import numpy as np
import sympy as sym
import matplotlib.pyplot as plt
from IPython.display import display,Math

print('Roots of unity')
display(Math('n^n = 1, \\quad z \\quad \\epsilon \\quad C'))
display(Math('z = e^{2 \\pi ik / n}'))
display(Math('k = 0,1, ..., n -1'))

Roots of unity

$\displaystyle n^n = 1, \quad z \quad \epsilon \quad C$

$\displaystyle z = e^{2 \pi ik / n}$

$\displaystyle k = 0,1, ..., n -1$

n = 5

for k in range(0,n):
    z = np.exp(2*np.pi*1j*k/n)
    print(z**n)

(1+0j)
(0.9999999999999999-1.1102230246251565e-16j)
(1-6.661338147750939e-16j)
(0.9999999999999997-4.440892098500626e-16j)
(1-1.1102230246251565e-15j)

n = 5

for k in range(0,n):
    z = sym.exp(2*sym.pi*sym.I*k/n)
    print(z**n)

1
1
1
1
1

n = 4

for k in range(0,n):
    z = sym.exp(2*sym.pi*sym.I*k/n)
    display(Math('(%s)^{%s} \\Rightarrow %s' %(sym.latex(z),n,sym.latex(z**n))))

$\displaystyle (1)^{4} \Rightarrow 1$

$\displaystyle (i)^{4} \Rightarrow 1$

$\displaystyle (-1)^{4} \Rightarrow 1$

$\displaystyle (- i)^{4} \Rightarrow 1$

n = 14

for k in range(0,n):
    z = np.exp(2*np.pi*1j*k/n)
    plt.plot([0,np.real(z)],[0,np.imag(z)])
    
plt.axis('square')
x = np.linspace(0,2*np.pi,100)
plt.plot(np.cos(x),np.sin(x),color='grey')

plt.show()

Exercise

display(Math('ke^{\\frac{k2 \\pi i}{n}}'))

$\displaystyle ke^{\frac{k2 \pi i}{n}}$

n = 200

color = np.linspace(0,.9,n)

for k in range(0,n):
    z = k*np.exp(2*np.pi*1j*k/n)
    plt.plot([0,np.real(z)],[0,np.imag(z)],linewidth=2,color=[color[k],color[k],color[k]])
    
plt.axis('square')
plt.axis('off')
plt.show()