我尝试在对数上使用 Python 的 matplotlib，这就是我得到的，一种星爆模式。由于角度在$\theta = 0$和$\theta = 2\pi$, 轮廓假设这两个点之间有一条轮廓线。这里实际上是$\bbox[#EEEDAA,1pt]{0=2\pi}$

或许可以用和语句来代替matplotlib.pyplot.contour轮廓线。meshgridnumpy.where

在上的等高线图$\log z$$z \in [-1,1] + i[-1,1]$

import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
plt.rcParams['figure.figsize']=4,4

plt.contour(
    np.arange(-1,1,0.01),
    np.arange(-1,1,0.01), 
    np.angle(t[...,None] + 1j*t[None,...]) , 
    levels = 2*np.pi*np.arange(-1,1,0.025)
)

等高线在负虚轴处累积$-i \, \mathbb{R}_{\geq 0}$

t = np.arange(-1,1,0.025)
plt.contour(t,t, np.angle(t[...,None] + 1j*t[None,...]) , levels = 2*np.pi*np.arange(-1,1,0.025))

L = 0.25
plt.xlim([-L,L])
plt.ylim([-L,L])

代码检查

import inspect
print inspect.getsource(plt.contour)

@_autogen_docstring(Axes.contour)
def contour(*args, **kwargs):
    ax = gca()
    # allow callers to override the hold state by passing hold=True|False
    washold = ax.ishold()
    hold = kwargs.pop('hold', None)
    if hold is not None:
        ax.hold(hold)
    try:
        ret = ax.contour(*args, **kwargs)
        draw_if_interactive()
    finally:
        ax.hold(washold)
    if ret._A is not None: sci(ret)
    return ret

print inspect.getsource(plt.Axes.contour)

def contour(self, *args, **kwargs):
    if not self._hold:
        self.cla()
    kwargs['filled'] = False
    return mcontour.QuadContourSet(self, *args, **kwargs)

最后可以调用包含实际源代码的函数：

import matplotlib
print inspect.getsource(matplotlib.contour.QuadContourSet)