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Fundamental Operations on Complex Numbers

Introduction

3 circles + 2 circles = 5 circles In mathematics, when a new mathematical object is introduced, defining operations on it is crucial to understanding its properties and how it interacts with other objects. This is evident in the creation and exploration of complex numbers, which extend the operations defined for real numbers.

Complex numbers expand the operations on real numbers to include an imaginary unit, \( i \), where \( i^2 = -1 \). This addition allows for a richer set of operations that are not possible within the real numbers alone. The operations on complex numbers, like addition, subtraction, multiplication, and division, are consistent with those on real numbers but are extended to accommodate the interaction between the real and imaginary parts.

You have seen many mathematical objects before and operations defined on them.

  • Real Numbers Operations: The operations on real numbers are closed, commutative, associative, and have elements of identity and inverses. They are defined for addition, subtraction, multiplication, and division (except division by zero).

  • Sets Operations: Operations on sets, such as union, intersection, and complement, follow different rules and laws, like commutativity, associativity, and distributivity, which dictate how sets combine and relate to each other.

  • Vector Operations: Operations on vectors include addition, scalar multiplication, dot product, and cross product. These operations are defined in the context of a vector space and have their own set of rules, such as the commutative and associative properties for vector addition and the distributive property for scalar multiplication.

Complex numbers, while they include an element (the imaginary unit) that vectors and sets do not, still maintain a structure with operations that resemble those of real numbers in many ways, albeit extended into two dimensions. This system of operations allows complex numbers to form a field, similar to real numbers, which supports a broad range of mathematical exploration and application.

Ordering Relations

Equality

Equality of complex numbers is defined based on the equality of their real and imaginary components. Two complex numbers are considered equal if, and only if, their corresponding real parts are equal and their corresponding imaginary parts are equal.

If we have two complex numbers, \( z_1 = a + bi \) and \( z_2 = c + di \), where \( a, b, c, \) and \( d \) are real numbers, and \( i \) is the imaginary unit (\( i^2 = -1 \)), then \( z_1 \) and \( z_2 \) are equal if:

\[ a = c \quad \text{and} \quad b = d \]

In other words, \( z_1 = z_2 \) if and only if \( a = c \) and \( b = d \).

Inequalities

In the real number line, which is a single-dimensional line, we can easily compare numbers using symbols like \(>\) and \(<\). For instance, if we have two real numbers, \(4\) and \(7\), we can say \(4 < 7\) because on the real number line, \(4\) is to the left of \(7\).

However, when we step into the realm of complex numbers and their representation on the Argand plane, this kind of simple comparison doesn't work anymore. The Argand plane is a two-dimensional space where the horizontal axis represents the real part of complex numbers and the vertical axis represents the imaginary part. So, a complex number is represented as a point on this plane, with its position determined by both its real and imaginary components.

For example, consider two complex numbers \(2 + 3i\) and \(4 + i\). On the Argand plane, these numbers are points in different positions, one not simply "greater" or "less" than the other in the way real numbers are on the real line. One might be further to the right (indicating a larger real part) or higher up (indicating a larger imaginary part), but these differences don't translate into a straightforward \(>\) or \(<\) comparison as they do on the real line.

Therefore, in the context of the Argand plane and complex numbers, we don't use \(>\) or \(<\) to compare numbers. Instead, we can talk about their magnitude (distance from the origin) or their angle with the real axis, but there's no direct way to say one complex number is greater or less than another like we can with numbers on the real line.

Operations on Complex Numbers

Addition of Complex Numbers:

Let \( z_1 = x_1 + iy_1 \) and \( z_2 = x_2 + iy_2 \), where \( x_1, x_2, y_1, \) and \( y_2 \) are real numbers and \( i \) is the imaginary unit. The sum of \( z_1 \) and \( z_2 \) is a complex number \( z_1 + z_2 \) found by adding the real parts and the imaginary parts separately:

\[ z_1 + z_2 \in \mathbb{C} \quad \text{such that} \quad z_1 + z_2 = (x_1 + x_2) + i(y_1 + y_2) \]

For example:

Let \( z_1 = 3 + 4i \) and \( z_2 = 1 - 2i \).

