Complex analysis review 1
2017-12-12 22:13
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Argument
Arga=arga+2πZStereographic Projection
x1=z+z¯1+|z|2x2=z−z¯1+|z|2x3=|z|2−1|z|2+1Jordan’s Theorem
Every Jordan curve divides the plane into an “interior” region bounded by the curve and an “exterior” region containing all of the nearby and far away exterior points, so that every continuous path connecting a point of one region to a point of the other intersects with that loop somewhere.Heine-Borel Theorem
Suppose that A is a compact set, G is an open covering of A, then there are finite open sets of G which can cover A.Bolzano-Weierstrass Theorem
There is at least an accumulating point in an infinit set.Cauchy-Riemann Identity
Suppose that f(z)=u(z)+iv(z),limz→z0f(z)−f(z0)z−z0=f′(z0)
For any path z→z0, the limits are equal.
Let z=x+iy0,x→x0,
f′(z0)=ux+ivx.
Let z=x0+iy,y→y0,
f′(z0)=vy−iuy.
Alltogether,
ux=vy,uy=−vx.∂f∂x=−i∂f∂y
Theorem 1
Complex function f(z)=u+iv is analytic on D if and only if u,v have continuous partial dirivatives and satisfy the Cauchy-Riemann identity.There is a fact that if f is an analytic function on a domain D, then f′ is also an analytic function. So u,v have continuous second order derivatives, then
∂2u∂x∂y=∂2u∂y∂x.
Which means that
∂2u∂2x+∂2u∂2y=0;∂2v∂2x+∂2v∂2y=0.
Two Important Operators
z=x+iy,z¯=x−iy∂f∂z=12(∂f∂x−i∂f∂y)∂f∂z¯=12(∂f∂x+i∂f∂y)
And after a simple calculation
df=∂f∂zdz+∂f∂z¯dz¯
If f is analytic, then ∂f∂z¯=0.
Conformality
Suppose that f(z) is analytic on D, and f′(z0)≠0, γ(t),(0≤t≤1) is a smooth curve which pass z0, and gamma(0)=z0. Let σ(t)=f(γ(t)), thenσ′(t)=f′(γ(t))γ′(t),σ′(0)=f′(γ(0))γ′(0)
Therefore,
argσ′(0)−argγ′(0)=argf′(z0)
Now suppose that there are two smooth curves pass through z0, then
argσ′1(0)−argγ′1(0)=argσ′2(0)−argγ′2(0)
Under the mapping w=f(z), the angles and the directions of rotation between two smooth curves where the derivatives are not zero, are invavirant.
On the other hand, since
limz→z0,z∈γ|w−w0||z−z0|=|f′(z0)|.
For any smooth curve through z0, the ratio of distance between image points and original points are the same, namely |f′(z0)|.
Integration of Complex Functions
Suppose that f(t)=u(t)+iv(t) defined on [a,b].∫baf(t)dt=∫bau(t)dt+i∫bav(t)dt.
γ is a rectifiable curve, f(z)=u(z)+iv(z),dz=dx+idy, then
∫γf(z)dz=∫γudx−vdy+i∫γvdx+udy.
∂f=∂f∂zdz,∂¯f=∂f∂z¯dz¯, then
dz∧dz¯=−2idA
Define
dw=∂w∧dz+∂¯w∧dz¯
Theorem 2
For any smooth (n−1) -form with compact support on the oriented n-dimensional manifold Ω,∫∂Ωω=∬Ωdw.
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