( t ) L ( C ) = ∫ a b x ′ ( t ) 2 + y ′ ( t ) 2 d t = ∫ a b ∣ z ′ ( t ) ∣ d t
piecewise smooth curve C joined from finite smooth parametric curve
z continuousz ′ piecewise continuousvertex, point where C not smooth length, orientation, simple closed like parametric curve simple closed contour separate C into
a bounded interior an unbounded exterior contour C : z ( t ) : [ a , b ] → C , function f ( z ) piecewise continuous on C
∫ C f ( z ) d z = ∫ a b f ( z ( t )) z ′ ( t ) d t
independent of parametrization
linearity
∫ C ( c f ( z ) + g ( z )) d z = c ∫ C f ( z ) d z + ∫ C g ( z ) d z
for contour C = C 1 ∪ C 2
∫ C f ( z ) d z = ∫ C 1 f ( z ) d z + ∫ C 2 f ( z ) d z
traversed in opposite direction
− C : t ↦ z ( − t ) : [ − b , − a ] → C ⇒ ∫ − C f ( z ) d z = − ∫ C f ( z ) d z
contour C of length L , f ( z ) piecewise continuous on C , f ( z ) bounded by M
⇒ ∫ C f ( z ) d z ≤ M L
f : D ⊆ C → C continuous in D
∃ antiderivative F s.t. ∀ C in D from z 1 to z 2
∫ C f ( z ) d z = F ( z 2 ) − F ( z 1 )
⇔ ∀ C closed,
∮ C f ( z ) d z = 0
on D ⊆ C , simple closed contour C in D , f : D → C holomorphic on and within C
⇒ ∫ C f ( z ) d z = 0
same as Cauchy-Goursat theorem, except f ′ ( z ) continuous
proof using Green's theorem and Cauchy-Riemann equations ∫ C f ( z ) d z = ∫ C ( u + i v ) ( d x + i d y ) = ∫ C ( u + i v ) d x + ( − v + i u ) d y = ∬ D ( ∂ x ∂ ( − v + i u ) − ∂ y ∂ ( u + i v ) ) d A = ∬ D ( − v x + i u x − u y + i v y ) d A = 0
on simply connected domain D ⊆ C , closed contour C in D , f : D → C holomorphic on D
⇒ ∫ C f ( z ) d z = 0
simply connected domain: every simple closed contour enclose only point in D multiply connected domain holomorphic function on simply connected domain has antiderivative entire function has antiderivative on multiply connected domain D ⊆ C , positively oriented contour C , negatively oriented simple closed contour C i , C i inside of and disjoint from C , f : D → C holomorphic on D and on the region between C , C i
⇒ ∫ C f ( z ) d z + ∫ C i f ( z ) d z = 0
for the following case, deformation of contour integral persist its value
positively oriented contour C 1 , C 2 , f holomorphic between C 1 , C 2
⇒ ∫ C 1 f ( z ) d z = ∫ C 2 f ( z ) d z
C positively oriented contour, f holomorphic inside and on C
∀ z 0 inside C
f ( z 0 ) = 2 πi 1 ∫ C z − z 0 f ( z ) d z
derivative
f ′ ( z 0 ) = 2 πi 1 ∫ C ( z − z 0 ) 2 f ( z ) d z
at z 0 , f holomorphic ⇒ ∀ n ∈ N , f ( n ) exist and holomorphic
f ( n ) ( z 0 ) = 2 πi n ! ∫ C ( z − z 0 ) ( n + 1 ) f ( z ) d z
f holomorphic ⇒ u , v have continuous partial derivative at all order
proof show using upper bound theorem ∫ C R z − z 0 f ( z ) − f ( z 0 ) d z = 0
on positively oriented contour C , evaluate integral
I = ∫ C g ( z ) d z
find f ( z ) , z 0 s.t. z 0 inside C and
g ( z ) = ( z − z 0 ) n f ( z )
calculate f ( n − 1 ) ( z 0 )
apply Cauchy’s integral formula
I = ( n − 1 )! 2 πi f ( n − 1 ) ( z 0 )
f continuous on D
∀ closed contour C
∫ C f ( z ) d z = 0
⇒ f holomorphic throughout D
bounded entire function is constant
non-constant polynomial of degree n has n root
non-constant holomorphic function f on open D ⇒ ∣ f ∣ has no maximum on D
for f on D ∪ ∂ D , ∣ f ∣ reach maximum always on ∂ D , never on interior f : U ε ( z 0 ) → C analytic ⇔
∀ z ∈ U ε ( z 0 ) , f ( z ) = n = 0 ∑ ∞ a n ( z − z 0 ) n
annual domain D , R 0 < ∣ z − z 0 ∣ < R 1 , any closed contour C in D , function f holomorphic in D ⇒
