(4.1): Contours
1. Find an admissible parametrization.
(a). the line segment from z = 1 + i to z =
2
3i.
z(t) = z1 (t) + t(z2 z1 )
z(t) = 1 + i + t( 2 3i (1 + i))
z(t) = 1 + i + t( 3 4i); 0 t 1
(b). the circle jz
2ij = 4 traversed once clockwise starting at z = 4 + 2i.
z(t) = 2i + 4e
it
;0
t
2
(c). the arc of the circle jzj = R lying in the second quadrant, from z = Ri to
z = R.
z(t) = Reit ;
=2
t
(d). the segment of the parabola y = x2 from the point (1; 1) to the point
(3; 9).
z(t) = t + it2 ; 1
2
3. Show that the ellipse xa2 +
admissible parametrization.
y2
b2
t
3
= 1 is a smooth curve by producing an
Let x = a cos t; y = b sin t. Substitute these values in the ellipse to obtain
cos2 t + sin2 t = 1, a true statement for all 0 t 2 .
5. Identify the interior of the curve. Is it positively oriented? Yes.
7. Parametrize the contour consisting of the perimeter of the square with
vertices 1 i; 1 i; 1 + i; and 1 + i. What is the length of this contour?
: z(t) = 1 i + 2t ; 0 t 1
i + 2i(t 1) ; 1 t 2
2 : z(t) = 1
:
z(t)
=
1
+
i 2(t 2) ; 2 t 3
3
:
z(t)
=
1
+
i 2i(t 3) ; 3 t 4
4
l( ) = 8.
1
8. Parametrize the contour . Also parametrize
1
.
The contour consists of of two smooth curves - the straight line piece
and the semi-circle 2 :
1
:
0
t
2
z1 (t) = ( 2 + 2i) + t( 1
1
: z2 (t) = e
i t
2
: z2 (t) = ei
1
: z1 (t) =
t
;1
t
2
;0
t
1
1 + (t
( 2 + 2i)) =
1)( 1 + 2i)
;1
t
2 + 2i + t(1
2i)
1
;
2
9. Parametrize the barbell-shaped contour with initial point
point 1.
1 and terminal
Circle 1 : z(t) = 2 + e 2 it ; 0 t 1
Straight line 2 : z(t) = 1 + 2(t 1) ; 1 t 2
Circle 3 : z(t) = 2 e 2 i(t 2) ; 2 t 3
Straight line 4 =
2:
You can reduce the above parametrization to the one given in the back of
the text. Namely,
8
2 + e 6 it
if
0 t 1=3
<
1 + 6(t 1=3) if 1=3 t 2=3
z(t) =
:
2 e6 i(t 2=3)
if
2=3 t 1:
Follow Wen’s solution discussed in class.
parametrized by z(t) = 5e3it ; 0
11. Find the length of the contour
z 0 (t) = R15e3it =) jz 0 (t)j = 15:
l( ) = 0 jz 0 (t)j dt = 15 :
(4.2): Contour Integrals
3. Evaluate the following integrals.
(a).
Z
1
(2t + it2 )dt
0
=
i
t2 + t3
3
2
1
=1+
0
i
3
t
.
(b).
Z
0
(1 + i) cos(it)dt
2
sin(it)
i
=
(1 + i)
=
(1 + i) sinh 2
0
= (1 + i)
2
sin(2i)
i
(c).
Z
1
(1 + 2it)5 dt
0
=
=
1
(1 + 2i)6
(1 + 2it)6
1
=
+
12i
12i
12i
0
11
29i
i
1 (1 + 2i)6 =
12
3
3
(d).
Z
2
0
=
=
1
2
Z
(t2
t
dt
+ i)2
4+i
u
2
dt ; u = t2 + i =) du = 2tdt limits: u = i to u = 4 + i
i
1
2u
5. Evaluate
4+i
1
1
2
+
=
2(4 + i) 2i
17
=
i
Z
C
where C : jz
7. Compute
1 + 2i.
6
(z
i)2
+
2
z
i
+1
8i
17
3(z
i)2 dz
ij = 4 traversed once counterclockwise.
Z
6
2
+
+ 1 3(z i)2 dz
2
(z
i)
z
i
C
= 6(0) + 2(2 i) + 0 + 0 = 4 i:
R
Re zdz along the directed line segment from z = 0 to z =
The line can be parametrized as z(t) = (1 + 2i)t = t + 2it for 0
z 0 (t) = (1 + 2i)dt
R
R1
Re zdz = 0 t (1 + 2i)dt = (1+2i)
[t2 ]10 = (1+2i)
:
2
2
t
1 =)
8. Let C be the perimeter of the square with vertices at the points
R z = 0; z =
1; z = 1+i; and z = i traversed once in that order. Show that C ez dz = 0.
