Complex numbers: OxRSE Training

YouTube lecture recording from October 2020

A recording of this lecture from October 2020; the material has changed a little since.

Why complex numbers?

Some equations, like

x^2+1=0

, have no solutions among the real numbers. Complex numbers give every polynomial a full set of roots.

We will meet them again when we study systems of differential equations. There, complex roots tell us that the system oscillates (e.g. populations of predators and prey that rise and fall in turn).

The link is

e^{i\theta}=\cos\theta+i\sin\theta

, which we show at the end of this lecture.

Imaginary numbers

The imaginary number $i$ :

The polynomial $x^2-1=0$ has two real roots: $1$ and $-1$
The polynomial $x^2+1=0$ has no real roots.

Consider solving:

x^2 = 1\qquad\text{and}\qquad x^2 = -1

We introduce an "imaginary" number

i

so that there are two solutions to

x^2 = -1

i

and

-i

. That is,

i

is a number with the property that

i^2=-1

Complex numbers

The complex numbers are the set of all expressions of the form

a + bi

where

i^2=-1

and

a

and

b

are real numbers:

\mathbb{C}=\left\{a + bi \mid a,\ b \in\mathbb{R}\right\}

For

z=a+bi\in\mathbb{C}

we define the real and imaginary parts of

z

to be

\Re(z) = a

and

\Im(z)=b

The imaginary number

i

has no home on the real number line. Instead, we locate it on the complex plane at the point

(0,1)

We can represent any complex number $z=a+bi$ as the point $(a,b)$ in the complex plane.
The coordinates $(a,b)$ are usually called the Cartesian coordinates for $z$ . (Named after the mathematician and philosopher René Descartes).
In this plane, the real numbers lie on the horizontal axis. We usually refer to the horizontal axis of $\mathbb{C}$ as the real axis.

The following plot shows a complex number with a real component of 18 and an imaginary component of 12:

Complex conjugates

The complex plane $\mathbb{C}$ contains all of the roots of every polynomial.

E.g., let's look at a quadratic equation. Recall the quadratic formula:

ax^2 +bx +c = 0 \iff x= \frac{-b \pm\sqrt{b^2-4ac}}{2a}

Now, let's consider this specific quadratic equation:

x^2-8x+25 = 0

which we can solve using the quadratic formula:

x=\frac{8\pm\sqrt{64-100}}{2}=\frac{8\pm\sqrt{-36}}{2}=\frac{8\pm6i}{2}=4\pm3i

Note that these two roots are reflections of one another through the real axis. They are conjugates of one another.
In general, let $z=a + bi$ . The conjugate of $z$ is the complex number $\bar{z}=a-bi$ .

We can also use SymPy's solve method to solve polynomials:

import sympy as sp
x = sp.symbols('x')
sp.solve(x**2 - 8*x + 25)

\left[ 4 - 3 i, \ 4 + 3 i\right]

Modulus (size) of a complex number

The distance to a point on the complex plane from 0 is called its modulus, and we find this by calculating the hypotenuse of the triangle with base

\Re(z)

and height

\Im(z)

E.g. The modulus of

4\pm3i

\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5

In general, the modulus of $z=a+bi$ is the real number $|z|=\sqrt{a^2+b^2}$ .
The modulus is connected to the conjugate by means of the formula $z\cdot \bar{z}=|z|^2$ . Indeed:

\begin{aligned} z\cdot\bar{z}&=(a+bi)(a-bi)=a^2-(bi)^2=a^2-b^2\cdot i^2\\ &=a^2-b^2(-1)=a^2+b^2=|z|^2 \end{aligned}

Complex numbers in Python

Note that, in Python, because i is often used as an index variable in loops, the language uses j, instead, to represent the imaginary unit:

x = 1 + 2j
print(f'x = {x}     Re(x) = {x.real}     Im(x) = {x.imag}     |x| = {abs(x)}')

x = (1+2j)     Re(x) = 1.0     Im(x) = 2.0     |x| = 2.23606797749979

In SymPy, the imaginary unit is sp.I:

x = 1 + 2 * sp.I
print(f'x = {x}     Re(x) = {sp.re(x)}     Im(x) = {sp.im(x)}     |x| = {sp.Abs(x)}')

x = 1 + 2*I     Re(x) = 1     Im(x) = 2     |x| = sqrt(5)

