Chapter 9: Cauchys integral formula

Question 1

Theorem 9.1

Deforming contours

Answer 1

Let γ be a simple contour described in the positive direction.
Let z0 be a point inside γ, and let C : z = z_0 + re^it (0 ≤ t ≤ 2π) where r is small
enough for C to lie inside γ.

Suppose f is analytic on a region D which contains γ and
C and all points in between (so D need not be simply-connected). Then

∫_γ f = ∫_C f

Question 2

Theorem 9.2 (Cauchy’s integral formula)

Answer 2

Let γ be a simple contour described in the
positive direction. Let w lie inside γ. Suppose that the function f is analytic on a simply connected
region D containing γ and its interior. Then

f(w) = (1/2πi) ∫_γ [f(z)/(z − w)] dz.

Question 3

CAUCHYS INTEGRAL FORMULA FORMS

Answer 3

f(w) = (1/2πi) ∫_γ [f(z)/(z − w)] dz.

∫_γ [f(z)/(z − w)] dz. = 2πi f(w)

value of f at w, a point lying strictly inside γ, in terms
of its values on γ.

Question 4

HOW TO use Cauchy’s Theorem and Cauchy’s Integral Formulae to evaluate
integrals of the form
∫_γ [f(z)/(z−w)] dz

Answer 4

• First draw a diagram showing the contour γ and the point w.
• If w is outside γ we use Cauchy’s Theorem (CT).

• If w is inside γ we use Cauchy’s Integral Formula (CIF).
The shape of γ is not otherwise important.

Question 5

EXAMPLE:
Let γ be the simple, positively oriented triangular contour from 0 to 2 −3i to 2 + 2i and
back to 0. Evaluate
∫_γ (e^z)/(z-1) .dz

Answer 5

We begin by drawing a diagram showing the contour γ and marking the
points at which the integrand is not analytic, γ is a simple contour described in the positive direction.

Let f(z) = e^z.  Then f is analytic in C which is a simply-connected region containing
γ. The point 1 is inside γ. By Cauchy’s integral formula,

∫_γ (e^z)/(z-1) .dz = ∫_γ f(z)/(z-1) .dz

= 2πif(1) = 2πie.

Question 6

EXAMPLE:
Let γ be the simple, positively oriented triangular contour from 0 to 2 −3i to 2 + 2i and
back to 0. Evaluate
∫_γ (e^z)/(z+1).dz

Answer 6

DIAGRAM with contour and points not analytic wrt to γ

γ is a simple contour described in the positive direction.

Let g(z) =  (e^z)/(z+1) here -1 lies outside  γ, and so we use Cauchy’s
Theorem. Now the function g is analytic in the half-plane H = {z ∈ C : Re z > −1},
which is a simply-connected region containing the contour γ. By Cauchy’s Theorem,

∫_γ (e^z)/(z+1).dz = ∫_γ g(z) .dz =0

Question 7

EXAMPLE:
Let γ be the simple, positively oriented triangular contour from 0 to 2 −3i to 2 + 2i and
back to 0. Evaluate
∫_γ (1/(z^2 -1)) .dz

Answer 7

DIAGRAM with contour and points not analytic wrt to γ

γ is a simple contour described in the positive direction.

PARTIAL FRACTIONS with CIF on first and CT =0 on 2nd
OR/……

Let h(z) = 1/ (z+1)
Then h is analytic in the half-plane H ={z ∈ C : Re z > −1},
Now the point 1 is inside γ. By Cauchy’s integral formula

∫_γ (1/(z^2 -1)) .dz = ∫_γ (h(z)/(z-1)) .dz = 2πih(1) = πi.

Question 8

EXAMPLE:
Let γ be the simple, positively oriented triangular contour from 0 to 2 −3i to 2 + 2i and
back to 0. Evaluate
∫_γ (ze^z^2)/((z-1)(2z-1)) .dz

Answer 8

DIAGRAM with contour and points not analytic wrt to γ

γ is a simple contour described in the positive direction.

(ze^z^2)/((z-1)(2z-1)) is analytic in C \ { 1/2, 1}
0.5 and 1 both lie
inside γ, so we must use partial fractions.

1/(z-1)(2z-1) = [1/(z-1)] - [1/(z-1/2)]

let k(z) = z e^z^2 . Then k is analytic in C, which is a simply-connected region
containing γ. Since the points 1/2, 1 lie inside γ, Cauchy’s integral formula give

∫_γ (ze^z^2)/((z-1)(2z-1)) .dz
= ∫_γ (k(z))/(2(z-1)(z-1/2)) .dz
= ∫_γ [(k(z))/(z-1)].dz - = ∫_γ [k(z)/(z-1/2)].dz

= = 2πik(1) − 2πik(1/2)
= 2πi (e −0.5e^(1/4))

Question 9

EXAMPLE:
Let γ be the simple, positively oriented triangular contour from 0 to 2 −3i to 2 + 2i and
back to 0. Evaluate
∫_γ (e^z^2)/ (z^2 -1) .dz

Answer 9

DIAGRAM with contour and points not analytic wrt to γ

γ is a simple contour described in the positive direction.
avoid the use of partial fractions

let m(Z) = (e^z^2)/(z+1)
let H={z ∈ C : Re z > −1} be the half-plane 

then m is analytic in H, which is a simply-connected region containing γ. Since the
point 1 lies in side γ, Cauchy’s integral formula:

