Welcome, future mathematicians! Today, we're going on a deep dive into the fascinating world of complex numbers, specifically how they can be rigorously defined and understood as ordered pairs of real numbers. This perspective is not just an academic exercise; it provides a solid foundation for all operations with complex numbers and beautifully connects them to geometry.
We've all met complex numbers in the form $a + bi$, where $a$ and $b$ are real numbers, and $i$ is the imaginary unit such that $i^2 = -1$. But have you ever wondered *why* $i^2 = -1$? Or how we can formally define arithmetic operations like multiplication in a way that doesn't just rely on "treating $i$ like a variable" and then substituting $i^2$? This is where the ordered pair representation shines!
### 1. The Fundamental Idea: Complex Numbers as Points in a Plane
Imagine the familiar Cartesian coordinate system, where every point is uniquely identified by an ordered pair $(x, y)$ of real numbers. Complex numbers can be seen in exactly the same way!
A
complex number $z$ can be formally defined as an
ordered pair of real numbers $(a, b)$.
* Here, 'a' is called the
real part of the complex number, denoted as $ ext{Re}(z) = a$.
* And 'b' is called the
imaginary part of the complex number, denoted as $ ext{Im}(z) = b$.
Why "ordered pair"? Because the order matters! The pair $(2, 3)$ is different from $(3, 2)$. Just like $2 + 3i$ is different from $3 + 2i$. This simple definition allows us to map every complex number to a unique point in a 2D plane (the Argand Plane), and vice-versa.
### 2. Equality of Complex Numbers
Just like with points in a Cartesian plane, two complex numbers represented as ordered pairs are equal if and only if their corresponding real and imaginary parts are equal.
Formally, if $z_1 = (a, b)$ and $z_2 = (c, d)$, then
$z_1 = z_2 iff a = c ext{ and } b = d$.
This seems intuitive, right? It means $(2, 3)$ is not equal to $(2, 4)$ or $(3, 3)$.
### 3. The Algebra of Complex Numbers (as Ordered Pairs)
Now, let's define how we perform basic arithmetic operations using this ordered pair representation.
#### 3.1. Addition of Complex Numbers
If $z_1 = (a, b)$ and $z_2 = (c, d)$, then their sum is defined as:
$z_1 + z_2 = (a, b) + (c, d) = mathbf{(a+c, b+d)}$
Intuition: This is exactly like vector addition in a 2D plane! You add the corresponding components. If you think about $a+bi$ form, $(a+bi) + (c+di) = (a+c) + (b+d)i$. The ordered pair definition perfectly mirrors this.
Example 1:
Let $z_1 = (2, 3)$ and $z_2 = (1, -5)$.
$z_1 + z_2 = (2+1, 3+(-5)) = (3, -2)$.
#### 3.2. Subtraction of Complex Numbers
Subtraction follows a similar pattern:
$z_1 - z_2 = (a, b) - (c, d) = mathbf{(a-c, b-d)}$
Example 2:
Let $z_1 = (4, 7)$ and $z_2 = (2, 3)$.
$z_1 - z_2 = (4-2, 7-3) = (2, 4)$.
#### 3.3. Multiplication of Complex Numbers - The Core Derivation!
This is where the ordered pair definition provides the most profound insight and mathematical rigor. If $z_1 = (a, b)$ and $z_2 = (c, d)$, then their product is defined as:
$z_1 cdot z_2 = (a, b) cdot (c, d) = mathbf{(ac - bd, ad + bc)}$
This definition might look a bit mysterious at first glance. Where does it come from? Let's derive it!
We know that complex numbers are usually written as $a+bi$. Let's assume for a moment that our ordered pair $(a,b)$ corresponds to $a+bi$.
Then, $(a+bi)(c+di)$
$= ac + adi + bci + bdi^2$
Since $i^2 = -1$, substitute this into the expression:
$= ac + adi + bci - bd$
Group the real parts and the imaginary parts:
$= (ac - bd) + (ad + bc)i$
Now, if we translate this back into the ordered pair notation, where the first component is the real part and the second is the imaginary part, we get:
$mathbf{(ac - bd, ad + bc)}$
This derivation shows that the seemingly abstract definition for multiplication of ordered pairs is perfectly consistent with the standard multiplication rule we use for $a+bi$ form, *provided* we accept $i^2 = -1$. The beauty of the ordered pair definition is that it allows us to *prove* $i^2 = -1$ from the definition itself, rather than assuming it! (We'll see this in the next section).
Example 3: Multiplication
Let $z_1 = (2, 3)$ and $z_2 = (1, -2)$.