3
Refer to the class notes for a solution that uses parametrization.
For an alternate solution consider the following:
Z
ez dz
C
Z 1
=
ez dz +
0
[ez ]10
=
=
=
+
(e
0:
Z
1+i
ez dz +
1
z 1+i
[e ]1 +
1+i
1) + (e
Z
i
ez dz +
1+i
z i
[e ]1+i + [ez ]0i
i
1+i
e) + (e
e
Z
0
ez dz
i
) + (1
ei )
R
9. Evaluate (x 2xyi)dz over the contour : z = t + it2 ; 0
z 0 = (1 + 2it)dt.
R
R1
R1
(x 2xyi)dz = 0 (t 2t3 i)(1 + 2it)dt = 0 [(t + 4t4 ) + i(2t2
h 2
i1
5
3
t4
13
= t2 + 4t5 + i 2t3
+ i 16 :
= 10
2
t
1 =)
2t3 )]dt
0
I
12. True or False:
I
zdz =
jzj=1
1
z dz.
jzj=1
The statement is true since we can use the fact that all z satisfy jzj = 1.
Indeed,
I
I
I
I
jzj2
1
zdz =
z zz dz =
dz
=
z
z dz.
jzj=1
jzj=1
13. Compute
t
.
R
jzj=1
(jz
Z
=
Z
2
1 + ij
(jz
z)dz along the semicircle z = 1
2
1 + ij
i + eit
(1
0
=
Z
(1
1+i
0
=
jzj=1
z)dz
2
(1 i + eit )) ieit dt
Z
it
it
e ) ie dt =
( eit ie2it )dt
1+i
0
eit
i
0
i 2it
e
2i
0
(4.3): Independence of Path
1. Calculate the following.
4
= i[ 1
1] =
2i.
i + eit ; 0
(a).
R
(3z 2
5z + i)dz
: line z = i to z = 1:
Z
Z
=
(3z 2
5z + i)dz
1
(3z 2
5z + i)dz
i
=
(b).
R
ez dz
=
R
1
[ez ]z=1 = e
2
sin2 z cos zdz
(e1
ez cos zdz
=
=
R
5
2
1)
(Log(z))2 dz
1
1
1:
e1 (only the end points matter)
)
=
2 sinh 1:
1
1
i
sin3 z z= = [sin i]3
3
3
i sinh3 1
:
3
0
: contour in the …gure.
Z
ez cos zdz (use integration by parts)
=
(h).
( i+
: contour in the …gure.
Z
sin2 z cos zdz
=
R
e
2
=
(f).
3
+i
2
: upper half of circle jzj = 1 from z = 1 to z =
Z
ez dz
=
(e).
5 2
z + iz]1i =
2
3 + 2i:
[z 3
=
1 z
[e (sin z + cos z)]iz=
2
1 i
[e (sin i + cos i) e (0 1)]
2
e
ei
+ (cosh 1 + i sinh 1):
2
2
: line z = 1 to z = i:
5
R
We …rst need to …nd the integral of (log(x))2 dx:
Z
I =
(log(x))2 dx ;
1
dx
x
= log xdx =) v = x[log x 1]
Z
= x log(x)[log x 1]
[log x
u
=
dv
I
log x =) du =
1]dx
= x log(x)[log x 1] x[log x 1] + x
= 2x 2x log(x) + x(log(x))2 :
Z
(Log(z))2 dz
=
2z
2zLog(z) + z(Log(z))2
=
2i
2iLog(i) + i(Log(i))2
i
z=1
[2]
2i[log jij + i ] + i[log jij + i ]2
2
2
=
2i
=
2i +
2
(i).
R
1
1+z 2 dz
i
4
2=
2 + i(2
2
2
=4):
: line z = 1 to z = 1 + i:
We will use the de…nition
of the complex inverse tangent function:
h
i
tan 1 (z) = 2i log ii+zz :
Z
=
1
dz
1 + z2
tan
1
z
1+i
z=1
tan 1 (1 + i) tan 1 (1)
i+1+i
i+1
i
i
log
log
=
2
i 1 i
2
i 1
i
i
log( 1 2i)
log( i)
=
2
2
p
i
3
=
log 5 + i( + tan 1 (2)) i
2
2
i
tan 1 (2)
=
log(5)
+ :
4
2
4
=
2. If P (z) is a polynomial and
0:
is any closed contour, explain why
P (z) Ris entire and therefore has an antiderivative. Thus, if
contour,
P (z)dz = 0:
6
R
P (z)dz =
is any closed
4. RTrue or False: If f is analytic at each point of a closed contour
f (z)dz = 0:
, then
This is false, Cauchy’s Theorem requires that f be analytic on and inside
1
.
