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714 CHAPTER 20 Complex Integration

The negative sign on the integral from σ − ib to σ + ib is due to the fact that counterclockwise
orientation on C requires that the integral over L be taken from σ +ib to σ −ib. Reversing these
limits of integration (as we have done) reverses the sign on the integral.
We will have the conclusion of the theorem if we can show that
f (z)

lim dz = 0.
b→∞
C ∗ z − z 0
A proof of this is outlined in Problem 15.

SECTION 20.3 PROBLEMS

z
In each of Problems 1 through 12, evaluate f (z)dz.All Hint: Consider (e /z)dz with γ as the unit cir-
γ γ
closed curves are positively oriented. These problems may cle about the origin. Evaluate the integral once using
involve Cauchy’s theorem, the integral formulas, and/or Cauchy’s integral formula and then directly by using
the deformation theorem. coordinate functions for γ .
4
1. f (z) = z /(z − 2i);γ is any closed path enclosing 2i. 14. Use the extended deformation theorem to evaluate
z − 4i
2
2. f (z)=sin(z )/(z − 5);γ isanyclosedpathenclosing dz
3
5. γ z + 4z
2 where γ is any closed path enclosing the origin, −2i
3. f (z) = (z − 5z +i)/(z − 1 + 2i);γ is the circle |z|=
3. and 2i.
3
2
4. f (z)=2z /(z −2) ;γ is the rectangle having vertices 15. Complete a proof of Theorem 20.9 by filling in the
4 ± i,−4 ± i. details of the following argument. Use a hypothesis of
the theorem to show that
2
z
5. f (z) = ie /(z − 2 + i) ;γ is the circle |z − 1|= 4.
n
3
6. f (z) = cos(z − i)/(z + 2i) ;γ is any closed path z f (z) ≤ M

enclosing −2i. z − z 0
for |z| to be sufficiently large. Do some manipulation
3
7. f (z) = z sin(3z)/(z + 4) ;γ is the circle |z − 2i|= 9.
to show that
8. f (z) = 2iz|z|;γ is the line segment from 1 to −i.
M

f (z) ≤
4
2
9. f (z) =−(2 + i)sin(z )/(z + 4) ;γ is any closed path n+1
z − z 0 |z| |1 − (z 0 /z)|
enclosing −4.
for |z| to be sufficiently large. Now require that b be
2
10. f (z)=(z −i) ;γ is the semicircle of radius 1 about 0 large enough that b > 2|z 0 | to show that
from i to −i.
M

11. f (z) = Re(z + 4);γ is the line segment from 3 + i to f (z) ≤ n+1 .

z − z 0 b
2 − 5i.
∗
Taking into account the length of C , show that
2
12. f (z) = 3z cosh(z)/(z + 2i) ;γ is the circle of radius
2
8 about 1. f (z) 2M
2σ

dz ≤ 4 − .
b n b
13. Evaluate C ∗ z − z 0
2π Finally, show that the right side of this inequality
e cos(θ) cos(sin(θ))dθ. approaches 0 as b →∞.
0
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October 14, 2010 15:32 THM/NEIL Page-714 27410_20_ch20_p695-714