The order relation is defined on the set of complex numbers.

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Q. The order relation is defined on the set of complex numbers.

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Concept understanding — Complex Number Arithmetic

Complex Number Arithmetic: A First Look

Imagine you're trying to solve $x^2 + 1 = 0$ . You know that no real number squared gives $-1$ . The square of any real number is either zero or positive. So this equation has no real solution. But what if we invent a number whose square is $-1$ ? That's exactly what mathematicians did — and that invention is the imaginary unit $i$ , defined by:

i^2 = -1

A complex number is any number of the form $a + bi$ , where $a$ and $b$ are real numbers. Here $a$ is called the real part, and $b$ is called the imaginary part. For example, $3 + 4i$ has real part $3$ and imaginary part $4$ .

Note

The name "imaginary" is unfortunate — these numbers are just as real (in the mathematical sense) as the numbers you already know. They're simply a different kind of number.

Why Bother?

Complex numbers let you solve equations that real numbers can't. Every polynomial equation — no matter how complicated — has a solution in the complex numbers. This is the Fundamental Theorem of Algebra, and it's one of the most important results in mathematics.

Arithmetic Operations

The rules are straightforward: treat $i$ like a variable, but remember that $i^2 = -1$ .

Addition and Subtraction

Add (or subtract) real parts with real parts, imaginary parts with imaginary parts.

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

(a + bi) - (c + di) = (a - c) + (b - d)i

Example: $(2 + 3i) + (4 - 5i) = (2 + 4) + (3 - 5)i = 6 - 2i$

Multiplication

Multiply like binomials, then replace $i^2$ with $-1$ .

(a + bi)(c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i

Example: $(2 + 3i)(4 - 5i) = 2(4) + 2(-5i) + 3i(4) + 3i(-5i)$

$= 8 - 10i + 12i - 15i^2$

$= 8 + 2i - 15(-1)$

$= 8 + 2i + 15 = 23 + 2i$

Watch out

The most common mistake: forgetting that $i^2 = -1$ , not $1$ . Always check your final step.

Division

Division is trickier. The key idea: multiply numerator and denominator by the complex conjugate of the denominator.

The complex conjugate of $a + bi$ is $a - bi$ . When you multiply a complex number by its conjugate, you get a real number:

(a + bi)(a - bi) = a^2 - (bi)^2 = a^2 - b^2 i^2 = a^2 + b^2

So to divide:

\frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)} = \frac{(a + bi)(c - di)}{c^2 + d^2}

Example: $\frac{2 + 3i}{4 - 5i} = \frac{(2 + 3i)(4 + 5i)}{(4 - 5i)(4 + 5i)} = \frac{8 + 10i + 12i + 15i^2}{16 + 25} = \frac{8 + 22i - 15}{41} = \frac{-7 + 22i}{41} = -\frac{7}{41} + \frac{22}{41}i$

Tip

To divide quickly: multiply top and bottom by the conjugate of the denominator, then simplify. The denominator always becomes $c^2 + d^2$ , a positive real number.

The Big Picture …

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Order relations require a total ordering that respects algebraic operations, which is impossible on the complex numbers because no ordering can make $\mathbb{C}$ compatible with its field structure.

The statement claims that an order relation is defined on the set of complex numbers. This is false, and understanding why reveals something deep about the nature of ordering and the complex number system.

An order relation on a field must satisfy certain properties to be useful in analysis and algebra. Specifically, for the real numbers $\mathbb{R}$ , we have a natural ordering $\leq$ that is:

Total: for any $a, b \in \mathbb{R}$ , either $a \leq b$ or $b \leq a$
Compatible with addition: if $a \leq b$ , then $a + c \leq b + c$
Compatible with multiplication: if $a \leq b$ and $0 \leq c$ , then $ac \leq bc$

The question is: can we extend such an ordering to $\mathbb{C}$ ?

Why complex numbers resist ordering

The fundamental obstruction comes from the imaginary unit $i$ . Let's see what happens if we try to order the complex numbers.

Assume an order exists. Suppose we could define $\leq$ on $\mathbb{C}$ satisfying the properties above.
Consider where $i$ sits relative to zero. We must have either $i > 0$ , $i < 0$ , or $i = 0$ . Since $i \neq 0$ , exactly one of the first two must hold.
Case 1: If $i > 0$ . Then by compatibility with multiplication, $i \cdot i > 0 \cdot i$ , which gives $i^2 > 0$ . But $i^2 = -1$ , so we'd have $-1 > 0$ . Adding $1$ to both sides: $0 > 1$ . This contradicts the fact that $1 = 1 \cdot 1 > 0$ (since any nonzero number squared must be positive in an ordered field).
Case 2: If $i < 0$ . Then $-i > 0$ . By the same multiplication argument, $(-i)^2 > 0$ , which gives $i^2 > 0$ , so again $-1 > 0$ . Same contradiction. …

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Q.The order relation is defined on the set of complex numbers.

Why complex numbers resist ordering

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