To add \( z_1 \) and \( z_2 \), we combine their real parts and their imaginary parts separately:

\( z_1 + z_2 = (3 + 4i) + (1 - 2i) = (3 + 1) + (4i - 2i) = 4 + 2i \)

So, the sum of \( z_1 \) and \( z_2 \) is \( 4 + 2i \).

Properties of addition for complex numbers are as follows:

  1. Closure Property: The sum of two complex numbers is a complex number, meaning if \( z_1 \) and \( z_2 \) are in the set of complex numbers \( \mathbb{C} \), then \( z_1 + z_2 \) is also in \( \mathbb{C} \).

  2. Commutative Property: The addition of complex numbers is commutative, which means that the order of addition does not affect the sum:

    \[ z_1 + z_2 = z_2 + z_1 \]
  3. Associative Property: The addition of complex numbers is associative, indicating that when three complex numbers are added, the sum is the same regardless of how the numbers are grouped:

    \[ z_1 + (z_2 + z_3) = (z_1 + z_2) + z_3 \]
  4. Additive Identity: The complex number \( 0 + 0i \), or simply \( 0 \), serves as the additive identity in the set of complex numbers, meaning adding zero to any complex number \( z \) will yield \( z \):

    \[ z + 0 = z \]
  5. Additive Inverse: Every complex number \( z = x + iy \) has an additive inverse \( -z = -x - iy \), such that their sum is the additive identity \( 0 \):

    \[ z + (-z) = (x + iy) + (-x - iy) = 0 \]

Geometrical Representation of \(-z\)

Geometrically on the Argand plane, \( -z \) is the reflection of \( z \) in the origin. Clearly, if \( \arg(z) = \theta \), then \( \arg(-z) = \theta + \pi \).

Geometrical representation of additive inverse of z

Subtraction of Complex Numbers:

Let \( z_1 = x_1 + iy_1 \) and \( z_2 = x_2 + iy_2 \), then we define subtraction as

\[ z_1 - z_2 = z_1 + (-z_2) \]

i.e., subtraction of \( z_2 \) from \( z_1 \) is equivalent to the addition of the additive inverse of \( z_2 \) to \( z_1 \).

For example:

Let \( z_1 = 6 + 8i \) and \( z_2 = 3 + 4i \).

To subtract \( z_2 \) from \( z_1 \), we first find the additive inverse of \( z_2 \), which is \( -z_2 = -3 - 4i \). Then, we add this additive inverse to \( z_1 \):

\( z_1 - z_2 = (6 + 8i) - (3 + 4i) = (6 - 3) + (8i - 4i) = 3 + 4i \)

So, the result of the subtraction \( z_1 - z_2 \) is \( 3 + 4i \).

Visualizing Addition and Subtraction of Complex Numbers

Addition

The addition of complex numbers can be represented geometrically using the Argand plane. Each complex number corresponds to a point on this plane, and their addition results in a new point that corresponds to the sum. If we have two complex numbers, \(z_1\) and \(z_2\), with \(z_1\) represented by point \(A\) and \(z_2\) by point \(B\), the addition is illustrated by plotting these points and completing a parallelogram with the origin. The point \(C\), which is the opposite vertex of the parallelogram from the origin, represents the sum \(z_1 + z_2\).

Visualizing Addition of Complex Numbers

  • The point \( A \) represents the complex number \( z_1 = x_1 + iy_1 \). This is shown in the diagram where \( OM = x_1 \) is the real component and \( AM = PL = y_1 \) is the imaginary component of \( z_1 \).

  • Similarly, point \( B \) represents the complex number \( z_2 = x_2 + iy_2 \), with \( ON = x_2 \) as the real part and \( BN = QL = y_2 \) as the imaginary part.

