Playing with complex numbers

Hello!

Another essay. This time let's play with complex numbers, with formulas and their visualization.

Idea

A complex number is a certain extension of a real number, in fact a vector for which a whole set of axioms is defined. Any complex (and hence real) number can be written as $$$z = a + bi$, where a is the real part, b is the imaginary part, i is the root of the equation $$$x ^ {2} = - 1$. For him, many operations are defined that are defined for a real number, for example, $$$z_1 + z_2 = a_1 + a_2 + (b_1 + b_2) i$. Interestingly, if you do various operations with them, raise to a power, multiply, etc., and then take $$$Re (z)$(real part) for the axis Ox, and $$$Im (z)$(imaginary part) for the Oy axis, you can get funny pictures.

By the way, I myself came up with all the following formulas.

Visualization function

Routine. The function, which according to this iterative function draws everything on the field:

import random import numpy as np def vis(A, f, step=1.0, c=None): x = [] y = [] for B in np.arange(0, A, step): v = f(A, B) x.append(v.real) y.append(v.imag) plt.figure(figsize=[8, 8]) mxabs = max([i[0] ** 2 + i[1] ** 2 for i in zip(x, y)]) ** 0.5 x = np.array(x) / mxabs y = np.array(y) / mxabs if c is None: plt.scatter(x, y) else: plt.scatter(x, y, color=[c(x[i], y[i]) for i in range(len(x))]) plt.show() 

All our functions depend on two parameters A and B. Moreover, we iterate over B inside vis (), and A is the global parameter of the function.

The Curl Function

$$

$$f (A, B) = Bsin (B) * e ^ {iBcos (A)}$$

Her declaration in python:

 def func_1(A, B): return math.sin(B) * B * math.e ** (1j * (B * math.cos(A))) 

And run

 A = 121.5 vis(A, func_1, step=0.1) 

And the result for A = 121.5:



And for A = 221.5:



Note that these numbers do not follow from the calculation of any definite integral on a smooth manifold and other smart words meaningless in this context. These are really random numbers, and there is still exactly infinity of different A, as a result of which beauty is obtained.

Need to paint

We declare a color function (such a function that returns a tuple of three numbers in coordinates):

 def sigm(x): #         0  1 return (1 / (1 + 1.2 ** (-x*50)) - 0.5) * 2 color_1 = lambda x, y: (0.2, sigm(x ** 2 + y ** 2) / 1.4, 1 - sigm(x ** 2 + y ** 2)) color_2 = lambda x, y: (sigm(x ** 2 + y ** 2), 0.5, 0.5) 

Choose random parameter A, let it be 149:

 vis(149, func_1, step=0.1, c=color_1) 

Geese Function

Geese are described as follows:

$$

$$f (A, B) = cos (B) sin (B) * B * e ^ {iBcos (A)}$$

Python declaration:

 def func_2(A, B): return math.cos(B) * math.sin(B) * B * math.e ** (1j * (B * math.cos(A))) 

Its result for A = 106:

Focaccia function

$$

$$f (A, B) = cos (B (A + 1)) * e ^ {iBcos (A)}$$

 def func_3(A, B): return math.cos((A + 1) * B) * math.e ** (1j * (B * math.cos(A))) 

 vis(246, func_3, step=0.1, c=color_2) 

 vis(246, func_3, step=0.1, c=color_2) 

Untitled Function

$$

$$f (A, B) = Bsin (A + B) * e ^ {iBsin (A)}$$

 color_3 = lambda x, y: (0.5, 0.5, sigm(x ** 2 + y ** 2)) vis(162, func_4, step=0.1, c=color_3) 

 vis(179, func_4, step=0.1, c=color_3) 

Beauty formula

$$

$$f (A, B) = cos (B (A + 1)) ^ {\ frac {3} {2}} * e ^ {iBcos (A)}$$

 def func_5(A, B): return math.cos((A + 1) * B) ** 1.5 * math.e ** (1j * (B * math.cos(A))) 

 color_4 = lambda x, y: (sigm(x ** 2 + y ** 2) / 2, 0.5, 1 - sigm(x ** 2 + y ** 2)) vis(345, func_5, step=0.1, c=color_4) 

 vis(350, func_5, step=0.1, c=color_4) 

That's all for now.