Using the multiplication formula $(ac - bd, ad + bc)$:
$a=2, b=3, c=1, d=-2$
Real part: $ac - bd = (2)(1) - (3)(-2) = 2 - (-6) = 2 + 6 = 8$
Imaginary part: $ad + bc = (2)(-2) + (3)(1) = -4 + 3 = -1$
So, $z_1 cdot z_2 = (8, -1)$.
Quick Check using $a+bi$ form:
$(2+3i)(1-2i) = 2(1) + 2(-2i) + 3i(1) + 3i(-2i)$
$= 2 - 4i + 3i - 6i^2$
$= 2 - i - 6(-1)$
$= 2 - i + 6 = 8 - i$.
This matches our ordered pair result $(8, -1)$. Fantastic!
### 4. Special Complex Numbers and Their Representations
#### 4.1. Real Numbers
A real number $a$ can be represented as a complex number with its imaginary part equal to zero, i.e., $(a, 0)$.
Let's check if the operations behave correctly for real numbers:
*
Addition: $(a, 0) + (c, 0) = (a+c, 0)$. This is standard real number addition.
*
Multiplication: $(a, 0) cdot (c, 0) = (ac - 0 cdot 0, a cdot 0 + 0 cdot c) = (ac, 0)$. This is standard real number multiplication.
This confirms that the set of real numbers is a subset of the complex numbers under this definition.
#### 4.2. The Imaginary Unit 'i'
The imaginary unit $i$ is represented by the ordered pair $mathbf{(0, 1)}$.
This is the complex number with real part 0 and imaginary part 1.
Now for the big reveal: Let's prove $i^2 = -1$ using only the ordered pair definition!
We need to calculate $i cdot i$, which is $(0, 1) cdot (0, 1)$.
Using our multiplication formula $(ac - bd, ad + bc)$:
$a=0, b=1, c=0, d=1$
Real part: $ac - bd = (0)(0) - (1)(1) = 0 - 1 = -1$
Imaginary part: $ad + bc = (0)(1) + (1)(0) = 0 + 0 = 0$
So, $(0, 1) cdot (0, 1) = (-1, 0)$.
Since $(-1, 0)$ represents the real number $-1$, we have rigorously shown that $i^2 = -1$ using the ordered pair definition.
This is a foundational concept that strengthens the entire structure of complex numbers!
#### 4.3. Connecting $(a, b)$ to $a + bi$
Now we can formally show how the ordered pair $(a, b)$ is equivalent to $a + bi$:
We know:
* $(a, b) = (a, 0) + (0, b)$ (from addition definition)
* We can write $(0, b)$ as the product of $(b, 0)$ and $(0, 1)$:
$(b, 0) cdot (0, 1) = (b cdot 0 - 0 cdot 1, b cdot 1 + 0 cdot 0) = (0, b)$.
(Think of $(b,0)$ as the real number $b$, and $(0,1)$ as $i$).
So, $(a, b) = (a, 0) + (b, 0) cdot (0, 1)$
If we identify $(a, 0)$ with $a$, $(b, 0)$ with $b$, and $(0, 1)$ with $i$, then:
$mathbf{(a, b) equiv a + bi}$
This completes the circle! The ordered pair definition provides the solid algebraic framework from which the familiar $a+bi$ form naturally emerges.
### 5. Multiplicative Inverse
For any non-zero complex number $z = (a, b)$, its multiplicative inverse, denoted $z^{-1}$, is a complex number $(x, y)$ such that $z cdot (x, y) = (1, 0)$ (where $(1,0)$ is the multiplicative identity, equivalent to the real number 1).
Let $(a, b) cdot (x, y) = (1, 0)$.
Using the multiplication formula: $(ax - by, ay + bx) = (1, 0)$.
This gives us a system of two linear equations:
1. $ax - by = 1$
2. $bx + ay = 0$
Let's solve for $x$ and $y$:
From (2), $bx = -ay implies x = -ay/b$ (if $b
eq 0$). Substitute into (1):
$a(-ay/b) - by = 1$
$-a^2y/b - b^2y/b = 1$
$-(a^2+b^2)y/b = 1$
$y = -b/(a^2+b^2)$
Now substitute $y$ back into $x = -ay/b$:
$x = -a/b cdot (-b/(a^2+b^2))$
$x = a/(a^2+b^2)$
So, the multiplicative inverse of $(a, b)$ is $mathbf{left(frac{a}{a^2+b^2}, frac{-b}{a^2+b^2}
ight)}$, provided $a^2+b^2
eq 0$ (i.e., $(a,b)$ is not $(0,0)$).
Connection to $a+bi$ form:
Recall that $frac{1}{a+bi} = frac{1}{a+bi} cdot frac{a-bi}{a-bi} = frac{a-bi}{a^2 - (bi)^2} = frac{a-bi}{a^2+b^2} = frac{a}{a^2+b^2} + frac{-b}{a^2+b^2}i$.