  • Observe that triangle \( OAM \) is congruent to triangle \( BQC \). They are congruent because they have the same side lengths (\( OM = QC = x_1 \) and \( AM = QB = y_1 \)) and their corresponding angles are equal (each pair of angles are vertical angles formed by intersecting lines, which are congruent).

  • Triangle \( OBN \) is congruent to triangle \( ACP \). These triangles are congruent for similar reasons: corresponding sides are equal (\( ON = AP = x_2 \), \( BN = CP = y_2 \)), and corresponding angles are equal.

Due to these congruencies, the horizontal and vertical components of the diagonals can be summed:

  • \( OL \) (the total real part of \( z_1 + z_2 \)) is equal to \( OM + ML \) which is the same as \( OM + AP \) because \( AP \) is equal to \( ON \). Therefore, \( OL = OM + ON = x_1 + x_2 \).

  • \( CL \) (the total imaginary part of \( z_1 + z_2 \)) is equal to \( CQ + QL \) which is equal to \( AM + BN \) since \( CQ = AM \) and \( QL = BN \). Hence, \( CL = AM + BN = y_1 + y_2 \).

The coordinates of point \( C \) are therefore \( (x_1 + x_2, y_1 + y_2) \), which means that point \( C \) represents the complex number \( z_1 + z_2 = (x_1 + x_2) + i(y_1 + y_2) \). Point \( C \) is the endpoint of the diagonal \( OC \) of parallelogram \( OACB \), which geometrically represents the sum of the complex numbers \( z_1 \) and \( z_2 \).

Subtraction

\[ z_1 - z_2 = z_1 + (-z_2) \]

To achieve this geometrically, we first draw \( OB' \) which is equal in length to \( OB \) but in the opposite direction. \( B' \) represents \( -z_2 \). And now we complete the parallelogram \( OAC'B' \), where the point \( C' \) represents \( z_1 - z_2 \).

Subtraction of Complex Numbers

Similarly, we plot \( z_2 - z_1 \). Clearly, \( z_2 - z_1 \) is the additive inverse of \( z_1 - z_2 \), and thus has the opposite direction of \( z_1 - z_2 \).

Addition and Subtraction together
Fig.1 - This figure demonstrates the geometric representation of addition and subtraction of complex numbers on the Argand plane. Parallelogram OACB shows addition where diagonal OC represents z1 + z2. Parallelograms OA'B'C' and OB'A'C'' show subtraction, where diagonals OC' and OC'' represent z1 - z2 and z2 - z1, respectively. The congruency of these parallelograms illustrates the relationship between addition and subtraction through additive inverses.

The above figure illustrates the geometric representation of addition and subtraction of complex numbers on the Argand plane.

  • Parallelogram \( OACB \) demonstrates the addition of \( z_1 \) and \( z_2 \). Point \( A \) represents \( z_1 \), and point \( B \) represents \( z_2 \). The diagonal \( OC \) represents the sum \( z_1 + z_2 \).
  • The parallelogram \( OA'B'C' \) shows the subtraction of \( z_2 \) from \( z_1 \), where \( A' \) represents \( -z_1 \) and \( B' \) represents \( -z_2 \). The diagonal \( OC' \) represents the difference \( z_1 - z_2 \).
  • The parallelogram \( OB'A'C'' \) represents the subtraction of \( z_1 \) from \( z_2 \), with diagonal \( OC'' \) representing \( z_2 - z_1 \).

Distance between two complex numbers

The modulus of a complex number is equivalent to its distance from the origin on the complex plane. Consequently, for the complex number \( z_1 + z_2 \), the distance from the origin to the point \( C \), denoted \( OC \), is given by \( |z_1 + z_2| \). Similarly, the modulus of the difference \( z_1 - z_2 \), represented by the length of \( OC' \), is \( |z_1 - z_2| \).

