Introduction To Complex Numbers

Introduction:

So far we have acquainted ourselves with different types of numbers, as follows:

$\mathbb{N}$ - representing the set of all natural numbers (positive integers), $\{1, 2, 3, ..., +\infty\}$

$\mathbb{Z}$ - representing the set of all integers, positive as well as negative $\{-\infty, ..., -2, -1, 0, 1, 2,..., +\infty \}$

$\mathbb{Q}$ - representing the set of all rational numbers, that can be represented as $\dfrac{p}{q}$, where $p \in \mathbb{Z}$ and $q \in \mathbb{N}$

$\mathbb{R}$ - representing the set of all real number including rational and irrational numbers.

Imaginary Number:

All the different types of numbers listed in the previous section does not answer the question as to how much is $\sqrt{-1}$.

Since, we know that any two negative numbers, when multiplied gives a positive result, threrefore $x \times x$ cannot be equal to $-y$ where $y$ is a positive quantity.

If we consider the quadratic equation:

$x^2 + 1 = 0$

$\Rightarrow x^2 = -1$

$\Rightarrow x = \pm \sqrt{-1}$

Since, it is not possible to assign a specific numeric value to $\sqrt{-1}$, it is denoted using $i$, with the property $i^2 + 1 = 0$.

Any number of the form $b \times i$ where $b \in \mathbb{R}$ and $i = \sqrt{-1}$, is considered as an imaginary number.

For example $\sqrt{-8} = \sqrt{8 \times( -1)} = \sqrt{8}\sqrt{-1} = 2\sqrt{2}\sqrt{-1} = 2\sqrt{2}i$

Therefore, any number of the form $\sqrt{-x}$ where $x > 0$ and $x \in \mathbb{R}$ is an imaginary number.

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Complex Numbers:

Any number of the form $a + bi$ where $i = \sqrt{-1}$ and $(a, b) \in \mathbb{R}$ is called a complex number.

In a complex number of the above form, $a$ is called the real part, and $b$ is called the imaginary part. Note, that the imaginary part is only $b$ and $NOT$ $bi$.

If $z$ is a complex number, then the real part is denoted by:

$\mathfrak{R}(z)$ or $\mathcal{Re}(z)$ or $\mathinner{Re}(z)$

while the imaginary part is denoted by:

$\mathfrak{I}(z)$ or $\mathcal{Im}(z)$ or $\mathinner{Im}(z)$

For example, if $z = 11 + 19i$ then,

$\mathcal{Re}(z) = 11$ and $\mathcal{Im}(z) = 19$

The set of all complex numbers is denoted using $\mathbb{C}$

Modulus Of Complex Numbers:

Given any complex number $z = a + bi$, where $(a, b) \in \mathbb{R}$, the modulus or absolute value of $z$ is given by:

$|z| = |\sqrt{a^2 + b^2}|$

In other words, the modulus is the non-negative square root of the sum of the squares of the real and imaginary part of the complex number.

Equality Of Complex Numbers:

Two complex numbers $z_1 = a + bi$ and $z_2 = c + di$ are considered to be equal if and only if, $a = c$ and $b = d$.

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Addition:

Given two complex numbers $z_1 = a + bi$ and $z_2 = c + di$,

$z_1 + z_2 = z_2 + z_1 = (a + c) + (b + d)i$

In other words, we can say that addition of complex numbers is commutative, and obtain the result by add the real and imaginary parts separately.

Subtraction, is the inverse operation, not commutative and is given by:

$z_1 - z_2 = (a - c) + (b - d)i$

Let us try the following problem:

--------- Reference to question: 5da6ef62-ffdf-4abb-b83a-30bad69f624d ---------

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Multiplication:

Multiplication of complex numbers is commutative as well as distributive over addition.

Given, two complex numbers $z_1 = a + bi$ and $z_2 = c + di$,

$z_1 \cdot z_2$

$= (a + bi) (c + di)$

$= a(c + di) + bi(c + di)$

$= ac + adi + bci + bdi^2$

$= (ac - bd) + (ad + bc)i$ $\because i^2 = 1$

Similarly, if we expand $z_2 \cdot z_1$, we will get the same expression:

$(ac - bd) + (ad + bc)i$

Therefore, multiplication is commutative. We leave it to the reader to prove that multiplication is distributive over addition/subtraction.

Conjugate:

Conjugate of a complex number $z = a + bi$, often called as Complex Conjugate, is the complex number $a - bi$ and is denoted by $\overline{z}$

We can see that:

$ z \cdot \overline{z} = (a + bi)(a - bi) = a^2 - (bi)^2 = a^2 - (b^2)(-1) = a^2 + b^2$

which is a real number.

The formal definition of conjugate of a complex number, $z$, is a complex number which has an equal real part and an imaginary part which is equal in magnitude but of opposite sign as that of $z$.

Therefore, for any complex number $z$ and its conjugate $\overline{z}$,

$|z|^2 = |\overline{z}|^2 = z \cdot \overline{z}$ and