Informal lecture notes for Dec. 4
Now that we have a solid background in the basics of complex differentiation,
it’s time to move on to complex integration.
1
Contour integrals
Consider a curve
C
in the complex plane parametrized by
z
(
t
) for
a
≤
z
≤
b
.
Of course, we can write this in terms of its real and imaginary parts as
z
(
t
) =
x
(
t
)+
iy
(
t
). Now suppose we have a (complex) function
f
(
z
) =
u
(
x, y
)+
iv
(
x, y
).
We want to integrate
f
along the curve
C
. In order to define such an integral
we start, as usual, with a Riemann sum approximation.
If we subdivide the
curve by subdividing the interval at points
a
=
t
0
, t
1
, . . . , t
n
=
b
, then the
corresponding Riemann sum looks like
n
j
=1
f
(
z
(
t
))
·
(
z
(
t
i
)

z
(
t
i

1
))
where the “dot” here refers, of course, to complex multiplication. (This is the
difference with the definition of a real line integral, in which both factors are
vectors and we take their dot product.) Then the
contour integral
of
f
along
C
, written
C
f
(
z
)
dz
, is defined to be the limit as
n
→ ∞
, assuming that this
limit exists.
This makes a good definition, but in order to compute with it, we will re
write it in terms of the parametrization and the real and imaginary parts of
f
and
z
. A little computation shows that
C
f
(
z
)
dz
=
b
a
[
u
(
x
(
t
)
, y
(
t
))
x
(
t
)

v
(
x
(
t
)
, y
(
t
))
y
(
t
)]
dt
+
i
b
a
[
v
(
x
(
t
)
, y
(
t
))
x
(
t
) +
u
(
x
(
t
)
, y
(
t
))
y
(
t
)]
dt
=
C
u dx

v dy
+
i
C
v dx
+
u dy.
An easy way to remember this is by writing the complex differential as
dz
=
dx
+
i dy
and then multiplying by
f
(
z
) written in terms of its real and imaginary
parts.
Example:
Let
C
be the quartercircle from 1 to
i
and let
f
(
z
) =
z
2
. Then
we can parametrize
C
as
z
(
t
) = cos
t
+
i
sin
t
for 0
≤
t
≤
π/
2. In terms of this
parametrization, we have that
f
(
z
(
t
)) = cos
2
t

sin
2
t
+
i
(2 cos
t
sin
t
). Further,
we have
dz
=
dx
+
i dy
= sin
t dt
+
i
cos
t dt
, and thus, after some algebra, we
find that
f
(
z
(
t
))
·
dz
=
sin
3
t

3 sin
t
cos
2
t
+
i
(
cos
3
t

3 cos
t
sin
2
t
)
dt
=
sin
t

4 sin
t
cos
2
t
+
i
(
cos
t

4 cos
t
sin
2
t
)
dt.
1
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Note that the last line is obtained using cos
2
t
= 1

sin
2
t
and vice versa, and
we have done this with an eye toward the fact that we’re about to integrate this
expression. We have
C
f
(
z
)
dz
=
π/
2
0
(
sin
t

4 sin
t
cos
2
t
)
dt
+
i
π/
2
0
(
cos
t

4 cos
t
sin
2
t
)
dt
=

cos
t
+
4
3
cos
3
t
π/
2
0
+
i
sin
t

4
3
sin
3
t
π/
2
0
=

1
3

1
3
i.
Before leaving this example, it is worth noting that these computations can
be streamlined by taking better advantage of complex functions. In particular,
this parametrization of the curve
C
can be rewritten as
z
(
t
) =
e
it
for 0
≤
t
≤
π/
2. Then we have
f
(
z
(
t
)) =
e
2
it
. Further, we see that
dz
=
ie
it
dt
, and thus
C
f
(
z
)
dz
=
i
π/
2
0
e
3
it
dt
=
i
π/
2
0
cos(3
t
)
dt

π/
2
0
sin(3
t
)
dt
=
i
1
3
sin(3
t
)
π/
2
0
+
1
3
cos(3
t
)
π/
2
0
=

1
3
i

1
3
.
Note that here we wrote
f
(
z
(
t
)) =
e
2
it
= cos(2
t
) +
i
sin(2
t
), while above we had
that this was equal to cos
2
t

sin
2
t
+
i
(2 cos
t
sin
t
). However, these can be seen
to be equal using the doubleangle formulas. In fact (to continue this digression
one more step), the relationship between the exponential and trig functions in
complex analysis provides an efficient way to derive and/or remember various
trig identities.
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 Spring '07
 NEEL
 Calculus, Integrals, Line integral, Complex number, Cr

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