mytut11s
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The University of Sydney
School of Mathematics and Statistics
Solutions to Tutorial 11: Complex integrals
MATH3068 Analysis Semester 2, 2009
Web Page: http://www.maths.usyd.edu.au:8000/u/UG/SM/MATH3068/Lecturer: Donald Cartwright
1. Evaluate the integral
C
|z| dz along the curve C , where C is
(a) the line segment from 0 to 1− i;
Solution: The curve C is parametrized by z (t) = (1 − i)t, 0 ≤ t ≤ 1. On this segment C ,|z| =
√ t2 + t2 =
√ 2 t (since t ≥ 0), and z ′(t) = (1 − i). So
C
|z| dz =
10
|z(t)|z′(t) dt =
10
√ 2 t(1− i) dt =
√ 2(1− i)
t2
2
10
=
√ 2
2 (1− i).
(b) the circle C 1(0) = {z ∈ C : |z| = 1}.
Solution: On C , |z| = 1 and so C |z| dz =
C
1 dz = 0 by Cauchy’s Theorem, because theconstant function 1 is differentiable. Alternatively, C is parametrized by z = eit, 0 ≤ t ≤ 2π,and z ′(t) = ieit, so
C
|z| dz =
2π0
|z(t)|z′(t) dt =
2π0
ieit dt = eit2π0
= 0.
2. Evaluate
C
ez dz, where C is
(a) the line segment from 1 to i;
Solution: Because ez
= F ′
(z) for all z, where F (z) = ez
, the integral of ez
along a curve C is F (β ) − F (α) = eβ − eα, where α and β are the start and finish points of C .
In both parts (a) and (b), C is a curve from 1 to i. The integral therefore has the samevalue in parts (a) and (b):
C
ez dz = ei − e = cos 1− e + i sin1.
(b) the circular arc from 1 to i along the circle |z| = 1;
Solution: See the solution of the previous part.
(c) the circle C 2(0) = {z ∈ C : |z| = 2}, taken once anti-clockwise.
Solution: The integral is around a closed curve, starting at α = 2 and ending at β = 2.So it equals eβ − eα = 0.
3. Evaluate
C
(2z + i) dz where C is
(a) the circular arc from 3 + 4i to −5 along the circle |z| = 5;
Solution: Since 2z + i = F ′(z) for F (z) = z2 + iz, the integral of 2z + i along a curve C isF (β )−F (α), where α and β are the start and finish points of C . So the value of
C
2z + i dzdepends only on the end-points of C .
C
2z + i dz = F (−5) − F (3 + 4i) =
z2 + iz
−5
3+4i
= 36− 32i.
This method would work with 2z + i replaced by any polynomial.
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(b) the straight line segment from −5 to 3 + 4i;
Solution: The curve has the same end-points as in part (a) but is directed from −5 to3 + 4i, so that α and β are the reverse of what they were in part (a). The value of theintegral is therefore −36 + 32i.
(c) the circle C 5(0) = {z ∈ C : |z| = 5}, taken once anti-clockwise.
Solution: The curve is closed, i.e., α = β , and so the integral is zero.
4. Evaluate C
f (z) dz , where f (z) = ez
z and C is the circle C 2(1) = {z ∈ C : |z− 1| = 2}, traversed
in the anti-clockwise direction.
Solution: Since ez differentiable on and inside C , we can put a = 0 in Cauchy’s Integral Formula,obtaining
C
ez
z dz = 2πie0 = 2πi.
5. Evaluate the following integrals C
f (z) dz . All the circles are traversed once in the anticlockwisedirection.
(a) f (z) = 2zz2+1
, C is the circle C 3(0) = {z ∈ C : |z| = 3}.
Solution:
1z + i
+ 1z − i
= z − i + x + i(z + i)(z − i)
= 2zz2 + 1
.
So C
2z
z2 + 1 dz =
C
1
z + i +
1
z − i
dz =
C
1
z + i dz +
C
1
z − i dz = 2πi + 2πi = 4πi
since C
1
z − a dz = 2πi
whenever C winds once around a, as shown in lectures.
(b) f (z) = 1z2+5z+6
, C is the unit circle C 1(0) = {z ∈ C : |z| = 1}.
Solution: Because z 2 + 5z + 6 = (z + 2)(z + 3), the only problem points are
−2 and
−3,
which are outside C . So the integrand is differentiable on and inside C , and therefore C
dzz2+5z+6
= 0 by Cauchy’s theorem.
(c) f (z) = cos zz(z2+1)
, C is the circle C 13
(0) = {z ∈ C : |z| = 13}.
Solution: The only problem points for this function are at 0 and ±i. Of these, only z = 0lies inside C . So we can apply the Cauchy Integral Formula to g(z) = cos z
z2+1 and a = 0,
getting C
cos z dz
z(z2 + 1) dz =
C
g(z)
z dz = 2πig(0) = 2πi.
(d) f (z) = z3+2z(z−1)3
, C is the circle C 12
(0) = {z ∈ C : |z| = 12}.
Solution: The integral is 0 by Cauchy’s Theorem, since f (z) is differentiable on andinside C .
6. Evaluate the integral ∞
−∞
1
x2 + 2x + 2 dx
by integrating 1z2+2z+2
around the closed curve consisting of the line segment from −R to R (where
R >√
2), followed by the semicircle above the x-axis going from R around to −R.
Solution:
•
−
1+i
R
2
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The denominator z2 + 2z + 2 factors into (z − α)(z − β ), where α = −1 + i and β = −1 − i. So
we can think of the integrand as f (z)z−α
, where f (z) = 1z−β
, which is differentiable on and inside C .So the integral around C is
2πif (α) = 2πi 1
α − β = π.
The horizontal part of the curve is parametrized by z (t) = t, −R ≤ t ≤ R. So the integral alongthat part is simply R−R
1
t2 + 2t + 2 dt.
On the semicircular part C ′ of C , z2 + 2z + 2 is dominated by the z2 term when R is large. Oneway to express this is by writing
|z2 + 2z + 2| = R21 +
2
z +
2
z2
≥ R2
1− 2
R − 2
R2
≥ R2
1 − 2
3 − 2
9
=
R2
9 once R ≥ 3.
Hence the integrand is in modulus at most 9R2 on C ′. Now C ′ has length π R, and so
C ′
1
z2 + 2z + 2 dz ≤
9
R2 · πR =
9π
R ,
which tends to 0 as R →∞. Therefore the second term on the right in the equation
π =
C
1
z2 + 2z + 2 dz =
R−R
1
t2 + 2t + 2 dt +
C ′
1
z2 + 2z + 2 dz
is thought of as an error term E (R), and has modulus at most 9π/R. Hence
R−R
1
t2 + 2t + 2 dt = π − E (R) → π as R →∞.
That is, ∞−∞
1t2 + 2t + 2
dt = π.
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