Complex Homework III
ˆ
1. Let f (z) = x2 + iy 2 . Evaluate
f (z) dz, where C is:
C
(a) the straight line joining 1 to 2 + i;
(b) the curve (1 + t) + t2 i, 0 ≤ t ≤ 1.
Are the results the same? Why might you expect this?
Solution. (a) The straight line joining 1 and 2+i may be parametrized as z(t) = 1(1−t)+(2+i)t =
(1 + t) + ti where 0 ≤ t ≤ 1. We have
z 0 (t) = 1 + i
f (z(t)) = (1 + t)2 + i(t2 ) = (1 + 2t + t2 ) + t2 i
f (z(t))z 0 (t) = (1 + 2t) + (1 + 2t + 2t2 )
Hence
ˆ
ˆ
ˆ
1
1
1 + 2t dt + i
1 + 2t + 2t2 dt
0
0
2 3 1
2 1
2
= t+t 0+i t+t + t
3 0
8
= 2 + i.
3
f (z) dz =
C
(b) We have
z 0 (t) = 1 + 2ti
f (z(t)) = (1 + t)2 + t4 i = (1 + 2t + t2 ) + t4 i
f (z(t))z 0 (t) = (1 + 2t + t2 − 2t5 ) + (2t + 4t2 + 2t3 + t4 )
Hence
ˆ
ˆ
2
5
1
1 + 2t + t − 2t dt + i
0
0
1
1
1
= t + t2 + t3 − t6 + i t2 +
3
3 0
91
= 2+ i
30
f (z) dz =
C
ˆ
1
2t + 4t2 + 2t3 + t4 dt
4 3 1 4 1 5 1
t + t + t
3
2
5 0
Since the function f (z) = x2 + iy 2 is not analytic, it is not surprising that we get two different
answers.
ˆ
2
2
2. Let f (z) = −2xy + i(x − y ). Evaluate
f (z) dz, where C is:
C
(a) the straight line from 0 to 1 + i;
(b) the curve t3 + ti, 0 ≤ t ≤ 1.
1
Are the results the same? Why might you expect this?
Solution. (a) Let z(t) = t + ti, 0 ≤ t ≤ 1 be the straight line joining 0 to 1 + i. We have
z 0 (t) = 1 + i
f (z(t)) = −2t2
f (z(t)) = −2t2 − 2t2 i
Hence,
ˆ
ˆ
ˆ
1
1
−2t2 dt
− 2t dt + i
0
0
2 3 1
2 3 1
= − t
+i − t
3 0
3 0
2 2
= − − i.
3 3
2
f (z) dz =
C
(b) We have
z 0 (t) = 3t2 + i
f (z(t)) = −2t4 + (t6 − t2 )i
f (z(t))z 0 (t) = (−7t6 + t2 ) + (3t8 − 5t4 )i
Hence,
ˆ
ˆ
ˆ
1
6
C
1
− 7t + t dt + i
3t8 − 5t4 dt
0
0
1
1
1
1
3
9
5
7
+i t −t
= −t + t
3 0
3
0
2 2
= − − i.
3 3
f (z) dz =
2
In this case, the function f (z) is analytic (in fact, we have f (z) = iz 2 ), so it is not surprising that
the value is the same (Cauchy’s Theorem implies the integral is path independent).
3. Compute
ˆ
2
zez dz,
C
where C is the curve parametrized by z(t) = t − t3 i, 0 ≤ t ≤ 1.
2
2
Solution. The function f (z) = zez is analytic on all of C, and has antiderivative F (z) = 12 ez ,
which is also analytic on all of C. Therefore,
ˆ
1 −2i
2
zez dz = F (z(1)) − F (z(0)) =
e
−1 .
2
C
4. Is the complex function
f (z) = zez
2
holomorphic or not? Be sure to explain your reasoning (saying “it has a z in it” is not sufficient!).
Solution. We write f (z) in terms of its real and imaginary parts:
f (z) = zez = (x − iy)(ex cos(y) + iex sin(y)) = (xex cos(y) + yex sin(y)) + i(xex sin(y) − yex cos(y))
We now have
∂u
= (xex + ex ) cos(y) + yex sin(y)
∂x
∂v
= xex cos(y) − ex (cos(y) − y sin(y))
∂y
∂v
−
= −(ex + xex ) sin(y) + yex cos(y)
∂x
∂u
= −xex sin(y) + ex (sin(y) + y cos(y))
∂y
Since the Cauchy-Riemann equations are not satisfied, the function cannot be holomorphic.
ˆ
1
5. Let C be a circle centered at 4+i of radius 1. Without any calculation, explain why
dz = 0.
C z
Solution. The function z1 only has a pole at z = 0, and is analytic otherwise. Since the curve C
does not enclose the singularity, Cauchy’s Theorem implies
ˆ
1
dz = 0.
C z
6. Let C be the curve defined parametrically as follows:
ˆ
z(t) = t(1 − t)et + cos(2πt3 )i,
0 ≤ t ≤ 1.
2
ez dz. Be sure to explain your reasoning!
Evaluate the integral
C
2
Solution. Note that z(1) = z(0). Since the curve is closed and the function ez is analytic on all
of C, Cauchy’s theorem again implies
ˆ
2
ez dz = 0.
C
7. Use Cauchy’s integral formula to evaluate
ˆ
C
ez
dz,
z−1
where C is the circle of radius 4 centered at 0, oriented counterclockwise.
Solution. We apply Cauchy’s Integral Formula with f (z) = ez , z0 = 1 (this is valid since z0 is
contained inside C). Hence
ˆ
ez
dz = 2πie1 = 2πie.
C z−1
8. Let C be the unit circle centered at 0 in C, oriented counterclockwise. Evaluate each of the
following integrals, and be sure that you can justify your answer by a calculation or a clear and
concise explanation if you use any theorem.
ˆ
ˆ
ˆ
2
e−z
4
dz;
(c)
z −5 dz;
(a)
z dz;
(b)
C
C z − i/2
C
3
ˆ
(d)
C
ˆ
z 2 − 1/3
dz;
z+5
(e)
C
1
dz;
(12z − 5)2
ˆ
(f)
C
e−2z
dz.
3z + 2
Solution. We will mostly use Cauchy’s Theorem (CT) or the Cauchy Integral Formula (CIF) to
evaluate. Note that sometimes we have to manipulate the integrand in order to get it to a form
where we can apply the Cauchy Integral Formula. We indicate where we use these theorems (be
sure you know why they are being used!).
(a)
ˆ
CT
z 4 dz = 0
C
(b)
ˆ
2
C
e−z
2
CIF
dz = 2πie−(i/2) = 2πie1/4
z − i/2
(c) The function f (z) = z −5 is analytic on all of C without the origin, and has antiderivative
F (z) = − 14 z −4 , which is analytic on the same set. Therefore
ˆ
z −5 dz = F (1) − F (1) = 0.
C
(d)
ˆ
C
(e) The function f (z) =
F (z) =
−1
12(12z−5) .
1
(12z−5)2
is analytic on all of C without the point
Therefore
ˆ
C
(f)
z 2 − 1/3
CT
dz = 0
z+5
ˆ
C
1
dz = F (1) − F (1) = 0.
(12z − 5)2
e−2z
1
dz =
3z + 2
3
ˆ
C
4
e−2z
CIF 2πi 4/3
e
2 dz =
3
z+3
5
12 ,
and has antiderivative