In short

No real number squared gives a negative result, so the equation x^2 = -1 has no solution on the number line. The fix is to invent a new number i with the defining property i^2 = -1. Every complex number is then a + bi where a and b are real. The number a is the real part, b is the imaginary part, and together they extend the one-dimensional number line into a two-dimensional number plane. The powers of i cycle with period four: i, -1, -i, 1, i, -1, \dots

Try to solve x^2 = -1. Pick any positive number and square it — the answer is positive. Pick any negative number and square it — still positive. Try zero — you get zero, not -1. No matter which real number you try, x^2 lands on zero or to the right of zero on the number line. It never goes left. The equation x^2 = -1 has no solution among the real numbers.

You have seen this pattern before. In Number Systems, every new layer of the number system was forced into existence by an equation the old layer could not solve. Subtraction forced the invention of negatives. Division forced the invention of fractions. The existence of \sqrt{2} forced the invention of irrationals. Each time, the response was the same: if the equation has no answer inside the current system, build a bigger system.

The equation x^2 = -1 is the next crack. And the patch follows the same pattern: invent a new number.

The need for complex numbers

The problem is not just the equation x^2 = -1 in isolation. It shows up naturally when you solve quadratic equations.

Take x^2 + 1 = 0. The quadratic formula gives

x = \frac{-0 \pm \sqrt{0^2 - 4(1)(1)}}{2(1)} = \frac{\pm\sqrt{-4}}{2}.

The discriminant is -4. You learned in Discriminant and Nature of Roots that a negative discriminant means no real roots. But the formula itself does not break — it produces \sqrt{-4}, a square root of a negative number. The formula is trying to give you an answer. It just needs a number system where \sqrt{-4} means something.

Or consider a more concrete quadratic: x^2 - 4x + 5 = 0. The discriminant is 16 - 20 = -4. No real roots. But this equation describes a parabola that sits entirely above the x-axis — it has a vertex at (2, 1) and never touches down. The parabola is real, the algebra is real, the question "where are the roots?" is real. The roots are the thing that is not real — in the literal, mathematical sense.

Parabola y equals x squared minus 4x plus 5 floating above the x-axisA coordinate plane with a parabola opening upward. The vertex is at the point (2, 1), marked with a red dot, and the parabola never crosses the x-axis. A dashed horizontal line at y equals 0 shows the x-axis. The label says discriminant equals negative 4 and no real roots.xy012341234vertex (2, 1)D = −4no real roots
The parabola $y = x^2 - 4x + 5$ has its vertex at $(2, 1)$, one unit above the $x$-axis. It never crosses down. The discriminant $D = -4$ is negative, confirming that the equation $x^2 - 4x + 5 = 0$ has no real solutions. The roots exist — but not on the real number line.

The situation is exactly the one that forced every previous extension of the number system. An equation you care about has no answer in the current system. Time to build a bigger one.

The imaginary unit

The imaginary unit

The number i is defined by the property

i^2 = -1.

Equivalently, i = \sqrt{-1}, where the square root symbol here means "the number whose square is -1." No real number has this property.

The name "imaginary" is unfortunate. It was coined centuries ago by mathematicians who were uncomfortable with the idea, and the name stuck. There is nothing imaginary about i — it is as precisely defined as \sqrt{2}, which is "the positive number whose square is 2." The number i is "the number whose square is -1." Both are defined by a single equation. Both extend the number system to handle something the previous system could not.

With i in hand, you can write the square root of any negative number. Since -4 = 4 \times (-1), you have

\sqrt{-4} = \sqrt{4} \times \sqrt{-1} = 2i.

More generally, for any positive real number a,

\sqrt{-a} = \sqrt{a}\; i.

The negative sign under the radical becomes an i sitting outside. Every square root of a negative number is just a real multiple of i.

Complex numbers: real part and imaginary part

Once i exists, you can add it to any real number. The number 3 + 2i makes sense: it is the real number 3 plus two copies of i. The number -1 + 7i makes sense. The number 0 + 4i, which you write as just 4i, makes sense. Every expression of the form a + bi — where a and b are real numbers — is a complex number.

