Why We Need Complex Numbers — Definition as an Ordered Pair

Every real number, when squared, gives a result that is zero or positive. That simple fact means an equation like $x^2 + 1 = 0$, i.e. $x^2 = -1$, has no solution among the real numbers — there is no real number whose square is negative. The same problem shows up whenever you solve a general quadratic $ax^2+bx+c=0$ and its discriminant $b^2-4ac$ turns out negative: the quadratic formula asks you to take the square root of a negative number, and the real number system simply has no answer to give.

Rather than declare such equations unsolvable, mathematicians enlarged the number system. Euler was the first to give the 'missing' square root of $-1$ a symbol, $i$. Later, Hamilton put the idea on a fully rigorous footing by defining a complex number not as a mysterious symbol but as an ordered pair of real numbers $(a, b)$. This sidesteps any hand-waving about what $\sqrt{-1}$ 'is' — a complex number is just a pair, exactly the way a point in a plane is a pair of coordinates.

Formally, the set of complex numbers is $\mathbf{C} = \{(a,b) \mid a \in \mathbf{R},\, b \in \mathbf{R}\} = \mathbf{R} \times \mathbf{R}$. Two complex numbers $z_1=(a,b)$ and $z_2=(c,d)$ are defined to be equal exactly when their corresponding entries match: $a=c$ and $b=d$. This is a stronger condition than 'equal in value' for a single real number — an ordered pair carries two independent pieces of information, so both must agree.

Once complex numbers exist as objects, we need rules for combining them. These are definitions, not derivations — we are choosing how addition and multiplication should behave so that the new system extends the real numbers sensibly:

• Addition: $(a,b) + (c,d) = (a+c,\ b+d)$ — add entry-wise.
• Negative: $-(a,b) = (-a,-b)$.
• Subtraction: $(a,b) - (c,d) = (a,b) + \big(-(c,d)\big) = (a-c,\ b-d)$.
• Multiplication: $(a,b)\cdot(c,d) = (ac-bd,\ ad+bc)$ — this one looks less obvious than addition, but it is exactly the rule that makes $(0,1)\cdot(0,1) = (-1,0)$, i.e. it builds in a square root of $-1$ from the start (more on this in the next section). …