f ( z ) = n = − ∞ ∑ ∞ c n ( z − z 0 ) n c n = 2 πi 1 ∫ C ( z − z 0 ) n + 1 f ( z ) d z
punctured disk, R 0 = 0 unique res f ( z 0 ) = c − 1 = 2 πi 1 ∫ C f ( z ) d z ⇒ ∫ C f ( z ) d z = 2 πi res f ( z 0 )
positively oriented contour C , f analytic on and within C except on finite number of singularities z k , k ∈ { 1 , 2 , ⋯ , n }
∫ C f ( z ) d z = 2 πi k = 1 ∑ n res f ( z k )
all singularity are inside negatively oriented contour C
∫ C f ( z ) d z = 2 πi res f ( ∞ )
Cauchy residue theorem can be rewritten as
k = 1 ∑ n res f ( z k ) + res f ( ∞ ) = 0
by replacing z with z 1
res ( z 2 1 f ( z 1 ) ) ( 0 ) = − res f ( ∞ )
z 0 is singularity ⇔
f holomorphic in U ^ ε ( z 0 ) and not at z 0
f analytic in deleted neighborhood U ^ ( z 0 )
z 0 is isolated singularity
f has Laurent series about z 0 not branch point isolated singular at infinity ⇐ ∃ R > 0 s.t. f analytic for R < ∣ z ∣ < ∞ ∃ m ∈ N + s.t. ∀ n < − m , c n = 0
z 0 is essential isolated singularity
Casorati-Weierstrass theorem
∀ ε , R > 0 , w ∈ C , ∃ z ∈ ∣ z − z 0 ∣ < R s.t ∣ f ( z ) − w ∣ < ε
∃ m ∈ N + s.t. ∀ n < − m , c n = 0
pole of order m at z 0
res f ( z 0 ) = { ( m − 1 )! 1 d z m − 1 d m − 1 [ ( z − z 0 ) m f ( z ) ] } ( z 0 )
f analytic at z 0 , f ( z 0 ) = 0 , ∃ m ∈ N + s.t.
∀ n < m , f ( n ) ( z 0 ) = 0
identically zero ⇐ m = ∞ otherwise, the zero is isolated p , q analytic, q has zero of order m at at z 0 , p ( z 0 ) = 0
⇒ q p has pole of order m at z 0
CP V ∫ − ∞ ∞ f ( x ) d x = R → ∞ lim ∫ − R R f ( x ) d x
if RHS exist
f is even ⇒ ∫ − ∞ ∞ f ( x ) d x = CP V ∫ − ∞ ∞ f ( x ) d x real, continuous, even, irreducible rational function f = q p , q ( x ) has finitely many zero z k above the real axis
⇒ f has pole z k above the real axis,
let C R be upper semicircle with missing side [ − R , R ] , then
∫ − R R f ( x ) d x + ∫ C R f ( x ) d x = 2 πi k ∑ res f ( z k )
if lim R → ∞ ∫ C R f ( x ) d x = 0 , then
∫ − ∞ ∞ f ( x ) d x = 2 πi k ∑ res f ( z k )
k > 0
∫ − ∞ ∞ f ( x ) sin ( k x ) d x or ∫ − ∞ ∞ f ( x ) cos ( k x ) d x
integrate instead
∫ − R R f ( x ) e ik x d x
f ( z ) analytic above the imaginary axis outside ∣ z ∣ < R 0 , semicircle contour C R : ∣ z ∣ = R > R 0 , 0 ≤ θ ≤ π
∣ f ( z ) ∣ ( C R ) ≤ M R , R → ∞ lim M R = 0 ⇒ ∀ k > 0 , R → ∞ lim ∫ C R f ( z ) e ik z d z = 0
proof by parametric integral and Jordan’s inequality
∫ 0 π e − k R s i n θ d θ < k R π
improper integral with singularity on real axis
circumvent each singularity x i on real axis by upper semicircle C i : ∣ z − x i ∣ = r i with r i → 0
f analytic on 0 < ∣ z − x i ∣ < r , clockwise upper semicircle C i : z = x i + r i e i θ , π ≥ θ ≥ 0 , r i < r , Laurent series of f about x i contain no even negative power
⇒ r i → 0 lim ∫ C i f ( z ) d z = − iπ res f ( x i )
proof by integrating Laurent series of parametric line integral term by term
circular segment of C r : z = r e i θ , 0 ≤ θ ≤ θ 1 , f ( z ) continuous on C r ⇒
z → 0 lim z f ( z ) = 0 ⇒ r → 0 lim ∫ C r f ( z ) d z = 0
proof by definition of limit and upper bound theorem
∫ θ 0 θ 1 f ( sin θ , cos θ ) d θ
define C : z = e i θ , θ 0 ≤ θ ≤ θ 1
sin θ = 2 i z − z − 1 , cos θ = 2 z + z − 1 , d θ = i z d z ⇒ ∫ θ 0 θ 1 f ( sin θ , cos θ ) d θ = ∫ C f ( 2 i z − z − 1 , 2 z + z − 1 ) i z d z
Laplace transform of f : R 0 + → R
f ~ ( s ) = ∫ 0 ∞ f ( t ) e − s t d t : C → C
inverse Laplace transform
γ ∈ R to the right of all sinularity of f ~ ( s ) , line contour Γ R from γ − i R to γ + i R
