Chapter 3: Simple Integrals Of Complex Valued Functions

Question 1

Parametrically defined curve

Answer 1

z’ is CONTINUOUS ON [a,b]

x= x(t) , y= y(t) for (a ≤ t ≤ b)
where x and y have continuous derivatives x’ and y’ on {a,b].

z=x+iy and
z(t) = x(t) + iy(t) (a ≤ t ≤ b)

Then define z’(t) to be
x’(t) + iy’(t) for (a ≤ t ≤ b)

z(t) is CONTINUOUSLY DIFFERENTIABLE ON [a.b]

when x’ and y’ are both CONTINUOUS on [a,b]

Question 2

(e^it)’

Answer 2

(e^it)’ = (cost + isin t)’

= − sin t + i cost = i(cost + isin t) = ie^it

Question 3

DEF Curve γ defined by a continuously differentiable complex-valued funct

Answer 3

A curve γ is defined by a continuously differentiable complex-valuedfunction z of a real variable t on [a, b] (a, b ∈ R).
So we write
γ : z = z(t) (a ≤ t ≤ b)

Question 4

ORIENTATION OF A CURVE

Answer 4

Curve has ORIENTATION defined by a parametrization, different parametrizations produce the same oriented subset of C.

The direction of the curve γ is the
direction in which the parameter t increases
from z(a) to z(b)

ARROW on diagram

Question 5

arc length

Answer 5

curve has length:
∫ over [a,b] of
|z’(t)| dt
which is finite as |z’| is continuous on [a, b].

element of arc length:
δs = [(δx)² + (δy)²]¹/²

Hence the length of γ is
∫ over γ of
δs =
∫ over  [a,b] of
[(dx/dt)² +(dy/dt)²] ] .dt =

∫ over [a, b] of
|z’(t)| dt

Question 6

DEF PATH

Answer 6

A path is a finite union of curves (joined successively at end points)

-γ has path γ described in opposite direction

Question 7

DEF CONTOUR

Answer 7

is a path whose final point is the same as its initial point.

figure 8

Question 8

DEF SIMPLE CONTOUR

Answer 8

A simple contour
is a contour without self-intersections.

figure 8 isnt, has intersections

Question 9

Example of path/contour:

Circle

Answer 9

z lies on a circle centre a and radius r, then |z − a| = r

modulus arg form z − a = reit for some t ∈ R, giving z = a + re^(it)

circle Cr, with centre a and radius r, described in the anticlockwise direction
can be given by:

z = a + re^(it) (0 ≤ t ≤ 2π).

Cr is a simple contour

Question 10

Example of path/contour:

The straight line segment from z₀ to z

Answer 10

suppose z₀≠ z₁

The straight line segment from z₀ to z₁ can be given by
z = z(t) = tz₁ + (1 − t)z₀
(0 ≤ t ≤ 1).

RHS expression is linear in t ,L, from z(0) = z₀ to z(1) = z₁
ARROW for direction

L is a path but it is not a contour

Question 11

Example of path/contour:

Semi circle

Answer 11

The semi-circle given by z = 2e^(it) (0 ≤ t ≤ π)

is a path but it is
not a contour.

(anticlockwise)

Question 12

Example of path/contour:

Answer 12

triangular contour is a simple contour.

3 line segments joined, forming a path

Question 13

Example of path/contour:

Answer 13

figure of eight
( two circles joined)
is a contour, but it is not a simple contour.

Question 14

DEF 3.3 continuous region containing path
integral over path of region

LINE INTEGRAL

Answer 14

Let f be continuous on a region containing the path γ. Let γ be given
by z = z(t), a ≤ t ≤ b. Then
∫ over  γ of 
f(z)dz =
∫ over  [a,b] of
f(z(t))z'(t) dt.

(on closed interval [a,b] is cont complex valued funct of a real var t , except for finitely many values of t corres to corners)

Question 15

Example: 
Evaluate 
∫ over  γ of 
zdz 
where γ is the union of the paths
γ_1 : z = x (0 ≤ x ≤ 1) and 
γ_2 : z = 1 + iy (0 ≤ y ≤ 1).

Answer 15

diagram along x to 1 then up to 1+i

z is continuous on γ

∫_γ z.dz = 0.5 + (i-0.5) = i

∫(γ_1) f(z).dz + ∫(γ_2) f(z)
both over [0,1]

Question 16

Example: 
Let R > 0 and γ : z = Re^it (0 ≤ t ≤ 2π). 
Evaluate 
∫ over  γ of 
1/z dz

Answer 16

Example: γ diagram circle radius R anti
∫ over γ of
1/z dz

=
∫ over [0,2π] of

(1/Re^it) Rie^it .dt = 2πi

as d/dt) (e^it) =(ie^it)

z(t)

Question 17

Example: 
Let R > 0 and γ : z = Re^it (0 ≤ t ≤ 2π). 
Let k be any integer with k ≥ 2.
Evaluate 
∫ over  γ of 
1/z^k dz

Answer 17

(1/R^ke^kit) Rie^it .dt

Example: 
Let R bigger than 0 and γ : z = Re^it (0 ≤ t ≤ 2π). 
Let k be any integer with k ≥ 2.
Evaluate 
∫ over  γ of 
1/z^k dz 

=
∫ over [0,2π] of

(1/R^(k-1)) *
∫ over [0,2π] of

ie^-(k-1)it .dt

=
(1/R^(k-1)) *

[(ie^(-(k-1)it)/ (-i(k-1))] from t=2π , t=0

=
0
as d/dt (e^it) =(ie^it)
e^-(k-1)i2π = e^0 and ∫ e^iat .dt = e^iat /ia

answer indep of both R and k

Question 18

Evaluate
∫_γ zbar.dz,
where γ = γ1 + γ2 + γ3

where
γ1 : z = x (−1 ≤ x ≤ 1)
γ2 : z = e^it (0 ≤ t ≤ π/2)
γ3 : z = i + (−1 − i)t (0 ≤ t ≤ 1)

Answer 18

diagram: from origin line along x to 1, arc to i then line to -1

zbar is continuous on γ (checked)

Hence:
∫_γ_1 zbar.dz = ∫ over [-1,1] of xdx =0

∫_γ_2 zbar.dz = ∫ over [0,π/2] of e^it dz/dt .dt =
∫ over [0,π/2] of e^it * ie^it .dt = iπ/2

∫_γ_3 zbar.dz = ∫ over [0,1] of (-i -t +it)(-1-i).dt
=-(1+i) ∫ over [0,1] of (-i -t +it).dt = i

SO

∫_γ zbar.dz= 0 + iπ/2 + i = i(π/2 +1)

Question 19

∫_γ f(z).dz

Answer 19

has the same value for any parameterization

Question 20

∫_γ (f_1(z) + f_2(z)).dz

Answer 20

∫_γ (f_1(z) + f_2(z)).dz = ∫_γ (f_1(z)).dz + ∫_γ (f_2(z)).dz

Question 21

∫_γ (cf(z) ).dz

Answer 21

∫_γ (cf(z) ).dz = c ∫_γ (f(z) ).dz c in complex plane

Question 22

∫_(γ_1 +γ_2) f(z).dz

Answer 22

∫(γ_1) f(z).dz + ∫(γ_2) f(z).dz

Question 23

-γ has path γ described in opposite direction

Answer 23

path γ from point z_0 to origin

then best to para by -
-γ has path γ described in opposite direction ie from origin to z_0