Addition and subtraction

Addition and subtraction of complex numbers work as you would expect:

(a+bi)\pm(c+di)=(a\pm c) + (b\pm d)i

and

-(a+bi)=-a-bi

Try adding:

(5+6i)+(1-i)

print((5 + 6j) + (1 - 1j))

(6+5j)

Try subtracting:

(5+6i)-(1-i)

print((5 + 6j) - (1 - 1j))

(4+7j)

Multiplication

Multiplication is not quite so convenient in Cartesian coordinates:

\begin{aligned} (a+bi)(c+di)&=ac + adi + bci + bidi \\ &= ac + adi + bci -bd \\ &= (ac-bd)+(ad+bc)i \end{aligned}

Try multiplying:

(5+6i)(1-i)

print((5 + 6j) * (1 - 1j))

(11+1j)

Division

Division is even more awkward in Cartesian coordinates: we have to multiply the numerator and the denominator by the complex conjugate of the denominator.

\begin{aligned} \frac{a+bi}{c+di}&=\frac{(a+bi)(c-di)}{(c+di)(c-di)}\\ &=\frac{(ac+bd)+(bc-ad)i}{c^2+d^2}=\left(\frac{ac+bd}{c^2+d^2}\right)+ \left(\frac{bc-ad}{c^2+d^2}\right)i \end{aligned}

Try dividing:

\frac{-4+7i}{2+3i}

print(sp.simplify((-4 + 7*sp.I) / (2 + 3*sp.I)))

1 + 2*I

Polar coordinates

It is often convenient to represent the complex number

z = a + bi

in terms of its polar coordinates

\langle r,\theta\rangle

The angle $\theta$ is called the argument of $z$ .
We take $\theta$ in the range $(-\pi, \pi]$ , so that each non-zero $z$ has exactly one argument. A round bracket leaves out its end point and a square bracket includes it: $\theta=\pi$ is allowed, but $\theta=-\pi$ is not.
The real number $r=|z|$ is sometimes denoted mod $(z)$ .

Representing a complex number in polar coordinates

Connection between Cartesian and polar

Let

z=x+iy

. If we are given the polar coordinates of

z

and want to express the Cartesian coordinates use

\begin{gathered} x=r\cos\theta\\ y=r\sin\theta\\ z=r\cos\theta + ri\sin\theta=r(\cos\theta + i\sin\theta) \end{gathered}

If we are given the Cartesian coordinates and want to find the polar coordinates, use:

r=\text{mod}(z)=|z|=\sqrt{x^2+y^2}

\theta=\text{arg}(z)= \begin{cases}\pi/2, & \text{if } x=0,y>0 \\ -\pi/2, & \text{if } x=0,y<0 \\ \arctan\left(\frac{y}{x}\right), & \text{if } x>0\\ \arctan\left(\frac{y}{x}\right)+\pi, & \text{if } x<0, y\geq 0\\ \arctan\left(\frac{y}{x}\right)-\pi, & \text{if } x<0, y<0 \end{cases}

Note: all of the fuss about the value of

\theta

in the formula above is to make sure that

z

gets into the proper quadrant. Beware of the sign of this tangent: it depends on which quadrant you are in.

The positive

x

axis is defined as having

\theta=0

and positive

\theta

goes in an anticlockwise sense around the

x-y

plane.