∫_γ (e^z^2)/ (z^2 -1) .dz =
∫_γ (m(z))/ (z -1) .dz
= 2πim(1) = πie

Question 10

Cauchy’s integral formula for the derivatives

Theorem 9

Answer 10

Let γ be a simple contour
described in the positive direction. Let w be any point inside γ. Suppose that the function
f is an analytic on a simply-connected region D containing γ and its interior. Then, for
all n ∈N

f^(n) (w)
= (n!/2iπ) ∫_γ (f(z))/ (z-w)^{n+1} .dz

equiv

∫_γ [f(z)/ (z-w)^{n+1}] .dz = (2iπ/n!) f^(n) (w)

Question 11

COROLLARY 1

derivatives

Answer 11

If f is analytic on a region D, then f has derivatives of all orders at each
point of D and each derivative is analytic on D.

Question 12

Corollary 2

derivatives

Answer 12

Suppose f = u + iv is analytic on a region D. Then u

and v are harmonic on D.

Question 13

EXAMPLE:
Let γ be the simple, positively oriented triangular contour from 0 to 2 −3i to 2 + 2i and
back to 0. Evaluate

∫_γ (sinz) / ((z-1)^[2n+1]).dz

Answer 13

We note that γ is a simple contour described in the positive direction.

Let f(z) = sin z . Then f is analytic in C, which is a simply-connected region
containing the contour γ.

1 lies inside γ, we use (CIF)^(2n) with w = 1. Now the (2n)th derivative of sin z is (−1)^n sin z

∫_γ (sinz) / ((z-1)^[2n+1]).dz
= ∫_γ f(z) / ((z-1)^[2n+1]).dz
= [2πi/(2n)!] f^(2n)(1) = 2πi/(2n)!*(−1)^n *sin 1.

Question 14

EXAMPLE:
Let γ be the simple, positively oriented triangular contour from 0 to 2 −3i to 2 + 2i and
back to 0. Evaluate

∫_γ(sinz) / ((z+1)^[2n+1]) .dz

Answer 14

We note that γ is a simple contour described in the positive direction.

−1 lies outside γ
(sinz) / ((z+1)^[2n+1] is analytic on the half-plane
H = {z ∈ C : Re z > −1} which is a simply connected region containing the
contour γ. Thus, by Cauchy’s Theorem

∫_γ(sinz) / ((z+1)^[2n+1]) .dz=0

Question 15

EXAMPLE:
Let γ be the simple, positively oriented triangular contour from 0 to 2 −3i to 2 + 2i and
back to 0. Evaluate

∫_γ (cos(z^2)) / ((2z-1)^2) .dz

Answer 15

We note that γ is a simple contour described in the positive direction.

let g(z) = cos(z^2)
g is analytic in C, which is a simply-connected region
containing γ
0.5 lies inside γ. Using Cauchy’s integral formula for the
first order derivative,

∫_γ (cos(z^2)) / ((2z-1)^2) .dz
=(1/4) ∫_γ (cos(z^2)) / ((z-1/2)^2) .dz

=(1/4) ∫_γ (g(z)) / ((z-1/2)^2) .dz

= (1/4) [2πi/1! ] g’(0.5)

=(πi/2)[[−2z sin(z^2)] @ z= 1/2
= −πi/2 * sin(1/4)

Question 16

Theorem 9.4 (Liouville’s Theorem)

Answer 16

A function which is analytic and bounded in the

complex plane is a constant.

Question 17

Corollary 9.5 (Fundamental Theorem of Algebra)

Answer 17

Let p(z) be a non-constant polynomial
with complex coefficients, then there is a point w ∈ C such that p(w) = 0.

Question 18

EXAMPLE:
Let γ be the simple, positively oriented triangular contour from 0 to 2 −3i to 2 + 2i and
back to 0. Evaluate

∫_γ (e^z) / ((z-1)^10).dz

Answer 18

We note that γ is a simple contour described in the positive direction.

Let h(z) = e^z
h is analytic in C, which is a simply- connected region containing
γ. Since the point 1 is inside γ, Cauchy’s integral formula for the 9th derivative
gives

∫_γ  (e^z) / ((z-1)^10).dz
=
∫_γ  h(z) / ((z-1)^10).dz
= 
(2πi) /(9!)  * h^(9) (1) = 2πie /(9!)

Question 19

EXAMPLE:
Let γ be the simple, positively oriented triangular contour from 0 to 2 −3i to 2 + 2i and
back to 0. Evaluate

∫_γ (e^z) / ((z^2-1)^2).dz

Answer 19

∫_γ (e^z) / ((z^2-1)^2).dz

We note that γ is a simple contour described in the positive direction.

Let k(z) = (e^z) / ((z+1)^2)

. Then k is analytic on the half-plane H = {z ∈ C : Re z > −1}
which is a simply connected region containing the contour γ. Since the point 1 is
inside γ, Cauchy’s integral formula for the first derivative

∫_γ (e^z) / ((z-1)^2(z+1)^2).dz

=
∫_γ k(z) / ((z-1)^2) .dz

= 2πi/1! k’(1) = 2πi (e/4−2e/8) = 0,

as k’(z) = [e^z/(z+1)^2] -[2e^z/ (z+1)^3]