Algebra Of The Complex Plane

addition

Given

a + b i, c + d i \in ℂ

, we define

(a + b i) + (c + d i) = (a + c) + (b + d) i

complex multiplication

given

a + b i, c + d i \in ℂ

, then:

(a + b i) \cdot (c + d i) = a b + i (a d + b c) - d b = (a d - b d) + i (a d + b c)

equality

Given

a + b i, c + d i \in ℂ

, they are equal when

a = c

and

b = d

conjugate

suppose

z = x + i y

, then the conjugate is denoted by

z : = x - i y

conjugate cancels

z = z

Let

z = x + i y

, then

z = x - i y = x + i (- y)

then

z = x - i (- y) = x + i y = z

as needed.

modulus

for any

z = x + i y \in ℂ

, we define

| z | : = \sqrt{x^{2} + y^{2}}

modulus ignores conjugates

| z | = | z |

Suppose that

z = x + i y

, then

| z | = | x - i y | = \sqrt{x^{2} + {(- y)}^{2}} = \sqrt{x^{2} + y^{2}} = | z |

complex number times it's conjugate equals the modulus squared

z \cdot z = {| z |}^{2}

suppose that

z = x + i y

, then

z \cdot z = (x + i y) \cdot (x - i y) = x^{2} + y^{2} + x y i - x y i = x^{2} + y^{2}

, but then again

| z | = | x + i y | = \sqrt{x^{2} + y^{2}}

thus

z \cdot z = {| z |}^{2}

inverse

the inverse of a complex number

z \in ℂ

is another

w \in ℂ

such that

z \cdot w = 1

, we denote

w

z^{- 1}

\frac{1}{z}

value of the inverse

Given

z \in ℂ

we have

z^{- 1} = \frac{z}{{| z |}^{2}}

We know that

z \cdot z = {| z |}^{2}

, therefore

z \cdot \frac{z}{{| z |}^{2}} = 1

, so by the definition of inverse

z^{- 1} = \frac{z}{{| z |}^{2}}

as needed.

Normalize Exercise

For any

z \in ℂ, | \frac{z}{z} | = 1

By the value of the inverse we have: $\frac{1}{z} = \frac{z}{{| z |}^{2}}$ since we know that two complex conjugates cancel and modulus ignores conjugates, then $\frac{z}{{| z |}^{2}} = \frac{z}{{| z |}^{2}}$

Now $| \frac{z}{z} | = | z (\frac{z}{{| z |}^{2}}) | = | {(\frac{z}{| z |})}^{2} | = | \frac{z}{| z |} | \cdot | \frac{z}{| z |} | = 1 \cdot 1 = 1$ as needed

real part of a complex number

given

z = x + i y \in ℂ

we say that

x

is the real part of

z

and define the function

ℛ (z) = x

imaginary part of a complex number

given

z = x + i y \in ℂ

we say that

y

is the imaginary part of

z

and define the function

ℐ (z) = y

extracting the real part of a complex number

suppose

z \in ℂ

, then

ℜ (z) = \frac{z + z}{2}

modulus is greater than it's components

for any

z \in ℂ

we have both

$| ℜ (z) | \leq | z |$
$| ℑ (z) | \leq | z |$

imaginary part distributes

ℑ (z + w) = ℑ (z) + ℑ (w)

modulus plus one of it's components is positive

Suppose that

f \in {ℛ, ℐ}

, then

| z | + f (z) \geq 0

Let

z = x + i y \in ℂ

then suppose without loss of generality that

f = ℛ

, then

- x \leq | x | = \sqrt{x^{2}} \leq \sqrt{x^{2} + y^{2}} = | x + i y |

therefore

0 \leq x + | x + i y |

which means

0 \leq ℛ (z) + | z |

creating squares

without use of the triangle inequality prove that

{| z - 1 |}^{2} = {| z |}^{2} + 1 - 2 ℛ (z)

, then conclude that

| z - 1 | \leq | z | + 1

${| z - 1 |}^{2}$ = $(z - 1) \cdot z - 1 = (z - 1) \cdot (z - 1) = z \cdot z - z - z + 1$ , now recall that $ℛ (z) = \frac{z + z}{2}$ so we can continue the equality with ${| z |}^{2} - 2 ℛ (z) + 1$ , showing the first part

for the second part, we can notice that ${(| z | + 1)}^{2} = {| z |}^{2} + 2 | z | + 1$ , so then ${| z - 1 |}^{2} = {| z |}^{2} + 1 - 2 ℛ (z) = {| z |}^{2} + 2 | z | - 2 | z | + 1 - 2 ℛ (z) = {(| z | + 1)}^{2} - 2 (| z | + ℛ (z))$ , then by this proposition we can see that the last term in the previous equality chain is less than or equal to ${(| z | + 1)}^{2}$ allowing us to conclude that ${| z - 1 |}^{2} \leq {(| z | + 1)}^{2}$ which implies that $| z - 1 | \leq | z | + 1$