9. Complex Numbers and Trigonometry#

9.1.1. Complex Numbers#

A complex number has a real part $x$ and a purely imaginary part $y$.

The Euclidean, polar, and trigonometric forms of a complex number $z$ are:

The second equality above is known as Euler’s formula

• Euler contributed many other formulas too!

The complex conjugate $\bar z$ of $z$ is defined as

The value $x$ is the real part of $z$ and $y$ is the imaginary part of $z$.

The symbol $| z |$ = $\sqrt{\bar{z}\cdot z} = r$ represents the modulus of $z$.

The value $r$ is the Euclidean distance of vector $(x,y)$ from the origin:

The value $\theta$ is the angle of $(x,y)$ with respect to the real axis.

Evidently, the tangent of $\theta$ is $\left(\frac{y}{x}\right)$.

Therefore,

Three elementary trigonometric functions are

We’ll need the following imports:

import matplotlib.pyplot as plt
plt.rcParams["figure.figsize"] = (11, 5)  #set default figure size
import numpy as np
from sympy import (Symbol, symbols, Eq, nsolve, sqrt, cos, sin, simplify,
                  init_printing, integrate)

9.1.2. An Example#

Let’s use Python to plot the trigonometric form of the complex number $z = 1 + \sqrt{3} i$.

# Abbreviate useful values and functions
π = np.pi


# Set parameters
r = 2
θ = π/3
x = r * np.cos(θ)
x_range = np.linspace(0, x, 1000)
θ_range = np.linspace(0, θ, 1000)

# Plot
fig = plt.figure(figsize=(8, 8))
ax = plt.subplot(111, projection='polar')

ax.plot((0, θ), (0, r), marker='o', color='b')          # Plot r
ax.plot(np.zeros(x_range.shape), x_range, color='b')       # Plot x
ax.plot(θ_range, x / np.cos(θ_range), color='b')        # Plot y
ax.plot(θ_range, np.full(θ_range.shape, 0.1), color='r')  # Plot θ

ax.margins(0) # Let the plot starts at origin

ax.set_title("Trigonometry of complex numbers", va='bottom',
    fontsize='x-large')

ax.set_rmax(2)
ax.set_rticks((0.5, 1, 1.5, 2))  # Less radial ticks
ax.set_rlabel_position(-88.5)    # Get radial labels away from plotted line

ax.text(θ, r+0.01 , r'$z = x + iy = 1 + \sqrt{3}\, i$')   # Label z
ax.text(θ+0.2, 1 , '$r = 2$')                             # Label r
ax.text(0-0.2, 0.5, '$x = 1$')                            # Label x
ax.text(0.5, 1.2, r'$y = \sqrt{3}$')                      # Label y
ax.text(0.25, 0.15, r'$\theta = 60^o$')                   # Label θ

ax.grid(True)
plt.show()

9.3.1. Example 1#

We can use de Moivre’s theorem to show that $r = \sqrt{x^2 + y^2}$.

We have

and thus

We recognize this as a theorem of Pythagoras.

9.3.2. Example 2#

Let $z = re^{i\theta}$ and $\bar{z} = re^{-i\theta}$ so that $\bar{z}$ is the complex conjugate of $z$.

$(z, \bar z)$ form a complex conjugate pair of complex numbers.

Let $a = pe^{i\omega}$ and $\bar{a} = pe^{-i\omega}$ be another complex conjugate pair.

For each element of a sequence of integers $n = 0, 1, 2, \ldots,$.

To do so, we can apply de Moivre’s formula.

Thus,

9.3.3. Example 3#

This example provides machinery that is at the heard of Samuelson’s analysis of his multiplier-accelerator model [Samuelson, 1939].

Thus, consider a second-order linear difference equation

whose characteristic polynomial is

or

has roots $z_1, z_1$.

A solution is a sequence $\{x_n\}_{n=0}^\infty$ that satisfies the difference equation.

Under the following circumstances, we can apply our example 2 formula to solve the difference equation

• the roots $z_1, z_2$ of the characteristic polynomial of the difference equation form a complex conjugate pair

• the values $x_0, x_1$ are given initial conditions

To solve the difference equation, recall from example 2 that

where $\omega, p$ are coefficients to be determined from information encoded in the initial conditions $x_1, x_0$.

Since $x_0 = 2 p \cos{\omega}$ and $x_1 = 2 pr \cos{(\omega + \theta)}$ the ratio of $x_1$ to $x_0$ is

We can solve this equation for $\omega$ then solve for $p$ using $x_0 = 2 pr^0 \cos{(\omega + n\theta)}$.

With the sympy package in Python, we are able to solve and plot the dynamics of $x_n$ given different values of $n$.

