Complex Numbers and Quadratic Equations

Written and reviewed by Devendra Dubey, Mathematics faculty at UniDrill (M.Sc. Mathematics, Osmania University · 10 years teaching Mathematics). Based on NCERT Mathematics Class XI, Chapter 4.

Is there a number that squares to −1?

Square any real number, 3 or −3 or 0.5, and the answer is never negative. So the equation $x^2 + 1 = 0$ seems to have no answer at all. For centuries mathematicians called such solutions "impossible". Then they tried something bold: if no such number exists, invent one, give it a name, and check that the usual rules of algebra still work. They did, and the result is one of the most useful ideas in mathematics. This chapter shows you how to calculate with it, where it lives on the page, and what it does to quadratic equations.

Why the real numbers are not enough

Each time a new kind of equation could not be solved, the number system grew.

Define the imaginary unit $i$ by

$$i^2 = -1, \qquad i = \sqrt{-1}$$

Now $x^2 + 1 = 0$ has the solution $x = i$. A complex number is any number of the form $z = a + ib$ where a and b are real. Here $a = \operatorname{Re} z$ is the real part and $b = \operatorname{Im} z$ is the imaginary part. The imaginary part is the real number b, without the i.

Two complex numbers are equal exactly when their real parts match and their imaginary parts match:

$$a + ib = c + id \iff a = c \text{ and } b = d$$

So one complex equation is secretly two real equations. This is the idea behind nearly every "find x and y" question.

Worked example: solving with equality. Find real x and y if $(2x + y) + i(x - 2y) = 5$.

Common wrong step. Treating the 5 as if it had no imaginary part to match, and writing only one equation. The right side is $5 + 0i$, so the imaginary side gives a second equation: it equals 0.

Adding, multiplying and dividing

Treat $i$ like a letter in algebra and replace $i^2$ by $-1$ whenever it appears.

The product rule is just expansion: $(a + ib)(c + id) = ac + iad + ibc + i^2 bd$, and $i^2 = -1$ turns the last term into $-bd$.

Complex numbers keep the familiar laws. Addition and multiplication are closed, commutative and associative, multiplication distributes over addition, and the identities and inverses are:

That is why the algebraic identities you know, such as $(z_1 \pm z_2)^2 = z_1^2 \pm 2z_1 z_2 + z_2^2$ and $z_1^2 - z_2^2 = (z_1 + z_2)(z_1 - z_2)$, still hold.

Worked example: multiply and divide. Find (a) $(2 + 3i)(4 - i)$ and (b) $\dfrac{4 + 3i}{2 - i}$ in the form $a + ib$.

Common wrong step. In (a), using $i^2 = +1$, which gives $5 + 10i$. The sign of $i^2$ is the entire point. In (b), multiplying by $2 - i$ itself instead of its conjugate. Then the denominator becomes $(2 - i)^2$, which is not real, and nothing has been simplified.

Powers of i: a cycle of four

Start multiplying: $i^1 = i$, $i^2 = -1$, $i^3 = i^2 \cdot i = -i$, $i^4 = (i^2)^2 = 1$. After that the pattern repeats, because $i^4 = 1$ means every extra block of four powers multiplies by 1.

So divide the exponent by 4 and look at the remainder. Negative powers are reciprocals: $i^{-1} = \dfrac{1}{i} = -i$, $i^{-2} = -1$, $i^{-3} = i$, $i^{-4} = 1$. The remainder rule works for negative exponents too, as long as you take the remainder properly (for example $-27 = 4 \times (-7) + 1$).

Worked example: a negative exponent. Find $i^{2023} + i^{-27}$.

Common wrong step. Reading off the remainder 3 for 27 and writing $i^{-27} = i^3 = -i$. That forgets the minus sign in the exponent, and the sum would come out as $-2i$.

Pause and predict: what is $1 + i + i^2 + i^3$, and what does that tell you about the sum of any four consecutive powers of $i$?

Check your answer: $1 + i - 1 - i = 0$. Any four consecutive powers add to $i^n(1 + i + i^2 + i^3) = 0$, so long sums of powers collapse quickly.

Square roots of negative numbers

For a positive real number a, $\sqrt{-a} = i\sqrt{a}$. The symbol $\sqrt{-1}$ means $i$ only. Both $i$ and $-i$ square to $-1$, but the root sign picks one, just as $\sqrt{4}$ means 2 and not $-2$.

Here is a trap that appears in many papers. The rule $\sqrt{a}\sqrt{b} = \sqrt{ab}$ is safe when at least one of a and b is non-negative. It fails when both are negative. If it worked, then $i \cdot i = \sqrt{-1}\sqrt{-1} = \sqrt{1} = 1$, which contradicts $i^2 = -1$. So convert each root to $i\sqrt{\ }$ first, and only then multiply.

Worked example: the root trap. Evaluate $\sqrt{-2} \cdot \sqrt{-8}$.

Common wrong step. Combining first: $\sqrt{-2}\sqrt{-8} = \sqrt{16} = 4$. The sign is wrong. Whenever both numbers under the roots are negative, the product of the roots is negative of what the combined root suggests.