Some examples

Example 1

Find the Cartesian coordinates for the complex number

z

with polar coordinates

r=2

and

\theta=\pi/6

\begin{aligned} \Re(z)&=x=r\cos\theta=2\cos(\pi/6)=2\left(\frac{\sqrt{3}}{2}\right)=\sqrt{3}\\ \Im(z)&=y=r\sin\theta=2\sin(\pi/6)=2\left(\frac{1}{2}\right)=1 \end{aligned}

Therefore,

z = \sqrt{3} + i

import cmath
import numpy as np
print(cmath.rect(2, np.pi/6))

(1.7320508075688774+0.9999999999999999j)

Example 2

Find the polar coordinates for the complex number

z= -3+4i

\begin{aligned} |z|=r &= \sqrt{(-3)^2+4^2}=\sqrt{25}=5 \\ \text{arg}(z)=\theta&=\arctan\left(\frac{y}{x}\right)+\pi \qquad \text{(as } x<0,\ y\geq 0\text{)}\\ &=\arctan(-4/3)+\pi = -0.93+\pi \approx 2.21 \text{ radians}\approx 127^\circ \end{aligned}

print(cmath.polar(-3 + 4j))

(5.0, 2.214297435588181)

Example 3

Find the polar coordinates for the complex number

z= -2i

\begin{aligned} \text{mod}(z)&=r = |z|=2\\ \text{arg}(z)&=\theta=-\frac{\pi}{2} \end{aligned}

print(cmath.polar(-2j))

(2.0, -1.5707963267948966)

Multiplication in polar coordinates

First a reminder of three useful and important identities:

\begin{aligned} \cos^2\theta + \sin^2\theta &= 1 \\ \cos(\theta_1+\theta_2)&=\cos\theta_1\cos \theta_2 - \sin\theta_1\sin\theta_2 \\ \sin(\theta_1+\theta_2)&=\sin\theta_1\cos \theta_2 + \sin\theta_2\cos\theta_1 \end{aligned}

Now, let

z_1=r_1\cos\theta_1+ir_1\sin\theta_1

and

z_2=r_2\cos\theta_2+ir_2\sin\theta_2

We first compute the real part of the product

z_1\cdot z_2

\begin{aligned} \Re(z_1\cdot z_2) &= r_1\cos\theta_1\cdot r_2\cos\theta_2 - r_1\sin\theta_1\cdot r_2\sin\theta_2\\ &=r_1r_2(\cos\theta_1\cos\theta_2 - \sin\theta_1\sin\theta_2)\\ &=r_1r_2\cos(\theta_1 + \theta_2) \end{aligned}

Note that for the real part the moduli have been multiplied and the arguments added.

Now we compute the imaginary part of

z_1\cdot z_2

\begin{aligned} \Im(z_1\cdot z_2) &= r_1\sin\theta_1\cdot r_2\cos\theta_2 + r_2\sin\theta_2\cdot r_1\cos\theta_1\\ &=r_1r_2(\sin\theta_1\cos\theta_2 + \sin\theta_2\cos\theta_1)\\ &=r_1r_2\sin(\theta_1 + \theta_2) \end{aligned}

For the imaginary part too, the moduli multiply while the arguments add.

This gives a relatively compact and highly geometric result for the product:

z_1\cdot z_2 = r_1r_2(\cos(\theta_1 + \theta_2)+i\sin(\theta_1 + \theta_2))

It is multiplicative in the modulus and additive in the argument:

\begin{aligned} |z_1z_2|&= |z_1|\cdot |z_2|\\ \arg(z_1z_2)&=\arg (z_1)+ \arg( z_2) \qquad \text{(adding or subtracting } 2\pi \text{ if needed, to stay in } (-\pi, \pi]\text{)} \end{aligned}

This means that when we multiply by

z

, we are rotating through the angle

\arg(z)

and radially stretching by a factor of

|z|

A remarkable connection with $e^{i\theta}$

First, think of

z=\cos\theta + i\sin\theta

as a function of

\theta

and differentiate with respect to

\theta

\frac{dz}{d\theta}=\frac{d}{d\theta}\left(\cos\theta+i\sin\theta\right)=-\sin\theta+i\cos\theta \qquad \text{(1)}

Next notice that the right-hand side is just the product

iz

iz=i(\cos\theta+i\sin\theta)=i\cos\theta+i^2\sin\theta=-\sin\theta+i\cos\theta \qquad \text{(2)}

Thus, from (1) and (2),

\frac{dz}{d\theta}=iz

This is a separable differential equation (we cover differential equations properly in the next lecture):

\int\frac{dz}{iz}=\int d\theta \qquad\Rightarrow\qquad \frac{1}{i}\ln z =\theta +c \qquad\Rightarrow\qquad \ln z =i\theta +ic

In exponential form:

z=e^{i\theta+ic} = e^{i\theta}\,e^{ic} = Ae^{i\theta} \qquad \text{with} \qquad A=e^{ic}

When

\theta=0

z=1

, giving

A=1

, so:

\cos\theta + i \sin\theta=e^{i\theta} \qquad \text{(3)}

Putting (3) into the polar form

r(\cos\theta + i\sin\theta)

, any complex number can now be written:

z=x+iy=r(\cos\theta + i\sin\theta)=r\,e^{i\theta}

Several important consequences

Any complex number can be written in the polar form $z = re^{i\theta}$ where $r=|z|$ and $\theta=\arg(z)$ .
The unit circle in $\mathbb{C}$ consists exactly of those complex numbers with modulus equal to 1, that is the numbers $e^{i\theta}$ .
Multiplication on the unit circle $r=1$ can be carried out by adding the angles:
$e^{i\theta_1}\cdot e^{i\theta_2} = e^{i(\theta_1+\theta_2)}$
Exponentiation on the unit circle $r=1$ can be done by multiplying the angle by the index (for integer $n$ ). Written with cos and sin, this is known as de Moivre's theorem.
$\left(e^{i\theta}\right)^n = e^{i\theta n}=e^{i(n\theta)}$
This makes powers quick. For example, $1+i=\sqrt{2}\,e^{i\pi/4}$ , so:
$(1+i)^8=\left(\sqrt{2}\right)^8\left(e^{i\pi/4}\right)^8=16\,e^{2\pi i}=16$
Finally, the famous identity by Leonhard Euler
$e^{\pi i}+1=0$

Introductory problems

For more practice, an AI tool can write new problems like the ones below. Ask it for problems of the same type and difficulty, or for problems set in your own field. For example, you could ask it to "write five physics problems on complex numbers, at the level of these ones". Work them on paper first, then ask the tool to check your answers and to explain any step you are unsure of.

Introductory problems 1

Simplify:

$2-2i\quad+\quad 3+i$
$4-6i\quad+\quad 19+4i$
$(1-i)^2$
$(2+i)^2$
$(4-3i)/(2+6i)$
$(1+2i)/(1-3i)$
$(2-i)^{-2} +(2+i)^{-2}$
$(5-i)^{-2} -(5+i)^{-2}$

Add the real and imaginary parts separately, use

i^2=-1

when multiplying, and divide by multiplying the numerator and the denominator by the complex conjugate of the denominator.

$2-2i + 3+i = 5-i$
$4-6i + 19+4i = 23-2i$
$(1-i)^2 = 1 - 2i + i^2 = -2i$
$(2+i)^2 = 4 + 4i + i^2 = 3+4i$
$\frac{4-3i}{2+6i} = \frac{(4-3i)(2-6i)}{(2+6i)(2-6i)} = \frac{8 - 30i + 18i^2}{4 + 36} = \frac{-10-30i}{40} = -\frac{1}{4} - \frac{3}{4}i$
$\frac{1+2i}{1-3i} = \frac{(1+2i)(1+3i)}{(1-3i)(1+3i)} = \frac{1 + 5i + 6i^2}{1+9} = \frac{-5+5i}{10} = -\frac{1}{2} + \frac{1}{2}i$
Using $(2-i)^2 = 3-4i$ and $(2+i)^2 = 3+4i$ :
$\frac{1}{3-4i} + \frac{1}{3+4i} = \frac{(3+4i) + (3-4i)}{(3-4i)(3+4i)} = \frac{6}{9+16} = \frac{6}{25}$
Using $(5-i)^2 = 24-10i$ and $(5+i)^2 = 24+10i$ :
$\frac{1}{24-10i} - \frac{1}{24+10i} = \frac{(24+10i) - (24-10i)}{(24-10i)(24+10i)} = \frac{20i}{576+100} = \frac{5}{169}i$
Note, in 7. the two numbers are complex conjugates of each other, so their sum is real, while in 8. their difference is purely imaginary.

Introductory problems 2

Find the sum, difference, product and quotient of the complex numbers

z_1=5+3i

and

z_2=-4+2i

\begin{aligned} z_1 + z_2 &= (5-4) + (3+2)i = 1+5i \\ z_1 - z_2 &= (5+4) + (3-2)i = 9+i \\ z_1 z_2 &= (5+3i)(-4+2i) = -20 + 10i - 12i + 6i^2 = -26-2i \\ \frac{z_1}{z_2} &= \frac{(5+3i)(-4-2i)}{(-4+2i)(-4-2i)} = \frac{-20 - 22i - 6i^2}{16+4} \\ &= \frac{-14-22i}{20} = -\frac{7}{10} - \frac{11}{10}i \end{aligned}

Main problems

Main problems 1

Solve the following equations for

z

$(7 + i)z - 3i = 6$
$(z-i)/(z+i)=2/3$
$z^2 + (1+4i)z + (15 + 27 i) = 0$

Hint for 3.: the quadratic formula needs the square root of

-75-100i

. To find it, find the real numbers

a

and

b

with

(a+bi)^2 = -75-100i

Rearrange, then divide by multiplying by the complex conjugate:
$\begin{aligned} z &= \frac{6+3i}{7+i} = \frac{(6+3i)(7-i)}{(7+i)(7-i)} = \frac{42 + 15i - 3i^2}{49+1} \\ &= \frac{45+15i}{50} = \frac{9}{10} + \frac{3}{10}i \end{aligned}$
Multiply both sides by $3(z+i)$ :
$3(z-i) = 2(z+i) \quad\Longrightarrow\quad 3z - 3i = 2z + 2i \quad\Longrightarrow\quad z = 5i$
The quadratic formula works for complex numbers too:
$z = \frac{-(1+4i) \pm \sqrt{(1+4i)^2 - 4(15+27i)}}{2}$
The number under the square root is $(1+4i)^2 - 4(15+27i) = -15 + 8i - 60 - 108i = -75-100i$ . Following the hint, $(a+bi)^2 = a^2 - b^2 + 2abi$ , so we need
$a^2 - b^2 = -75 \qquad\text{and}\qquad 2ab = -100$
The second equation gives $b = -50/a$ . Substituting into the first gives $a^4 + 75a^2 - 2500 = 0$ , which is a quadratic in $a^2$ with roots $a^2 = 25$ and $a^2 = -100$ . Since $a$ is real, $a^2 = 25$ , so $a=5$ and $b=-10$ (or $a=-5$ and $b=10$ , the same root with the opposite sign). So, the square root is $\pm(5-10i)$ , and
$\begin{aligned} z &= \frac{-1-4i + (5-10i)}{2} = 2-7i \\ \text{or}\qquad z &= \frac{-1-4i - (5-10i)}{2} = -3+3i \end{aligned}$
We can check by multiplying out: $(z - (2-7i))(z - (-3+3i)) = z^2 + (1+4i)z + (15+27i)$ .

Main problems 2

Represent the complex numbers

z_1= 5-2i

and

z_2=-2 +4i

on a sketch/plot of the complex plane (an Argand diagram).

Write down $z_1$ and $z_2$ in polar form (either give $r$ and $\theta$ for each, or write them as exponentials).
What is the product of $z_1$ and $z_2$ ?

z_1

is the point

(5,-2)

, below the real axis on the right, and

z_2

is the point

(-2,4)

, above the real axis on the left.

from matplotlib import pyplot as plt

z1 = 5 - 2j
z2 = -2 + 4j

for z in [z1, z2]:
    plt.plot([0, z.real], [0, z.imag], marker='o')
plt.axhline(0, color='black', linewidth=0.5)
plt.axvline(0, color='black', linewidth=0.5)
plt.xlabel('Real axis')
plt.ylabel('Imaginary axis')
plt.gca().set_aspect('equal')

For $z_1$ , which is in the fourth quadrant:
$r_1 = \sqrt{5^2 + 2^2} = \sqrt{29} \approx 5.39, \qquad \theta_1 = \arctan\left(\frac{-2}{5}\right) \approx -0.381$
For $z_2$ , which is in the second quadrant, we add $\pi$ to the arctangent:
$\begin{aligned} r_2 &= \sqrt{2^2 + 4^2} = \sqrt{20} \approx 4.47, \\ \theta_2 &= \arctan\left(\frac{4}{-2}\right) + \pi \approx -1.107 + \pi \approx 2.034 \end{aligned}$
So, $z_1 \approx 5.39\,e^{-0.381i}$ and $z_2 \approx 4.47\,e^{2.034i}$ .
In Cartesian form:
$z_1 z_2 = (5-2i)(-2+4i) = -10 + 20i + 4i - 8i^2 = -2+24i$
In polar form, the moduli multiply and the arguments add:
$z_1 z_2 = \sqrt{29}\sqrt{20}\,e^{(-0.381 + 2.034)i} \approx 24.1\,e^{1.65i}$
which is the same number, since $\sqrt{2^2 + 24^2} = \sqrt{580} \approx 24.1$ .

Main problems 3

Find the sum, difference, product and quotient of the complex numbers

z_1=5e^{4i}

and

z_2=3e^{-2i}

. Can you represent the results on an Argand diagram?

The product and quotient are easiest in polar form, where the moduli multiply (or divide) and the arguments add (or subtract):

z_1 z_2 = 15\,e^{2i}, \qquad \frac{z_1}{z_2} = \frac{5}{3}\,e^{6i}

For the sum and difference we need Cartesian form, using

e^{i\theta}=\cos\theta+i\sin\theta

\begin{aligned} z_1 &= 5\cos(4) + 5i\sin(4) \approx -3.268 - 3.784i \\ z_2 &= 3\cos(2) - 3i\sin(2) \approx -1.248 - 2.728i \end{aligned}

So:

\begin{aligned} z_1 + z_2 &\approx -4.517 - 6.512i \\ z_1 - z_2 &\approx -2.020 - 1.056i \\ z_1 z_2 &= 15\cos(2) + 15i\sin(2) \approx -6.242 + 13.64i \\ \frac{z_1}{z_2} &= \frac{5}{3}\cos(6) + \frac{5}{3}i\sin(6) \approx 1.600 - 0.466i \end{aligned}

(An argument of 6 radians is the same as

6 - 2\pi \approx -0.283

radians, just below the positive real axis.)

import numpy as np
from matplotlib import pyplot as plt

z1 = 5 * np.exp(4j)
z2 = 3 * np.exp(-2j)

for z in [z1, z2, z1 + z2, z1 - z2, z1 * z2, z1 / z2]:
    plt.plot(z.real, z.imag, marker='o')
plt.axhline(0, color='black', linewidth=0.5)
plt.axvline(0, color='black', linewidth=0.5)
plt.xlabel('Real axis')
plt.ylabel('Imaginary axis')
plt.gca().set_aspect('equal')

Main problems 4

What are the

(x,y)

coordinates of the complex number

5e^{4i}

Main problems 5

z=1+i

, mark on an Argand diagram the four points

A

B

C

and

D

representing

z

z^2

z^3

and

z^4

respectively.