This again perfectly matches our ordered pair derivation!
Example 4: Multiplicative Inverse
Find the inverse of $z = (3, 4)$.
Here $a=3, b=4$.
$a^2+b^2 = 3^2+4^2 = 9+16 = 25$.
Inverse is $left(frac{3}{25}, frac{-4}{25}
ight)$.
Check: $(3,4) cdot (frac{3}{25}, frac{-4}{25}) = (3 cdot frac{3}{25} - 4 cdot frac{-4}{25}, 3 cdot frac{-4}{25} + 4 cdot frac{3}{25})$
$= (frac{9}{25} + frac{16}{25}, frac{-12}{25} + frac{12}{25}) = (frac{25}{25}, 0) = (1, 0)$. This is correct!
### 6. Geometric Interpretation: The Argand Plane
The representation of complex numbers as ordered pairs $(a, b)$ naturally leads to their geometric interpretation. We can plot $(a, b)$ as a point in the Cartesian plane, which we call the
Argand Plane (or Complex Plane).
* The horizontal axis is called the
real axis (representing the real part $a$).
* The vertical axis is called the
imaginary axis (representing the imaginary part $b$).
Complex Number Form |
Ordered Pair Form |
Geometric Interpretation |
|---|
$z = a+bi$ |
$z = (a, b)$ |
A point $(a, b)$ in the Argand Plane, or a position vector from origin $(0,0)$ to $(a,b)$. |
$z_1 + z_2$ |
$(a+c, b+d)$ |
Vector addition (parallelogram law) in the Argand Plane. |
$z_1 - z_2$ |
$(a-c, b-d)$ |
Vector subtraction. The vector from $z_2$ to $z_1$. |
Real number $a$ |
$(a, 0)$ |
A point on the real axis. |
Purely imaginary number $bi$ |
$(0, b)$ |
A point on the imaginary axis. |
This visual representation is incredibly powerful for understanding complex number operations. For instance, multiplication by $i$ (which is $(0,1)$) corresponds to a counter-clockwise rotation of $90^circ$ about the origin. If $z=(a,b)$, then $i cdot z = (0,1) cdot (a,b) = (0 cdot a - 1 cdot b, 0 cdot b + 1 cdot a) = (-b, a)$. Geometrically, mapping $(a,b)$ to $(-b,a)$ is exactly a $90^circ$ counter-clockwise rotation!
### 7. Properties of Complex Numbers (as ordered pairs)
The set of complex numbers, $mathbb{C}$, with the addition and multiplication operations defined above, forms a field. This means they satisfy several fundamental algebraic properties:
*
Closure: $(a,b)+(c,d)$ is a complex number; $(a,b)cdot(c,d)$ is a complex number.
*
Commutativity of Addition: $(a,b)+(c,d) = (c,d)+(a,b)$
*
Associativity of Addition: $((a,b)+(c,d))+(e,f) = (a,b)+((c,d)+(e,f))$
*
Additive Identity: $(0,0)$ such that $(a,b)+(0,0)=(a,b)$.
*
Additive Inverse: For every $(a,b)$, there is $(-a,-b)$ such that $(a,b)+(-a,-b)=(0,0)$.
*
Commutativity of Multiplication: $(a,b)cdot(c,d) = (c,d)cdot(a,b)$
*
Associativity of Multiplication: $((a,b)cdot(c,d))cdot(e,f) = (a,b)cdot((c,d)cdot(e,f))$
*
Multiplicative Identity: $(1,0)$ such that $(a,b)cdot(1,0)=(a,b)$.
*
Multiplicative Inverse: For every non-zero $(a,b)$, there exists $(x,y)$ such that $(a,b)cdot(x,y)=(1,0)$. (We derived this above).
*
Distributivity: $(a,b)cdot((c,d)+(e,f)) = (a,b)cdot(c,d) + (a,b)cdot(e,f)$
JEE Focus: While you typically work with complex numbers in $a+bi$ form, understanding their ordered pair representation is crucial for a deeper conceptual grasp.
1. It provides the
rigorous mathematical foundation, proving that complex numbers are not just a "trick" but a consistent algebraic structure.
2. It lays the groundwork for
geometric interpretations in the Argand plane, which is heavily tested in JEE Advanced (e.g., transformations, loci of points, geometry problems involving complex numbers).
3. Occasionally, problems might explicitly mention $(a,b)$ form, or require you to reason about complex numbers purely from their real and imaginary components. For example, questions about the algebraic properties listed above often stem from this fundamental definition.
By understanding complex numbers as ordered pairs, you move beyond just memorizing formulas for $a+bi$ and gain a powerful insight into *why* these formulas work and *how* complex numbers behave, both algebraically and geometrically. This robust understanding is what differentiates a good student from a great one in JEE!