Chapter 3: Simple Integrals Of Complex Valued Functions Flashcards
Parametrically defined curve
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]
(e^it)’
(e^it)’ = (cost + isin t)’
= − sin t + i cost = i(cost + isin t) = ie^it
DEF Curve γ defined by a continuously differentiable complex-valued funct
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)
ORIENTATION OF A CURVE
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
arc length
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
DEF PATH
A path is a finite union of curves (joined successively at end points)
-γ has path γ described in opposite direction
DEF CONTOUR
is a path whose final point is the same as its initial point.
figure 8
DEF SIMPLE CONTOUR
A simple contour
is a contour without self-intersections.
figure 8 isnt, has intersections
Example of path/contour:
Circle
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
Example of path/contour:
The straight line segment from z₀ to z
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
Example of path/contour:
Semi circle
The semi-circle given by z = 2e^(it) (0 ≤ t ≤ π)
is a path but it is
not a contour.
(anticlockwise)
Example of path/contour:
triangular contour is a simple contour.
3 line segments joined, forming a path
Example of path/contour:
figure of eight
( two circles joined)
is a contour, but it is not a simple contour.
DEF 3.3 continuous region containing path
integral over path of region
LINE INTEGRAL
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)
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).
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]
Example: Let R > 0 and γ : z = Re^it (0 ≤ t ≤ 2π). Evaluate ∫ over γ of 1/z dz
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)
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
(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
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)
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)
∫_γ f(z).dz
has the same value for any parameterization
∫_γ (f_1(z) + f_2(z)).dz
∫_γ (f_1(z) + f_2(z)).dz = ∫_γ (f_1(z)).dz + ∫_γ (f_2(z)).dz
∫_γ (cf(z) ).dz
∫_γ (cf(z) ).dz = c ∫_γ (f(z) ).dz c in complex plane
∫_(γ_1 +γ_2) f(z).dz
∫(γ_1) f(z).dz + ∫(γ_2) f(z).dz
-γ has path γ described in opposite direction
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