f ( t ) = 2 πi 1 R → ∞ lim ∫ Γ R f ~ ( s ) e s t d s
f is analytic except possibly for poles
simple closed contour C , f non-zero and analytic on C , meromorphic within C
2 πi 1 ∫ C f ( z ) f ′ ( z ) d z = Z − P
Z , number of zero of f within C counted with orderP , number of pole of f within C counted with order closed contour C : z = z ( t ) , a ≤ t ≤ b , z ( a ) = z ( b ) = z 0 , f ( z ) = ρ e i ϕ = 0 on C , Γ : f ( z ( t ))
difference in argument
Δ C arg f = ϕ ( b ) − ϕ ( a ) = 2 πk
winding number of Γ with respect to 0
ν ( Γ , 0 ) = 2 π 1 Δ C arg f = k
Z − P = 2 πi 1 ∫ C f ( z ) f ′ ( z ) d z = 2 π 1 Δ C arg f = ν ( Γ , 0 )
f , g analytic on and within C , ∣ g ∣ < ∣ f ∣ on C
⇒ f , f + g have the same number of zero counted with order inside C
constrain location of polynomial root using Rouché’s theorem split polynomial P = f + g where f ( z ) = z n C : ∣ z ∣ = 1 + ∣ a n − 1 ∣ + ⋯ + ∣ a 0 ∣ ⇒ ∣ g ∣ < ∣ f ∣ on C ⇒ root of P are within C D : ∣ z ∣ ≤ 1 , in t D : ∣ z ∣ < 1 , g : D → in t D analytic
⇒ ∃ ! z 0 ∈ D s.t. g ( z 0 ) = z 0 (fixed point)
when changing w continuously without crossing Γ , ν ( Γ , w ) is topological invariant closed curve Γ , Γ ~ , one can be continuously deformed into another without crossing point w
⇔ ν ( Γ , w ) = ν ( Γ ~ , w )
vector field V : z ↦ f ( z ) , simple closed contour C enclose singularity z i , i = 1 , 2 , ⋯ of V with index I V ( C i ) ⇒
I V ( C i ) = ν ( Γ , 0 ) = Z − P = i ∑ I V ( C i )
zero or pole of f
positively oriented contour C i enclose singularity z i of V
vector field index is integer number of rotation of V traversing C i
I V ( C i ) = ν ( Γ i , 0 )
T ( z ) = cz + d a z + b : C ˉ → C ˉ , a d − b c = 0
extended complex plane C ˉ := C ∪ { ∞ }
only singularity is simple pole − c d
inverse
T − 1 ( w ) = c w − a b − d w
derivative non-zero
T ′ ( z ) = ( cz + d ) 2 a d − b c = 0
analytic except at pole non-constant mapping is composition of linear transformation and inversion
f 1 ( z ) = cz + d , f 2 ( z ) = z 1 , f 3 ( z ) = c a − c a d − b c z ⇒ T = f 1 ∘ f 2 ∘ f 3
they are a group
map circle to circle in C ˉ
line are circle of infinite radius through ∞ need 3 point to specify
A z z ˉ + ( B − i C ) z + ( B + i C ) z ˉ + D = 0 A , B , C , D ∈ R
map 3 distinct point z 1 , z 2 , z 3 to w = T ( z ) by z 1 ↦ w 1 , z 2 ↦ w 2 , z 3 ↦ w 3
( w − w 3 ) ( w 2 − w 1 ) ( w − w 1 ) ( w 2 − w 3 ) = ( z − z 3 ) ( z 2 − z 1 ) ( z − z 1 ) ( z 2 − z 3 )
f : D → C
analytic ∀ z ∈ D , f ′ ( z ) = 0 analytic f at z 0 , f ′ ( z 0 ) = 0
⇒ ∃ U ( z 0 ) , ∃ ! f − 1 analytic
derivative
( f − 1 ) ′ ( w ) = f ′ ( z ) 1
f conformal, smooth C 1 , C 2 at z 0 and intersect with acute angle ψ
⇒ f ( C 1 ) , f ( C 2 ) intersect at f ( z 0 ) with acute angle ψ
proof by taking f ′ ( z 0 ) through C 1 and C 2 and their equality
proof by constructing f s.t. ϕ = R e f
T ( C ) = k on smooth C ⇒ Φ ( T ( C )) = k h : D ⊆ R 2 → R with continuous second partial derivative satisfy Laplace’s equation
h xx + h yy = 0
hormonic ϕ : R → R , prescribed ϕ ( ∂ R ) , find ϕ
Neumann problem, prescribed ∇ ϕ ( ∂ R ) simply connected R bounded by simple closed C , ϕ harmonic on R
⇒ ϕ attain extremum on boundary
proof by constructing f s.t. ϕ = R e f and using maximum modulus principle
simply connected R bounded by simple closed C
⇒ ϕ is unique
proof by assuming both ϕ 1 , ϕ 2 are solutions and arguing their difference is 0 using maximum principle