In this example, we set the initial values: - $r = 0.9$ - $\theta = \frac{1}{4}\pi$ - $x_0 = 4$ - $x_1 = r \cdot 2\sqrt{2} = 1.8 \sqrt{2}$.

We first numerically solve for $\omega$ and $p$ using nsolve in the sympy package based on the above initial condition:

# Set parameters
r = 0.9
θ = π/4
x0 = 4
x1 = 2 * r * sqrt(2)

# Define symbols to be calculated
ω, p = symbols('ω p', real=True)

# Solve for ω
## Note: we choose the solution near 0
eq1 = Eq(x1/x0 - r * cos(ω+θ) / cos(ω), 0)
ω = nsolve(eq1, ω, 0)
ω = float(ω)
print(f'ω = {ω:1.3f}')

# Solve for p
eq2 = Eq(x0 - 2 * p * cos(ω), 0)
p = nsolve(eq2, p, 0)
p = float(p)
print(f'p = {p:1.3f}')

ω = 0.000
p = 2.000

Using the code above, we compute that $\omega = 0$ and $p = 2$.

Then we plug in the values we solve for $\omega$ and $p$ and plot the dynamic.

# Define range of n
max_n = 30
n = np.arange(0, max_n+1, 0.01)

# Define x_n
x = lambda n: 2 * p * r**n * np.cos(ω + n * θ)

# Plot
fig, ax = plt.subplots(figsize=(12, 8))

ax.plot(n, x(n))
ax.set(xlim=(0, max_n), ylim=(-5, 5), xlabel='$n$', ylabel='$x_n$')

# Set x-axis in the middle of the plot
ax.spines['bottom'].set_position('center')
ax.spines['right'].set_color('none')
ax.spines['top'].set_color('none')
ax.xaxis.set_ticks_position('bottom')
ax.yaxis.set_ticks_position('left')

ticklab = ax.xaxis.get_ticklabels()[0] # Set x-label position
trans = ticklab.get_transform()
ax.xaxis.set_label_coords(31, 0, transform=trans)

ticklab = ax.yaxis.get_ticklabels()[0] # Set y-label position
trans = ticklab.get_transform()
ax.yaxis.set_label_coords(0, 5, transform=trans)

ax.grid()
plt.show()

9.3.4. Trigonometric Identities#

We can obtain a complete suite of trigonometric identities by appropriately manipulating polar forms of complex numbers.

We’ll get many of them by deducing implications of the equality

For example, we’ll calculate identities for

$\cos{(\omega + \theta)}$ and $\sin{(\omega + \theta)}$.

Using the sine and cosine formulas presented at the beginning of this lecture, we have:

We can also obtain the trigonometric identities as follows:

Since both real and imaginary parts of the above formula should be equal, we get:

The equations above are also known as the angle sum identities. We can verify the equations using the simplify function in the sympy package:

# Define symbols
ω, θ = symbols('ω θ', real=True)

# Verify
print("cos(ω)cos(θ) - sin(ω)sin(θ) =",
    simplify(cos(ω)*cos(θ) - sin(ω) * sin(θ)))
print("cos(ω)sin(θ) + sin(ω)cos(θ) =",
    simplify(cos(ω)*sin(θ) + sin(ω) * cos(θ)))

cos(ω)cos(θ) - sin(ω)sin(θ) = cos(θ + ω)
cos(ω)sin(θ) + sin(ω)cos(θ) = sin(θ + ω)

9.3.5. Trigonometric Integrals#

We can also compute the trigonometric integrals using polar forms of complex numbers.

For example, we want to solve the following integral:

Using Euler’s formula, we have:

and thus:

We can verify the analytical as well as numerical results using integrate in the sympy package:

# Set initial printing
init_printing(use_latex="mathjax")

ω = Symbol('ω')
print('The analytical solution for integral of cos(ω)sin(ω) is:')
integrate(cos(ω) * sin(ω), ω)

The analytical solution for integral of cos(ω)sin(ω) is:

print('The numerical solution for the integral of cos(ω)sin(ω) \
from -π to π is:')
integrate(cos(ω) * sin(ω), (ω, -π, π))

The numerical solution for the integral of cos(ω)sin(ω) from -π to π is:

9.3.6. Exercises#

# Import symbolic π from sympy
from sympy import pi

print('The analytical solution for the integral of cos(ω)**2 \
from -π to π is:')

integrate(cos(ω)**2, (ω, -pi, pi))

The analytical solution for the integral of cos(ω)**2 from -π to π is:

print('The analytical solution for the integral of sin(ω)**2 \
from -π to π is:')

integrate(sin(ω)**2, (ω, -pi, pi))

The analytical solution for the integral of sin(ω)**2 from -π to π is: