Source video: Lesson 3: Complex conjugates and dividing complex numbers - Khan Academy

Given a complex number

where $2$ is a real number, $3i$ is an imaginary number. The real part of z $Re(z)$ is 2, while the imaginary part of z $Im(z)$ is 3.

Complex conjugate

The complex conjugate of z is

where the real part Re(z*) remains the same, while the imaginary part Im(z*) has the opposite sign.

This act like mirror reflecting over Real axis.

The sum of a complex number and its complex conjugate is two times of its real part, 2Re(z):

Graphically, vector addition

The product of a complex number and its complex conjugate is a real number, and it’s equal to the magnitude of the complex number squared:

which is useful in the division of comple numbers: multiply the numerator and denominator by the conjugate of the denominator to convert the division to one complex number

For example:

Factoring sum of squares

We can factor a difference of squares as:

But the sum of squares x² + y² cannot be factorized if without considering imaginary unit i.

Modulus of complex value

Source page Lesson 5

The absolute value (modulus) of a number is the distance away from zero.

A complex number $3-4i$ plotted on the complex plane:

The absolute value of 3-4i is the hypotenuse of the right triangle.

Based on the Pythagorean theorem, |3-4i|= √(3²+4²) = 5. (Because distance is positive, so only take the positive square root).

polar & rectangular forms

Source page: Lesson 5

A complex number can be represented in two forms:

1. Read number + imaginary number: a + bi
2. Exponential form

Both of them have the same diagram, and just are described in different coordinates:

The two arguments of complex number z can be (r, φ) or (a,b)

• Magnitude: r = |z| = √(a²+b²)

• Argument (Polar angle): φ = arctan(b/a)

• a = r⋅cosφ

• b = r⋅sinφ

• z = a+bi = r⋅cosφ + r⋅sinφi = r(cosφ + sinφ i) = $r e^{iφ}$

  where $cosφ + sinφ i = e^{iφ}$ can be derived by using Taylor series.

Multiplying complex number

Source page Lesson 7

Given z,

• 3z has the same direction as z, but three times it’s magnitude;

• -3z is in the opposite direction, and has three times modulus of z

• -3iz = $1e^{iπ} ⋅ 3e^{i⋅π/2} ⋅ re^{iφ} = 3r e^{i(π+π/2+φ)}$ which turns into the opposite direction then rotates another 90 degrees in counter-clock wise.

  Or this can be derived from the loop of 1 -> i -> -1 -> -i with multiplying i each time.

• z⋅(-1-i) : the angle and modulus of $-1-i$ take effect on z separately.

  • The angle of -1-i is 225 degrees (observed from the complex plane), so it will rotate the z by 225 degrees.

  • The magnitude of -1-i is √2, so z will be scaled by √2.

  • (I think it as a vector addition: z⋅(-1-i) = -z-zi,, but there’re too many steps).

Operations in polar form

Source page: Lesson 8

Multiplication

Given:

What is w₁⋅w₂ ?

$3 e^{i⋅330°} ⋅ 2 e^{i⋅120°} = 6 e^{i(450°)}$

w₁⋅w₂ can be viewed as w₁ transforming w₂, i.e., w₂ is transformed by multiplying w₁

1. The modulus of w₂ is scaled by the modulus of w₁ (2), so |w₁⋅w₂| = 3⋅2 = 6;

2. The argument (angle) of w₁ is rotated by the argument of w₁ (330°), so arg(w₁⋅w₂) = 120° + 330° = 450° = 90°

So w₁⋅w₂ = 6 (cos(90°) + i sin(90°)) = 6i

Division

What is $\frac{w₁}{w₂}$ ?

$8e^{ i\frac{4π}{3} } / (2e^{ i\frac{7π}{6} }) = 4 e^{i(π/6)}$

Another way to think about it is that w₁ is transformed by w₂:

1. the modulus of w₁ is divided by the modulus of w₂;

2. the argument of w₁ is rotated clock-wise by the argument of w₂.

So $\frac{w₁}{w₂} = 4(cos(π/6) + i sin(π/6)) = 4(√3/2 + i/2)$

Powers of complex number

Consider the complex number $z = -1 + i \sqrt 3$. Find $z^4$ in polar and rectangular form.

1. Modulus of z is 2, so its modulus is times itself four times: 2⁴ = 16
2. Argument of z is φ = $\rm arctan(\sqrt 3)$=60°=120°, so its angle rotate by 4 times of its argument: φx4 = 480° = 120°

The polar form is $z⁴ = 16 (cos(120°) + i sin(120°))$, so the rectangular form is $z⁴ = 16(1/2 + i √3/2) = 8 + i8√3 $

Complex number equations

Given equation x³=1, find all of the real and/or complex roots of this equation.

For a real number: $z= 1 = 1 + 0i = 1e^{i0°}$, its argument arg(z) can be 0, 2π, 4π, …, i.e., $1 = e^{i0} = e^{i2π} = e^{i4π} = e^{i6π}$ …

Plug these exponential form into x³=1:

Fundamental theorem of Algebra

Source page: Lesson 9

The Fundamental theorem of Algebra: a n-th degree polynomial has n roots.

$P(x) = ax^n + bx^{n-1} + ... + K$