Conjugate and modulus

For $z = a + ib$, the conjugate flips only the sign of the imaginary part, and the modulus is the length $\sqrt{a^2 + b^2}$:

$$\bar{z} = a - ib, \qquad \lvert z \rvert = \sqrt{a^2 + b^2}$$

The modulus is always a non-negative real number. The product of a number and its conjugate is real, because the cross terms cancel:

$$z\,\bar{z} = (a + ib)(a - ib) = a^2 + b^2 = \lvert z \rvert^2$$

This is exactly why division works. To find $\dfrac{1}{z}$ multiply top and bottom by $\bar{z}$, and the denominator becomes a real number:

$$z^{-1} = \dfrac{1}{z} = \dfrac{\bar{z}}{\lvert z \rvert^2} = \dfrac{a - ib}{a^2 + b^2}$$

Worked example: inverse and check. Find the multiplicative inverse of $3 - 4i$.

Common wrong step. Dividing by $\lvert z \rvert = 5$ instead of $\lvert z \rvert^2 = 25$, which gives $\dfrac{3 + 4i}{5}$. The check in step 4 would catch it: that number times $z$ is 5, not 1.

Pause and predict: for which complex numbers is $\bar{z} = z$?

Check your answer: exactly the real numbers. $a - ib = a + ib$ forces $b = 0$. So the real axis is the set of points that conjugation leaves alone.

The Argand plane: complex numbers as points

Each complex number $x + iy$ matches exactly one point $(x, y)$ of the plane, and each point matches one number. This plane is called the Argand plane (or complex plane). Its horizontal axis is the real axis (numbers $a + 0i$) and its vertical axis is the imaginary axis (numbers $0 + ib$).

Two facts turn algebra into geometry. The modulus $\lvert z \rvert = \sqrt{x^2 + y^2}$ is the distance of the point from the origin (Pythagoras). And the conjugate $x - iy$ is the mirror image of $x + iy$ in the real axis.

What does multiplying by $i$ do? If $z = x + iy$ then $iz = ix + i^2 y = -y + ix$, so the point $(x, y)$ moves to $(-y, x)$. This is a rotation through $90^\circ$ anticlockwise about the origin. For $z = 3 + 2i$ we get $iz = -2 + 3i$, so the point $(3, 2)$ moves to $(-2, 3)$. In terms of parts, $\operatorname{Re}(iz) = -\operatorname{Im}(z)$ and $\operatorname{Im}(iz) = \operatorname{Re}(z)$.

Pause and predict: if you multiply a point by $i$ twice, where does it land?

Check your answer: at the opposite point through the origin. Two quarter turns make a half turn, and the algebra agrees because $i \cdot i = -1$ sends $z$ to $-z$.

Quadratic equations with negative discriminant

A quadratic $ax^2 + bx + c = 0$ with real a, b, c and $a \ne 0$ has the roots

$$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

The discriminant $D = b^2 - 4ac$ decides what kind of roots you get. Before this chapter, $D < 0$ meant "no roots". Now it means complex roots:

Worked example: a negative discriminant. Solve $2x^2 + x + 1 = 0$.

Common wrong step. Dropping the $i$ and writing $\sqrt{-7}$ as $\sqrt{7}$, which pretends the roots are real. The roots come as a conjugate pair because the coefficients are real, and a missing $i$ breaks that pairing.

Beyond the textbook. For the exam: complex numbers are introduced so that $x^2 + 1 = 0$ has a solution. In reality: the real push came from cubic equations. In the 1500s, a formula for solving certain cubics produced square roots of negative numbers on the way to perfectly ordinary real answers. Bombelli showed that if you simply followed the rules of algebra, these "impossible" quantities cancelled and the real roots appeared. Why it matters: complex numbers were accepted because they worked, long before the Argand plane gave them a picture.

Beyond the textbook. For the exam: multiplying by $i$ rotates a point by $90^\circ$. In reality: this is why engineers describe alternating current with complex numbers. A phase shift of a wave becomes a multiplication, and engineers write $j$ instead of $i$ because $i$ already stands for current. Why it matters: a quarter turn is a multiplication by the same $i$ whose square is $-1$.

Beyond the textbook. For the exam: never compare complex numbers with $<$ or $>$ unless they are real. In reality: it is impossible to order $\mathbb{C}$ so that the usual rules hold. If $i > 0$ then $i^2 > 0$, so $-1 > 0$, which is false. If $i < 0$ then $-i > 0$ and $(-i)^2 > 0$, so again $-1 > 0$. Why it matters: to compare complex numbers you compare their moduli, which are real. The next chapter, Linear Inequalities, lives entirely in $\mathbb{R}$, where order exists.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Previous: Trigonometric Functions gave you the quarter turn of $\pi/2$, which here shows up as multiplication by $i$.
• Next: Linear Inequalities returns to the real line, where order exists and inequalities make sense.
• Binomial Theorem: Binomial Theorem expansions such as $(1 + i)^n$ use the powers of $i$ you learned here to collapse into a real and imaginary part.
• Sets: Sets places $\mathbb{C}$ at the top of the chain $\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R} \subset \mathbb{C}$.