Storing roots of a complex function in an array in SciPy

Some answers:

(i) Use a list. Arrays have fixed size. Appending to a list is a very cheap option. When you append a new root to the list, check that the previous root is not in the list by, e.g., calculating np.amin(np.abs(np.array(a)-b)) where a is the list of existing roots and b is the new root. If this value is very small, you have arrived at an existing root. (How small depends on the function. It cannot be 0.0, as then you are usually not recognizing the same root due to floating point and iteration inaccuracies.)

If you have a very large number of roots (thousands), you may want to sort them as soon as you receive them. This makes searching the matching roots faster. On the other hand, most probably >90% of the time is spent in iterating the roots, and you do not need to worry about other performance issues. Then you just compile the list, sort it (list sorting is easy and fast) and convert into an array, if you need.

(ii) Yes. Two examples below: (For countour stuff, thank you belongs to Warren Weckesser and his very good answer!)

import numpy as np
from scipy.special import jv, hankel1, jvp, h1vp
import matplotlib.pyplot as plt


nr = 2
m = 20

# create a 2000 x 2000 sample complex plane between -2-2i .. +2+2i
x = np.linspace(-2, 2, 2000)
y = np.linspace(-2, 2, 2000)
X, Y = np.meshgrid(x, y)
C = X + 1j * Y 
z = 1-C**2

# draw a contour image of the imaginary part (red) and real part (blue)
csr = plt.contour(x, y, z.real, 5, colors='b')
plt.clabel(csr)
csi = plt.contour(x, y, z.imag, 5, colors='r')
plt.clabel(csi)
plt.axis('equal')
plt.savefig('contours.png')

# draw an image of the absolute value of the function, black representing zeros
plt.figure()
plt.imshow(abs(z), extent=[-2,2,-2,2], cmap=plt.cm.gray)
plt.axis('equal')
plt.savefig('absval.png')

This gives countours.png:

and absval.png:

Note that if you want to zoom the images, it is usually necessary to change the limits and recalculate the complex values z to avoid missing details. The images can of course be plotted on top of each other, the image colour palette can be changed, countour has a million options. If you only want to draw the zeros, replace the number 5 (number of contours) with [0] (plot only the listed contour) in the countour calls.

Of course, you will replace my (1-C^2) by your own function. The only thing to take care is that if the function receives an array of complex numbers, it returns an array of results in the same shape calculated pointwise. Imshow needs to receive an array of scalars. For more information, please see the imshow documentation.

(iii) There may be, but there are no general methods for finding all minima/maxima/zeros of arbitrary functions. (The function may even have an infinite number of roots.) Your idea of first plotting the function is a good one. Then you'll understand its behaviour easier.