Given that the parallelograms \( OACB \) and \( OAC'B' \) are congruent, we can ascertain that the lengths of \( OC' \) and \( AB \) are identical, hence \( OC' = AB \). This equality reveals that the diagonal \( AB \) of parallelogram \( OACB \) is equal to \( |z_1 - z_2| \), which is the modulus of the difference of \( z_1 \) and \( z_2 \).

Furthermore, the segment \( AB \) represents the distance between the points \( A \) and \( B \), which correspond to the complex numbers \( z_1 \) and \( z_2 \), respectively. Therefore, the distance between \( z_1 \) and \( z_2 \) on the complex plane is expressed by \( |z_1 - z_2| \).

Directly finding differnce of complex numbers on Argand plane

How can one directly find out \( z_1 - z_2 \) on the complex plane?

Let \( z_1 \) and \( z_2 \) be represented by \( A \) and \( B \). Draw an arrow \( BA \) with tail at \( B \) and head at \( A \) joining \( z_1 \) and \( z_2 \). Translate this arrow parallelly (without rotation) such that the tail is now at the origin, then the location of the head now is nothing but \( z_1 - z_2 \). Similarly, thing you do with \( z_2 - z_1 \). This time head of the arrow is at \( z_2 \), and tail is at \( z_1 \). And we translate the arrow parallelly such that tail is at the origin. The new location of the arrow is \( z_2 - z_1 \).

Subtraction of complex numbers directly

Multiplication of Complex Numbers

When multiplying two complex numbers \( z_1 = x_1 + iy_1 \) and \( z_2 = x_2 + iy_2 \), we follow the distributive and commutative laws of algebra, similar to multiplying real numbers, while treating \( i \) as a coefficient that also follows the special property \( i^2 = -1 \).

The multiplication process is as follows:

\[\begin{align} (x_1 + iy_1)(x_2 + iy_2) &= x_1x_2 + x_1(iy_2) + (iy_1)x_2 + (iy_1)(iy_2)\\ &= x_1x_2 + ix_1y_2 + ix_2y_1 + i^2y_1y_2 \\ &= x_1x_2 + i(x_1y_2 + x_2y_1) + (-1)y_1y_2 \quad(since, i^2 = -1)\\ &= (x_1x_2 - y_1y_2) + i(x_1y_2 + x_2y_1) \end{align}\]

This multiplication produces another complex number where \( (x_1x_2 - y_1y_2) \) is the real part and \( (x_1y_2 + x_2y_1) \) is the imaginary part.

Further observations include:

  1. \( k(x + iy) = (k + i 0)(x + iy) = kx + kiy \) where \( k \in \mathbb{R} \)

  2. \( 0(x + iy) = 0 \)

  3. \[(x + iy)(x + iy) = (x + iy)^2 = x^2 + ixy + ixy + i^2y^2 = (x^2 - y^2) + i(2xy) \]
  4. \( (x - iy)^2 = (x^2 - y^2) - i(2xy) \)

  5. \( (x + iy)(x - iy) = x^2 - (iy)^2 = x^2 + y^2 \)

  6. \[ (x + iy)^3 = x^3 + i3x^2y + 3xiy^2 + i^3y^3 = x^3 + i3x^2y + i3xy^2 - y^3 = (x^3 - 3xy^2) + i(3x^2y - y^3) \]
  7. \[ (x - iy)^3 = x^3 - i3x^2y - 3xiy^2 - i^3y^3 = x^3 - i3x^2y - i3xy^2 + y^3 = (x^3 - 3xy^2) + i(y^3 - 3x^2y) \]

Properties of Multiplication

  1. Closure: The product of any two complex numbers is always another complex number. This ensures that complex numbers are closed under multiplication within the set of complex numbers.

  2. Commutativity: For two complex numbers \( z_1 \) and \( z_2 \), the product is the same regardless of the order in which they are multiplied, i.e., \( z_1z_2 = z_2z_1 \).

