Complex Numbers - a MaS Tutor.

\sqrt{-1} = i

i^{2} = -1

A complex number is written in the form $a + b i$ , where $a, b \in ℝ$ . $a$ is the real part of the complex number and $b i$ is the imaginary part.

If $b = 0$ , that is $z = a$ , then $z$ is a real number.

If $a = 0$ , that is $z = b i$ , then $z$ is an imaginary number.

Complex numbers can be added or subtracted by adding or subtracting their real parts, and adding or subtracting their imaginary parts.

For example:
If $z_{1} = 4 + 3 i$ and $z_{2} = -3 + i$ ,

then

z_{1} + z_{2} = 4 + 3 i + -3 + i

= 4 - 3 + 3 i + i

= 1 + 4 i

and

z_{2} - z_{1} = -3 + i - (4 + 3 i)

= -3 + i - 4 - 3 i

= -3 - 4 + i - 3 i

= -1 - 2 i

A real number can be multiplied by a complex number by multiplying out the brackets in the usual way.

Complex numbers can be multiplied together by multiplying out the brackets in the usual way simplifying terms using $i^{2} = -1$ .

If $b^{2} - 4 a c < 0$ then the quadratic equation $a x^{2} + b x + c = 0$ has two distinct complex roots, which are not real.

For any complex number, $z = a + b i$ , the complex conjugate of the number is defined as $z * = a - b i$ .

Complex numbers can be divided by forming a quotient and multiplying the numerator and denominator by the complex conjugate of the demoninator. Simplifying terms with $i^{2} = -1$ .

For real numbers $a$ , $b$ and $c$ , if the roots of the quadratic equation $a z^{2} + b z + c = 0$ are non-real complex numbers, then they occur as a conjugate pair.

If the roots of a quadratic equation are $α$ and $β$ , then you can write the equation as $(z - α) (z - β) = 0$ or $z^{2} - (α + β) z + α β = 0$ .

If $f (z)$ is a polynomial with real coefficients, and $z_{1}$ is a root of $f (z) = 0$ , then $z_{1} *$ is also a root.

An equation of the form $a z^{3} + b z^{2} + c z + d = 0$ has three roots. For a cubic equation with real coefficients, either:

all three roots are real, or
one root is real and the other two roots form a complex conjugate pair.

An equation of the form $a z^{4} + b z^{3} + c z^{2} + c z + d = 0$ has four roots. For a quartic (order 4 polynomial) equation with real coefficients, either:

all four roots are real, or
two roots are real and the other two roots form a complex conjugate pair, or
two roots form a complex conjugate pair and the other two roots also form a complex conjugate pair.

You can use Euler’s relation, $e^{i θ} = cos θ + i sin θ$ , to write a complex number $z$ in exponential form: $z = r e^{i θ}$ where $r = | z |$ and $θ = arg z$ .

For any two complex numbers $z_{1} = r_{1} e^{i θ_{1}}$ and $z_{2} = r_{2} e^{i θ_{2}}$

$z_{1} z_{2} = r_{1} r_{2} e^{i (θ_{1} + θ_{2})}$
$\frac{z_{1}}{z_{2}} = \frac{r_{1}}{r_{2}} e^{i (θ_{1} - θ_{2})}$

De Moivre’s theorem: For any integer $n$ , ${(r (cos θ + i sin θ))}^{n} = r^{n} (cos n θ + i sin n θ)$

$z + \frac{1}{z} = 2 cos θ$
$z^{n} + \frac{1}{z^{n}} = 2 cos n θ$
$z - \frac{1}{z} = 2 i cos θ$
$z^{n} - \frac{1}{z^{n}} = 2 i sin n θ$

For $w, z \in ℂ$ ,

$Σ_{r = 0}^{n - 1} w z^{r} = w + w z + w z^{2} + ... + w z^{n - 1} = \frac{w (z^{n} - 1)}{z - 1}$
$Σ_{r = 0}^{∞} w z^{r} = w + w z + w z^{2} + ... + w z^{n - 1} = \frac{w}{1 - z}, | z | < 1$

If $z$ and $w$ are non-zero complex numbers and $n$ is a positive integer, then the equation $z^{n} = w$ has $n$ distinct solutions.

Any complex number $z = r (cos θ + i sin θ)$ , can be written $z = r (cos (θ + 2 k π) + i sin (θ + 2 k π))$ , where $k$ is any integer.

In general, the solutions to $z^{n} = 1$ are $z = cos (\frac{2 π k}{n}) + i sin (\frac{2 π k}{n}) = e^{\frac{2 π i k}{n}}$ for $k = 1, 2, \dots, n$ and are known as the $n^{th}$ roots of unity. If $n$ is a positive integer, then there is an $n^{th}$ root of unity $ω = e^{\frac{2 k i}{n}}$ such that:

The $n^{th}$ roots of unity are $1, w, w^{2}, \dots, w^{n - 1}$ ;
$1, w, w^{2}, \dots, w^{n - 1}$ form the vertices of a regular n-gon;
$1 + w + w^{2} + \dots + w^{n - 1} = 0$ .

The $n^{th}$ roots of any complex number $s$ lie on the vertices of a regular $n$ -gon with its centre at the origin.

If $z_{1}$ is one root of the equation $z^{n} = s$ , and 1, ω, ω2, …, ω n−1 are the $n^{th}$ roots of unity, then the roots of zn = s are given by z1, z1ω, z1ω2, …, z1ωn−1.