R 2 + 1(R 2 1)(R 2 4)
(length of C R )
= R (2R 2 + 1)(R 2 1)(R 2 4)
.
As R ,
0 limR C R 2z2 1
z4 + 5 z2 + 4 dz limR
R (2R 2 + 1)(R 2 1)(R 2 4)
= 0
=
limR
C R
2z2 1z4 + 5 z2 + 4
dz = 0
2.3 p. 149: 1, 2
1. Use an antiderivative to show that for every contour C extending from a point z1 to a pointz2,
C z n dz = 1n + 1 z n +12 z n +11 (n = 0 , 1, 2, . . . ).
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Solution:
C z n dz = z2
z 1z n dz
= zn +1
n + 1
z 2
z 1
= zn +12n + 1
zn +11n + 1
= 1n + 1
z n +12 z n +11 .
2. By nding an antiderivative, evaluate each of these integrals, where the path is any contourbetween the indicated limits of integration.
(a)
i2
iez dz
Solution:
i
2
iez dz =
1
ezi2
i
= 1
ei(2 ) ei
= 1
(i + 1) .
(b) +2 i
0cos
z2
dz
Solution:
+2 i
0cos
z2
dz = 2sinz2
+2 i
0
= 2sin + 2 i
2 2sin0= 2sin
2
+ i
= 2 sin2
cos i + cos2
sin i
= 2 cos i
= 2e i
2
e i
2
2
= e 1 + e.
(c) 3
1 (z 2)3 dzSolution:
3
1(z 2)3 dz =
14
(z 2)43
1
= 14
(3 2)4 14
(1 2)4= 0 .
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3 HW #7(c)
3.1 p. 160: 1(b), (c), (f)
Apply the Cauchy-Goursat Theorem to show that C f (z) dz = 0, where the contour C is the unitcircle |z| = 1, in either direction, and when(b) f (z) = ze z
Solution: f (z) = ze z is entire = f is analytic on and inside the contour C . Therefore,
C ze z dz = 0.(c) f (z) =
1(z2 + 2 z + 2)
Solution: f (z) is analytic for all z except where z2 + 2 z + 2 = 0 =
z =
1 + i,
1
i .
Neither of these points is on or inside C . Therefore, f is analytic on and inside the contourC
= C z(z2 + 2 z + 2) dz = 0 .(f) f (z) = Log ( z + 2)
Solution: Log (z) is analytic everywhere except the set {(x, y )|x 0 and y = 0}, andw = z + 2 is entire. Note that w = ( x +2)+ iy . Therefore, Log ( z + 2) is analytic everywhereexcept the set
{(x, y )|x + 2 0 and y = 0}= {(x, y )|x 2 and y = 0}.The set where f is not analytic is not on or inside C =
f is analytic on and inside C
= C Log (z + 2) dz = 0 .3.2 p. 161: 2(a), (c)
Let C 1 denote the positively oriented boundary of the square whose sides lie along the line x =
1, y = 1 and let C 2 be the positively oriented circle |z| = 4. Show that
C 1 f (z) dz = C 2 f (z) dz,where
(a) f (z) = 1
3z2 + 1Solution:
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1
-1
-4
4C 2
C 1
f (z) = 1
3z2 + 1 is analytic for all z except where 3 z2 + 1 = 0 = f is analytic for all
z= 1 3.
So, f is analtyic on C 1 and C 2 and between C 1 and C 2. Therefore, by the Path DeformationPrinciple,
C 1 13z2 + 1 dz = C 2 13z2 + 1 dz.(c) f (z) =
z1 ez
Solution: f (z) = z1 ez
is analytic for all z such that
1 ez = 0=
ez
= 1
But,
ez = 1
= ex eiy = 1 e i0
= x = ln1 and y = 0 + 2 n, n Z .
Therefore, f is analytic for
z= 2 ni, n Z .
So, f is analytic for z = 0 , 2i, 4 i , . . . = f is anlytic on C 1 and C 2 and between C 1and C 2. Therefore, C 1 z1 ez dz = C 2
z1 ez
dz.
3.3 Problem 3
Evaluate the following
(a) C (z3
+ 3) dz , where C is the upper half of the unit circle traversed counterclockwise.
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C 1
-1 1 -1 1
So,
C (z3 + 3) dz = 1
11(z3 + 3) dz
= z4
4 + 3 z
1
1
= 6 .
(b) C (z3 + 3) dz , where C is the unit circle traversed clockwise.Solution: f (z) = z3 + 3 is entire, and so is analytic on and inside C . Therefore, by theCauchy-Goursat theorem,
C (z3 + 3) dz = 0 .(c)
C
e1z dz , where C is the circle of radius 3 centered at 1 + 5 i traversed counterclockwise.
e1z is analytic for all z= 0, because
1z
is analytic for all z= 0 and ez is entire. z0 = 0 is not on
or inside C , so e1z is analytic on and inside C . Therefore,
C e 1z dz = 0by the Cauchy-Goursat theorem.
(d) C cos 3 + 1z 3 dz , where C is the unit square with corners at 0, 1, 1 + i, and i, traversedclockwise.Solution:
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1
1 C
32
f (z) = cos 3 + 1z 3
is analytic for all z= 3 = f is analytic on and inside C . So, by the
Cauchy-Goursat theorem,
C cos 3 + 1z 3 dz = 0 .
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