  3. Associativity: Multiplication of complex numbers is associative, meaning that when multiplying three complex numbers together, the product is the same no matter how the numbers are grouped. If \( z_1 \), \( z_2 \), and \( z_3 \) are complex numbers, then \( z_1(z_2z_3) = (z_1z_2)z_3 \).

  4. Existence of Multiplicative Identity: For every complex number \( z \), there exists a multiplicative identity which is \( 1 \) (or \( 1 + 0i \)) such that \( z \cdot 1 = z \). Thus, \( 1 \) (or \( 1 + 0i \)) is the multiplicative identity in the set of complex numbers.

  5. Multiplicative Inverse: Every complex number \( z \) that isn’t zero has a multiplicative inverse, called \( z^{-1} \), and it’s a number that, when multiplied by \( z \), gives 1. If \( z = a + ib \), then the inverse would be some \( x + iy \).

    To find this \( x + iy \), you multiply \( z \) by \( x + iy \) and set it to equal 1:

    \[ (a + ib)(x + iy) = 1 \]

    This equation expands to:

    \[ (ax - by) + i(ay + bx) = 1 \]

    For the equation to hold true, the real parts must add up to 1 and the imaginary parts must cancel out:

    \( ax - by = 1 \) and \( ay + bx = 0 \)

    Solving these gives you \( x = \frac{a}{a^2 + b^2} \) and \( y = -\frac{b}{a^2 + b^2} \). Therefore, the inverse of \( z \) is:

    \[ z^{-1} = \frac{a}{a^2 + b^2} - i\frac{b}{a^2 + b^2} \]

    And another way to write this is:

    \[ \frac{1}{z} = \frac{a - ib}{a^2 + b^2} \]

    This is how you find the multiplicative inverse of a complex number—it’s just a matter of switching the sign of the imaginary part and dividing by the sum of the squares of both parts.

    Finding mulitplicative inverse

    Obtaining the multiplicative inverse of \( z \), i.e., \( \frac{1}{z} \) using the above formula is equivalent to multiplying the numerator and denominator of the fraction \( \frac{1}{z} \) with the conjugate of \( z \).

    e.g., The multiplicative inverse of \( 1 + 2i \) is \( \frac{1}{1 + 2i} \) which is equal to

    \[ \frac{1 \times (1 - 2i)}{(1 + 2i)(1 - 2i)} = \frac{1 - 2i}{1 + 4} = \frac{1 - 2i}{5} = \frac{1}{5} - \frac{2i}{5} \]

    Here, by multiplying both the numerator and the denominator by the conjugate of the denominator, we can eliminate the imaginary unit from the denominator, thereby finding the complex number's multiplicative inverse in standard form. This technique is widely used because it simplifies the division of complex numbers into an algebraic expression involving only real numbers in the denominator.

  6. Distributive: The multiplication of complex numbers is distributive over addition.

    For all \( z_1, z_2, z_3 \in \mathbb{C} \), \( z_1 (z_2 + z_3) = z_1z_2 + z_1z_3 \)

Division of complex numbers

Let \( z_1 = x_1 + iy_1 \) and \( z_2 = x_2 + iy_2 \neq 0 \). Then the division of complex number \( z_1 \) by \( z_2 \), i.e., \( \frac{z_1}{z_2} \), is a complex number such that \( \frac{z_1}{z_2} = z_1 \left( \frac{1}{z_2} \right) \), i.e.it is a product of \( z_1 \) with multiplicative inverse of \( z_2 \).

Thus, to obtain the quotient \( \frac{z_1}{z_2} \), we multiply the numerator and denominator of the fraction with the conjugate of \( z_2 \).

\[ \frac{z_1}{z_2} = \frac{x_1 + iy_1}{x_2 + iy_2} = (x_1 + iy_1)(x_2 - iy_2) \]
\[ = \frac{(x_1x_2 + y_1y_2) + i(y_1x_2 - x_1y_2)}{x_2^2 + y_2^2} \]

In this way, by multiplying by the conjugate, we rationalize the denominator, transforming the division of complex numbers into an easier addition of two complex numbers in standard form, where the real and imaginary parts are separated.

