这几乎和@Hooked代码一样，但是要快得多。

import numpy as np
from numpy import pi
import pylab as plt
from colorsys import hls_to_rgb

def colorize(z):
    r = np.abs(z)
    arg = np.angle(z) 

    h = (arg + pi)  / (2 * pi) + 0.5
    l = 1.0 - 1.0/(1.0 + r**0.3)
    s = 0.8

    c = np.vectorize(hls_to_rgb) (h,l,s) # --> tuple
    c = np.array(c)  # -->  array of (3,n,m) shape, but need (n,m,3)
    c = c.swapaxes(0,2) 
    return c

N=1000
x,y = np.ogrid[-5:5:N*1j, -5:5:N*1j]
z = x + 1j*y

w = 1/(z+1j)**2 + 1/(z-2)**2
img = colorize(w)
plt.imshow(img)
plt.show()

修改mpmath的绘图代码，即使你不知道numpy和matplotlib的原始函数**，你也可以用它来绘制numpy数组**，如果你知道这个函数，see my original answer可以使用mpmath.cplot。

from colorsys import hls_to_rgb

def colorize(z):
    n,m = z.shape
    c = np.zeros((n,m,3))
    c[np.isinf(z)] = (1.0, 1.0, 1.0)
    c[np.isnan(z)] = (0.5, 0.5, 0.5)

    idx = ~(np.isinf(z) + np.isnan(z))
    A = (np.angle(z[idx]) + np.pi) / (2*np.pi)
    A = (A + 0.5) % 1.0
    B = 1.0 - 1.0/(1.0+abs(z[idx])**0.3)
    c[idx] = [hls_to_rgb(a, b, 0.8) for a,b in zip(A,B)]
    return c

从这里，你可以绘制一个任意的复数numpy数组：

N = 1000
A = np.zeros((N,N),dtype='complex')
axis_x = np.linspace(-5,5,N)
axis_y = np.linspace(-5,5,N)
X,Y = np.meshgrid(axis_x,axis_y)
Z = X + Y*1j

A = 1/(Z+1j)**2 + 1/(Z-2)**2

# Plot the array "A" using colorize
import pylab as plt
plt.imshow(colorize(A), interpolation='none',extent=(-5,5,-5,5))
plt.show()

库mpmath使用matplotlib生成复杂平面的美丽图像。在复杂平面上，您通常关心极点，因此函数的参数给出颜色（因此极点将形成螺旋）。极大或极小值的区域由饱和度控制。来自文档：
默认情况下，复数参数（相位）显示为颜色（色调），幅值显示为亮度。您还可以提供自定义颜色函数（* color *）。此函数应将复数作为输入，并返回包含范围为0.0 - 1.0的浮点数的RGB三元组。
示例：

import mpmath
mpmath.cplot(mpmath.gamma, points=100000)

显示zeta function、平凡零点和临界条带的另一个示例：

import mpmath
mpmath.cplot(mpmath.zeta, [-45,5],[-25,25], points=100000)

您可以使用matplotlib.colors.hsv_to_rgb代替colorsys.hls_to_rgb。matplotlib函数大约快10倍！请查看以下结果：

import numpy as np
import matplotlib.pyplot as plt
from matplotlib.colors import hsv_to_rgb
import time

def Complex2HSV(z, rmin, rmax, hue_start=90):
    # get amplidude of z and limit to [rmin, rmax]
    amp = np.abs(z)
    amp = np.where(amp < rmin, rmin, amp)
    amp = np.where(amp > rmax, rmax, amp)
    ph = np.angle(z, deg=1) + hue_start
    # HSV are values in range [0,1]
    h = (ph % 360) / 360
    s = 0.85 * np.ones_like(h)
    v = (amp -rmin) / (rmax - rmin)
    return hsv_to_rgb(np.dstack((h,s,v)))

下面是@nadapez挑选答案的方法：

from colorsys import hls_to_rgb
def colorize(z):
    r = np.abs(z)
    arg = np.angle(z) 

    h = (arg + np.pi)  / (2 * np.pi) + 0.5
    l = 1.0 - 1.0/(1.0 + r**0.3)
    s = 0.8

    c = np.vectorize(hls_to_rgb) (h,l,s) # --> tuple
    c = np.array(c)  # -->  array of (3,n,m) shape, but need (n,m,3)
    c = c.swapaxes(0,2) 
    return c

用1024*1024二维阵列对两种方法的结果进行检验：

N=1024
x, y = np.ogrid[-4:4:N*1j, -4:4:N*1j]
z = x + 1j*y

t0 = time.time()
img = Complex2HSV(z, 0, 4)
t1 = time.time()
print "Complex2HSV method: "+ str (t1 - t0) +" s"

t0 = time.time()
img = colorize(z)
t1 = time.time()
print "colorize method: "+ str (t1 - t0) +" s"

在我的旧笔记本电脑上显示的结果是：

Complex2HSV method: 0.250999927521 s
colorize method: 2.03200006485 s

您也可以使用PIL.image来转换它

import PIL.Image

def colorize(z):
  z = Zxx
  n,m = z.shape

  A = (np.angle(z) + np.pi) / (2*np.pi)
  A = (A + 0.5) % 1.0 * 255
  B = 1.0 - 1.0/(1.0+abs(z)**0.3)
  B = abs(z)/ z.max() * 255
  H = np.ones_like(B)
  image = PIL.Image.fromarray(np.stack((A, B, np.full_like(A, 255)), axis=-1).astype(np.uint8), "HSV") # HSV has range 0..255 for all channels
  image = image.convert(mode="RGB")

  return np.array(image)