Complex number

A complex number is any expression of the form

z = a + bi

where a, b \in \mathbb{R} and i^2 = -1. The number a is called the real part of z, written \operatorname{Re}(z). The number b is called the imaginary part of z, written \operatorname{Im}(z).

Notice that the imaginary part is the real number b, not bi. If z = 3 + 2i, then \operatorname{Re}(z) = 3 and \operatorname{Im}(z) = 2, not 2i. This is a convention that catches people the first time, so note it now.

The set of all complex numbers is written \mathbb{C}. It contains every real number — because any real a can be written as a + 0i — and it contains the purely imaginary numbers like 3i = 0 + 3i. The hierarchy of number systems extends one more layer:

\mathbb{N} \;\subset\; \mathbb{Z} \;\subset\; \mathbb{Q} \;\subset\; \mathbb{R} \;\subset\; \mathbb{C}
Number systems nested from naturals to complex numbersFive nested rounded rectangles. The outermost is the complex numbers C. Inside it sit the reals R, then the rationals Q, then the integers Z, then the naturals N at the centre. The band between the reals and the complex numbers contains example complex numbers like 3 plus 2i, negative 1 plus 7i, and 4i. Each inner band shows example numbers from that layer.ℂℝℚℤℕ1 2 3 5 42−7 −1 01/2 −3/4 22/7√2 π e3 + 2i −1 + 7i 4icountingnegativesfractionsirrationalscomplex numbers
The complex numbers $\mathbb{C}$ wrap around the entire tower of number systems. Every real number is a complex number with $b = 0$. The new outer band holds numbers like $3 + 2i$ and $4i$ — numbers with a non-zero imaginary part — which have no place on the real number line.

Powers of i

One of the first things to nail down is what happens when you repeatedly multiply by i. Start from the definition and keep going.

i^1 = i
i^2 = -1
i^3 = i^2 \cdot i = (-1) \cdot i = -i
i^4 = i^3 \cdot i = (-i) \cdot i = -i^2 = -(-1) = 1

And now i^5 = i^4 \cdot i = 1 \cdot i = i. The cycle repeats. The powers of i rotate through exactly four values:

i, \quad -1, \quad -i, \quad 1, \quad i, \quad -1, \quad -i, \quad 1, \quad \dots

The period is 4. To find i^n for any positive integer n, divide n by 4 and look at the remainder:

Remainder when n \div 4 i^n
1 i
2 -1
3 -i
0 1

This works for large powers too. What is i^{2026}? Divide 2026 by 4: 2026 = 4 \times 506 + 2. Remainder 2. So i^{2026} = -1.

What about negative exponents? Since i^4 = 1, you can write i^{-1} = i^{-1} \cdot \dfrac{i^4}{i^4} = \dfrac{i^3}{i^4} = \dfrac{-i}{1} = -i. Or more directly: i^{-1} = \dfrac{1}{i} = \dfrac{1}{i} \cdot \dfrac{i}{i} = \dfrac{i}{i^2} = \dfrac{i}{-1} = -i. The negative powers cycle through the same four values in reverse order.

The four-cycle of powers of i shown as a circular diagramA circle with four points equally spaced around it, labelled clockwise starting from the right: 1 at the 3 o'clock position, i at the 12 o'clock position, negative 1 at the 9 o'clock position, and negative i at the 6 o'clock position. Red curved arrows go clockwise from each point to the next, labelled multiply by i. Below the circle, a small table shows the remainder rule: remainder 0 gives 1, remainder 1 gives i, remainder 2 gives negative 1, remainder 3 gives negative i.1i⁰, i⁴, i⁸…ii¹, i⁵, i⁹…−1i², i⁶, i¹⁰…−ii³, i⁷, i¹¹…×i×i×i×i
The powers of $i$ cycle through four values. Starting at $1$ (right) and multiplying by $i$ each time, you move counterclockwise: $1 \to i \to -1 \to -i \to 1 \to \dots$ The cycle has period $4$, so $i^n$ depends only on the remainder when $n$ is divided by $4$.