For
R 1 example, the function f (z) = z is analytic on jzj = 1; but not inside and
z dz = 2 i 6= 0:
1
z
has no antiderivative in the punctured plane Crf0g:
R 1
Suppose f (z) = z1 has an antiderivative in C r f0g. Then
z dz = 0 for
every closed curve
2 C r f0g. But this is notR true for any loop
which
encloses the pole z = 0: For instance if : jzj = 1, jzj=1 z1 dz = 2 i.
5. Explain why f (z) =
(4.4): Cauchy’s Integral Theorem
1. Let D be the domain given on page 199 and the contour shown. Decide
if the following given contours can be deformed to .
(a). Yes
(b). No
(c). Yes
(d). No
3. Let D : 1 < jzj < 5 and : jz 3j = 1 traversed once positively starting at
z = 4. Decide which of the following contours are continuously deformable
to in D.
(a). Yes
(b). Yes
(c). No
(d). Yes
(e). Yes
10. Determine the domain of analyticity and explain why
(a). f (z) =
z
z 2 +25
RD = fz j z 6= 5ig.
f (z)dz = 0 since f is analytic on and inside
jzj=2
(b). f (z) = e
z
(2z + 1)
This is an entire function, so
R
jzj=2
f (z)dz = 0.
7
R
jzj=2
: jzj = 2.
f (z)dz = 0:
(c). f (z) =
cos z
z 2 6z+10
RD = fz j z 6= 3 ig.
f (z)dz = 0 since f is analytic on and inside
jzj=2
: jzj = 2.
D
R = fz j z 6= (2n 1) ; n = 0; 1; 2; g.
f (z)dz = 0 since f is analytic on and inside
jzj=2
: jzj = 2.
z
2
(d). f (z) = sec
0; 1; 2;
=
1
cos(z=2)
=)
z
2
6= (2n
1) 2 =) z 6= (2n
1) ; n =
(e). f (z) = Log(z + 3)
z + 3 = x + iy + 3 = x + 3 + iy. We need to avoid y = 0 and x+3 0.
D
3g.
R = C r fx + iy j y = 0; x
f
(z)dz
=
0
since
f
is
analytic
on and inside : jzj = 2.
jzj=2
R 1
13. Evaluate z2 +1 dz along each of the three contours.
i
R
R h i=2
R
i=2
1
i
dz
=
(a). 1 z21+1 dz = 1 (z+i)(z
i)
(z+i)
(z i) dz = 0
22 i = :
1
(b). Deform into a barbell.
R
R
R
1
1
dz = 1 (z+i)(z
2
i) dz =
1 z +1
(c).
R
1
1
z 2 +1 dz
=
R
1
1
(z+i)(z i) dz
=
1
R
h
i=2
(z+i)
1
h
i=2
(z+i)
i=2
(z i)
i
dz = 2i 2 i 2i 2 i = 0:
i
i=2
i
0=
:
(z i) dz = 2 2 i
R
z
15. Evaluate
: jzj = 4 traversed twice in the clockwise
(z+2)(z 1) dz; where
direction.
i
h
R
R
2=3
1=3
z
2 23 2 i + 13 2 i = 4 i:
(z+2)(z 1) dz = jzj=4 (z+2) + (z 1) dz =
17. Evaluate
R
2z 2 z+1
(z+1)(z 1)2 dz;
where
is the contour shown in the …gure.
First get the partial fractions.
2z 2 z + 1
(z + 1)(z 1)2
(2)
(1)
(2)
A0
A0
A1
+
+
(z + 1) (z 1)2
(z 1)
=
(1)
(2)
Cover-Up rule gives, A0 = 1; A0 = 1. To …nd
(2)
A1
1 d 2z 2 z + 1
z!1 1! dz
(z + 1)
d (z + 1)(4z 1) (2z 2
= lim
z!1 dz
(z + 1)2
6 2
=
= 1:
4
=
lim
8
z + 1)
R
2z 2 z+1
(z+1)(z 1)2 dz
=
R h
1
(z+1)
+
1
(z 1)2
+
1
(z 1)
i
dz = [2 i + 0
2 i] = 0:
(4.5): Cauchy’s Integral Formula and Its Consequences
1. Let f be analytic inside and on the simple closed contour . What is the
value of
Z
f (z)
1
dz
2 i
z z0
when z0 lies outside ?
Since f is analytic inside and on , the interior of the closed loop is a
simply-connected domain (call it D) in which f is analytic. Given that z0 lies
outside , the function zf (z)
z0 is analytic in D as well. By Cauchy’s Integral
R f (z)
Theorem,
z z0 dz = 0.
R f (z)
Therefore, 21 i
z z0 dz = 0.
2. Let f and g be analytic inside and on the simple loop . Prove that if
f (z) = g(z) for all z on , then f (z) = g(z) for all z inside .
