Download MATH 417 Homework 5: Integration of Complex Functions - Prof. David S. Cabrera and more Assignments Mathematics in PDF only on Docsity!
MATH 417 Homework 5 Instructor: D. Cabrera Due Friday, July 16
- Show that (^) ∫
C
f (z) dz = 0
where C is the circle |z| = 2 oriented clockwise for each function below:
(a) f (z) = ze−z
(b) f (z) =
z^2 + 9
- If C is the unit circle |z| = 1 oriented clockwise, then is ∫
C
Log (z + 3) dz = 0?
Why or why not? Recall that Log z is the principal logarithm where |z| > 0 and −π < arg z < π.
- Evaluate (^) ∫
C
dz z^2 − 1 where C is the circle |z| = 2 oriented counterclockwise.
- Evaluate (^) ∫
C
cos z z(z + 2)
dz
where C is the square of side 6 centered at z = 0 and oriented counterclockwise.
- Evaluate (^) ∫
C
ez (z − π)^3
dz
where C is the square of side 4 centered at z = 0 oriented counterclockwise.
- Evaluate (^) ∫
C
2 z + 1 z^4 − 2 z^2 + 1
dz
where C is the circle |z| = 10 oriented clockwise.
MATH 417 Homework 5 Instructor: D. Cabrera Due July 16
- Show that (^) ∫
C
f (z) dz = 0
where C is the circle |z| = 2 oriented clockwise for each function below:
(a) f (z) = ze−z
(b) f (z) =
z^2 + 9
Solution:
(a) The function f (z) = ze−z^ is entire and C is a simple closed contour. By the Cauchy-Goursat Theorem, the value of the integral is 0.
(b) The function f (z) =
z^2 + 9
has singular points at x = 3i and x = − 3 i. Each of these points lies outside of C. Therefore, f (z) is analytic everywhere on and inside C so, by the Cauchy-Goursat Theorem, the value of the integral is 0.
- If C is the unit circle |z| = 1 oriented clockwise, then is ∫
C
Log (z + 3) dz = 0?
Why or why not? Recall that Log z is the principal logarithm where |z| > 0 and −π < arg z < π.
Solution: The function Log (z +3) has sin- gular points at all points on the branch cut which starts at z = −3 and extends along the negative real axis. These points lie out- side of the unit circle so, by the Cauchy- Goursat Theorem, the value of the integral is 0.
x
y
C
branch cut 1
Solution: The function
cos z z(z + 2)
has singularities at z = − 2 , 0 and both points lie inside the contour C. Again, we can evaluate the integral in a few different ways. One way is to start with a partial fraction decomposition of the function:
cos z z(z + 2)
cos z z
cos z z + 2
Then we have (^) ∫
C
cos z z(z + 2)
C
cos z z
dz −
C
cos z z + 2
dz
Now we can evaluate each integral using the Cauchy Integral Formula as C is a simple closed contour. In the first integral, we have f (z) = cos z, which is analytic on and inside C, and z 0 = 0 so ∫
C
cos z z
dz = 2πi cos 0 = 2πi
In the second integral, we have f (z) = cos z, which is analytic on and inside C, and z 0 = −2 so ∫
C
cos z z + 2
dz = 2πi cos(−2) = 2πi cos 2
The value of the integral is then:
∫
C
cos z z(z + 2)
dz =
(2πi) −
(2πi cos 2) = πi(1 − cos 2)
- Evaluate (^) ∫
C
ez (z − π)^3
dz
where C is the square of side 8 centered at z = 0 oriented counterclockwise.
Solution: First, we recognize that z = π is a singular point of the integrand and lies inside the simple closed contour C. Now let f (z) = ez^ which is analytic on and inside C. We can then use the extended Cauchy Integral Formula: ∫
C
f (z) (z − z 0 )n+^
dz =
2 πi n!
f (n)(z 0 )
where n = 2 and z 0 = π. The value of the integral is then: ∫
C
ez (z − π)^3
dz =
2 πi 2!
d^2 dz^2
ez
z=π
= πieπ
- Evaluate (^) ∫
C
2 z + 1 z^4 − 2 z^2 + 1
dz
where C is the circle |z| = 10 oriented clockwise.
Solution: First, we’ll factor the integrand into: 2 z + 1 z^4 − 2 z^2 + 1
2 z + 1 (z − 1)^2 (z + 1)^2
The singular points of the function are z = 1, −1. Instead of performing the par- tial fraction decomposition, we will deform C into two simple closed contours oriented counterclockwise, each of which encloses a singular point. This is possible because the function is analytic on C, the new con- tours, and everywhere in between (the re- gion in gray). Let C 1 be the contour enclos- ing z = 1 and C 2 be the contour enclosing z = −1. Then
x
y
C
-1 1
C 2 C 1
C
2 z + 1 z^4 − 2 z^2 + 1
dz =
C 1
2 z+ (z+1)^2 (z − 1)^2
dz +
C 2
2 z+ (z−1)^2 (z + 1)^2
dz
In the first integral, we have f (z) =
2 z + 1 (z + 1)^2
which is analytic everywhere on and inside C 1 , z 0 = 1, and n = 1. Using the extended Cauchy Integral Formula we have: ∫
C 1
2 z+ (z+1)^2 (z − 1)^2
dz =
2 πi 1!
d dz
2 z + 1 (z + 1)^2
z=
= 2πi
πi 2
In the second integral, we have f (z) =
2 z + 1 (z − 1)^2
which is analytic everywhere on and inside C 2 , z 0 = −1, and n = 1. Using the extended Cauchy Integral Formula we have: ∫
C 2
2 z+ (z−1)^2 (z + 1)^2
dz =
2 πi 1!
d dz
2 z + 1 (z − 1)^2
z=− 1
= 2πi
πi 2
Thus, the value of the integral is ∫
C
2 z + 1 z^4 − 2 z^2 + 1
dz = −
πi 2
πi 2
