复变函数可视化傅里叶展开复反演

青旅半醒 2022-01-28 23:21 335阅读 0赞

**泰勒展开&傅里叶展开**  
 f ( z ) = 1 1 − z = 1 + z + z 2 + z 3 + . . . = 1 + r e θ + r 2 e 2 θ + r 3 e 3 θ + . . . \\begin\{aligned\} f(z)& = \\frac\{1\}\{1-z\}\\\\ &=1+z+z^2+z^3+...\\\\ &=1 + re^\{\\theta\} + r^\{2\}e^\{2\\theta\}+ r^\{3\}e^\{3\\theta\}+... \\end\{aligned\} f(z)=1−z1=1\+z\+z2\+z3\+...=1\+reθ\+r2e2θ\+r3e3θ\+...

**复反演**  
 f ( z ) = 1 z f(z) = \\frac\{1\}\{z\}\\\\ f(z)=z1  
**模长变成倒数，角度变成相反数**

一个过原点的直线经过复反演之后还是一条直线
    一个直线不经过原点，反演之后变成一个圆
    
    一个经过原点的圆反演之后是一个直线
    一个不经过原点的圆反演之后是一个圆

**两个例子：**

**复变函数的积分**  
 ∫ z 0 z 1 f ( z ) d z = ∑ f ( Δ z ) Δ z \\int\_\{ z\_\{0\} \}^\{ z\_\{1\} \}f(z)dz\\\\ =\\sum f(\\Delta z) \\Delta z ∫z0z1f(z)dz=∑f(Δz)Δz

**每一小段旋转拉伸之后的加和**  
**向量加法，最终的位置减去起始位置就是积分函数值**

f ( z ) = 1 z f ( a + b i ) = 1 a + b i = a − b i a 2 + b 2 \\begin\{aligned\} f(z) &= \\frac\{1\}\{z\} \\\\f(a+bi) &= \\frac\{1\}\{a+bi\} \\\\ \\\\&= \\frac\{a-bi\}\{a^2+b^2\} \\end\{aligned\} f(z)f(a\+bi)=z1=a\+bi1=a2\+b2a−bi

**1个例子**

%matplotlib inline
    import numpy as np
    import matplotlib.pyplot as plt
    
    class shape:
        x = []
        y = []
        def __init__(self, x, y):
            self.x = x.copy()
            self.y = y.copy()
            
    def draw(shapetemp):
        plt.plot(shapetemp.x,shapetemp.y)
    
    def fanyan(shape1):
        temp = []
        for x,y in zip(shape1.x,shape1.y):
            temp.append( (x/(x*x+y*y),-y/(x*x+y*y)) )
    # print((x/(x*x+y*y),-y/(x*x+y*y)) )
        return shape(list(list(zip(*temp))[0]),list(list(zip(*temp))[1]))
    
    
    x = np.arange(-10,10,0.001)
    y = x+ 0.1
    y1 = -x+ 0.1
    shape1 = shape(x,y)
    shape1 = shape(x,y1)
    
    draw(shape1)
    draw(shape2)
    sf1 = fanyan(shape1)
    sf2 = fanyan(shape2)
    draw(sf1)
    draw(sf2)
    plt.grid()
    plt.axis('equal')
    plt.show()

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