Let z0 be any point inside . We need to show that f (z0 ) = g(z0 ). Applying
Cauchy’s Integral Formula we obtain
Z
Z
f (z)
1
g(z)
1
dz
dz
f (z0 ) g(z0 ) =
2 i
z z0
2 i
z z0
Z
1
f (z) g(z)
=
dz
2 i
z z0
= 0; since f (z) = g(z) for all z on .
3. Let C : jzj = 2 traversed once positively. Compute the following.
(a).
=
Z
sin 3z
dz
z
=2
C
2 i[sin 3z]z= =2 =
2 i:
(b).
Z
=
=
zez
dz
3
C 2z
Z
1
zez
dz
2 C z 3=2
3
3 i 3=2
1
2 i[zez ]z=3=2 = i e3=2 =
e :
2
2
2
9
(c).
Z
cos z
dz
3 + 9z
z
ZC
Z
cos z
cos z
dz
=
dz
=
2 + 9)
z(z
z(z
+
3i)(z 3i)
C
C
cos z
cos 0
2 i
= 2 i
=2 i
=
:
(z + 3i)(z 3i) z=0
9
9
(d). Let f (z) = 5z 2 + 2z + 1.
Z
5z 2 + 2z + 1
dz
(z i)3
C
2 i (2)
[f (i)] = 10 i:
2!
=
(e). Let f (z) = e
z
.
Z
e z
dz
2
C (z + 1)
2 i (1)
[f ( 1)] =
1!
=
(f). Let f (z) =
sin z
z 4.
Z
sin z
dz
z 2 (z 4)
2 i (1)
(z
[f (0)] = 2 i
1!
i
:
2
2 ie:
C
=
=
4. Compute
Z
C
4) cos z sin z
(z 4)2
z=0
z+i
dz;
z 3 + 2z 2
where C is
(a). circle jzj = 1 traversed once counterclockwise.
Z
z+i
dz
+ 2z 2
ZC
z+i
=
dz
2 (z + 2)
z
C
=
=
z3
2 i z+i
1! z + 2
i(2 i)
:
2
0
=2 i
z=0
10
(z + 2) (z + i)
(z + 2)2
z=0
(b). circle jz + 2
ij = 2 traversed once counterclockwise.
Z
z+i
dz
3 + 2z 2
z
ZC
z+i
=
dz
2 (z + 2)
z
C
2+i
z+i
=2 i
= 2 i
z 2 z= 2
4
=
(c). circle jz
i( 2 + i)
:
2
2ij = 1 traversed once counterclockwise.
Z
z+i
dz
3 + 2z 2
z
C
Z
z+i
=
dz
2
C z (z + 2)
= 0:
6. Evaluate
Z
(z 2
eiz
dz;
+ 1)2
where is the circle jzj = 3 traversed once counterclockwise.
Sketch the contour jzj = 3 and reduce the contour to a vertical barbell
consising of:
ij =
1
2
a circle traversed counter clockwise C2 : jz + ij =
1
2
a circle traversed counter clockwise C1 : jz
straight line C3 connecting
straight line
Let f (z) =
i
2
C3 connecting
eiz
(z+i)2
and g(z) =
to
i
2
i
2
to
eiz
(z i)2 .
Z
and
i
2
Then
eiz
dz
(z 2 + 1)2
Z
Z
Z
Z
eiz
eiz
g(z)
f (z)
=
dz +
dz +
dz +
dz
2
2
2
2
2
2
i)
C1 (z
C3 (z + 1)
C3 (z + 1)
C2 (z + i)
Z
Z
f 0 (i) 2 i
eiz
g 0 (i) 2 i
eiz
=
+
dz
dz
+
2
2
2
2
1!
1!
C3 (z + 1)
C3 (z + 1)
i cosh 1
i sinh 1
= 2 i
+2 i
= (cosh 1 sinh 1)
2
2
11
7. Compute
Z
cos z
dz;
z 2 (z 3)
where is the indicated contour.
The poles of the integrand are z = 0 (mul = 2); and z = 3. Since the
small loop does not enclose the poles we can deform to a circle, say jzj = 1.
Therefore,
Z
cos z
dz
z 2 (z 3)
2 i
d
cos z
(z 3) sin z cos z
=
lim
= 2 i lim
z!0
1! z!0 dz (z 3)
(z 3)2
2 i
=
:
9
11. Let f = u + iv be analytic in a domain D. Explain why
in D.
@2u
@x2
is harmonic
By Theorem 16, all derivatives of f exist and are analytic. In particular,
f (1) (z) = ux + ivx ; and f (2) (z) = uxx + ivxx .
2
Since f (2) (z) = uxx + ivxx is analytic, its real part, @@xu2 ; is harmonic in D.
12