Find by calculation or from your diagram, the moduli and arguments of the complex numbers

z^2-1

and

z+z^4

In polar form,

z = 1+i = \sqrt{2}\,e^{i\pi/4}

. So, each power of

z

multiplies the modulus by

\sqrt{2}

and rotates through another

\pi/4

\begin{aligned} z^2 &= 2\,e^{i\pi/2} = 2i \\ z^3 &= 2\sqrt{2}\,e^{3i\pi/4} = -2+2i \\ z^4 &= 4\,e^{i\pi} = -4 \end{aligned}

So,

A

B

C

and

D

are the points

(1,1)

(0,2)

(-2,2)

and

(-4,0)

, spiralling outwards anticlockwise.

Then

z^2 - 1 = -1+2i

, which is in the second quadrant:

|z^2-1| = \sqrt{1+4} = \sqrt{5} \approx 2.24, \qquad \arg(z^2-1) = \pi - \arctan(2) \approx 2.03

And

z + z^4 = -3+i

, also in the second quadrant:

\begin{aligned} |z+z^4| &= \sqrt{9+1} = \sqrt{10} \approx 3.16, \\ \arg(z+z^4) &= \pi - \arctan\left(\frac{1}{3}\right) \approx 2.82 \end{aligned}

Main problems 6

Find the complex numbers represented by the vertices of a square if one vertex represents

3+3i

and the centre of the square represents

1+2i

From the centre to the given vertex is

(3+3i) - (1+2i) = 2+i

. The other vertices are the same distance from the centre, rotated by a right angle each time. Multiplying by

i

rotates through

\pi/2

, so the steps from the centre to the other vertices are

(2+i)i = -1+2i, \qquad (2+i)i^2 = -2-i, \qquad (2+i)i^3 = 1-2i

Adding each to the centre

1+2i

gives the vertices

3+3i, \qquad 4i, \qquad -1+i, \qquad 2

Extension problems

Extension problems 1

Use Python's complex numbers to check your answers to Introductory problems 1 and Main problems 1.

Python writes

i

as j, so we can type each expression in directly:

print(2 - 2j + 3 + 1j)
print(4 - 6j + 19 + 4j)
print((1 - 1j)**2)
print((2 + 1j)**2)
print((4 - 3j) / (2 + 6j))
print((1 + 2j) / (1 - 3j))
print((2 - 1j)**-2 + (2 + 1j)**-2)
print((5 - 1j)**-2 - (5 + 1j)**-2)

(5-1j)
(23-2j)
-2j
(3+4j)
(-0.25-0.75j)
(-0.5+0.49999999999999994j)
(0.24+0j)
0.029585798816568046j

These agree with our answers:

6/25 = 0.24

and

5/169 \approx 0.0296

. Python works with decimals, so a few answers carry a tiny rounding error, like

0.49999999999999994

for

1/2

To solve the equations in Main problems 1, SymPy keeps the answers exact (it writes

i

as I):

import sympy as sp

z = sp.symbols('z')
print(sp.solve((7 + sp.I) * z - 3 * sp.I - 6, z))
print(sp.solve((z - sp.I) / (z + sp.I) - sp.Rational(2, 3), z))
print(sp.solve(z**2 + (1 + 4 * sp.I) * z + (15 + 27 * sp.I), z))

[9/10 + 3*I/10]
[5*I]
[-3 + 3*I, 2 - 7*I]

Complex numbers

YouTube lecture recording from October 2020

Why complex numbers?

Imaginary numbers

Complex numbers

Complex conjugates

Modulus (size) of a complex number

Complex numbers in Python

Addition and subtraction

Multiplication

Division

Polar coordinates

Connection between Cartesian and polar

Some examples

Example 1

Example 2

Example 3

Multiplication in polar coordinates

A remarkable connection with eiθe^{i\theta}eiθ

Several important consequences

Introductory problems

Introductory problems 1

Introductory problems 2

Main problems

Main problems 1

Main problems 2

Main problems 3

Main problems 4

Main problems 5

Main problems 6

Extension problems

Extension problems 1

A remarkable connection with $e^{i\theta}$