Example

Let's say we have two complex numbers:

\[ z_1 = 3 + 4i \quad \text{and} \quad z_2 = 1 - 2i \]

We want to find the quotient \( \frac{z_1}{z_2} \).

First, we multiply the numerator and denominator by the conjugate of the denominator:

\[ \frac{z_1}{z_2} = \frac{3 + 4i}{1 - 2i} \times \frac{1 + 2i}{1 + 2i} \]

Now, calculate the product in the numerator and the denominator separately:

Numerator:

\[ (3 + 4i)(1 + 2i) = 3 + 6i + 4i + 8i^2 \]

Since \( i^2 = -1 \), this simplifies to:

\[ 3 + 6i + 4i - 8 = -5 + 10i \]

Denominator:

\[ (1 - 2i)(1 + 2i) = 1 + 2i - 2i - 4i^2 \]

Again, using \( i^2 = -1 \), this simplifies to:

\[ 1 - 4(-1) = 1 + 4 = 5 \]

So, putting it together, we have:

\[ \frac{z_1}{z_2} = \frac{-5 + 10i}{5} \]

We can split this into real and imaginary parts and simplify:

\[ \frac{z_1}{z_2} = \frac{-5}{5} + \frac{10i}{5} = -1 + 2i \]

Therefore, the quotient of \( z_1 \) by \( z_2 \) is the complex number \( -1 + 2i \).

Powers of iota

The imaginary unit \( i \) is defined such that \( i^2 = -1 \). Using this definition, the powers of \( i \) can be determined:

  1. \( i^2 = -1 \): This is the definition of the imaginary unit.
  2. \( i^3 = i \cdot i^2 = i \cdot (-1) = -i \): Here, we multiply \( i^2 \) by \( i \) to get \( i^3 \).
  3. \( i^4 = (i^2)^2 = (-1)^2 = 1 \): Squaring \( i^2 \) results in 1 because the square of -1 is 1.

Thus, in general, \( i^n = i^{4q+r} = (i^4)^q \cdot i^r = i^r \) where \( n \) is an integer, and \( n = 4q + r \) where \( 0 \leq r < 4 \).

Here’s what the general formula means:

  • \( n \) is an integer representing the exponent to which \( i \) is raised.
  • \( q \) is the quotient when \( n \) is divided by 4. It represents how many full cycles of 4 (the period of the powers of \( i \)) there are in \( n \).
  • \( r \) is the remainder when \( n \) is divided by 4. It determines the actual power of \( i \) since \( i^4 = 1 \), and therefore \( (i^4)^q = 1^q = 1 \) for any integer \( q \), the value of \( i^n \) ultimately depends only on \( r \).

This leads to the simplified form \( i^n = i^r \), where \( r \) can only be 0, 1, 2, or 3.

For example, \( i^{58} = i^{4 \cdot 14 + 2} = i^2 = -1 \).

  • \( n = 58 \) is divided by 4 giving \( q = 14 \) and \( r = 2 \).
  • We only need to consider the remainder \( r \) because \( (i^4)^q = 1 \) regardless of \( q \).
  • Since \( r = 2 \), we use the base power \( i^2 = -1 \).

Sum of four integral powers of iota is equal to zero, i.e., \( i^n + i^{n+1} + i^{n+2} + i^{n+3} = 0 \).

This statement asserts that the sum of any four consecutive powers of \( i \) is zero. It’s valid because the powers of \( i \) are cyclical with a period of 4:

  • \( i^0 = 1 \)
  • \( i^1 = i \)
  • \( i^2 = -1 \)
  • \( i^3 = -i \)

And since \( i^4 = 1 \), the sequence repeats. Thus, the sum of these four quantities is \( 1 + i - 1 - i = 0 \).