Writing square roots of negative numbers

With the powers of i settled, you can now express \sqrt{-a} for any positive a.

\sqrt{-a} = i\sqrt{a}

A few examples:

\sqrt{-9} = 3i \qquad \sqrt{-2} = i\sqrt{2} \qquad \sqrt{-25} = 5i

There is a subtlety here. The identity \sqrt{ab} = \sqrt{a}\sqrt{b} that you learned for positive reals does not extend to negative numbers. If you carelessly write \sqrt{-4} \cdot \sqrt{-9} = \sqrt{(-4)(-9)} = \sqrt{36} = 6, you get the wrong answer. The correct computation is \sqrt{-4} \cdot \sqrt{-9} = (2i)(3i) = 6i^2 = -6. Always convert to the i form first, then multiply.

Returning to the quadratic

Now go back to x^2 - 4x + 5 = 0, the equation whose discriminant was -4.

x = \frac{4 \pm \sqrt{-4}}{2} = \frac{4 \pm 2i}{2} = 2 \pm i

The two roots are 2 + i and 2 - i. They come in a pair: same real part, opposite imaginary parts. This always happens when the coefficients of the quadratic are real — the complex roots appear in conjugate pairs.

Check by expanding (x - (2+i))(x - (2-i)):

(x - 2 - i)(x - 2 + i) = (x-2)^2 - (i)^2 = x^2 - 4x + 4 - (-1) = x^2 - 4x + 5.

The original polynomial is recovered. The complex roots work — they satisfy the equation in the same way real roots would, just with arithmetic that passes through i.

The two complex roots 2 plus i and 2 minus i of the quadratic x squared minus 4x plus 5A vertical number line on the left shows the real axis; a horizontal extension to the right shows the imaginary axis, forming a cross. Two red dots are placed at coordinates (2, 1) and (2, negative 1), representing the roots 2 plus i and 2 minus i. A vertical dashed line connects them, showing they are reflections of each other across the real axis. The vertex of the parabola at (2, 1) on the real graph corresponds to the real part of both roots being 2.Real axisImaginary axis01234i−i2 + i2 − iconjugate pair
The two roots $2 + i$ and $2 - i$ sit at mirror-image positions above and below the real axis. Their real part $2$ matches the $x$-coordinate of the parabola's vertex. The dashed line between them shows the conjugate symmetry: when a quadratic with real coefficients has complex roots, they always come in conjugate pairs.

Equality of complex numbers

Two complex numbers a + bi and c + di are equal if and only if their real parts are equal and their imaginary parts are equal:

a + bi = c + di \quad \Longleftrightarrow \quad a = c \;\text{ and }\; b = d.

This is stronger than equality of real numbers. A single real equation gives you one condition. A single complex equation gives you two — one from the real parts, one from the imaginary parts. This is why complex equations are often equivalent to a pair of simultaneous real equations.

For instance, if (x + y) + (x - y)i = 4 + 2i, then matching real parts gives x + y = 4 and matching imaginary parts gives x - y = 2. Adding: 2x = 6, so x = 3 and y = 1.

Try this directly. Drag the red point around the complex plane below. The readouts show the real part, imaginary part, and the value of z^2 — computed using the rule (a + bi)^2 = (a^2 - b^2) + 2abi. Watch how the square changes as you move through different quadrants.

Interactive complex number explorerA coordinate plane representing the complex plane with the real axis horizontal and imaginary axis vertical, both running from negative four to four. A red draggable point represents a complex number z. Readouts above the plane show the real part, imaginary part, and the value of z squared, all updating as the reader drags the point.ReIm123−1−2−3↔ drag the red point
Drag the red point to any position $z = a + bi$. The top readouts show the real and imaginary parts. The bottom readouts show $z^2$: its real part is $a^2 - b^2$ and its imaginary part is $2ab$. Place $z$ at the point $i$ (coordinates $(0,1)$) and confirm that $z^2 = -1$.

Worked examples

Example 1: Simplify $i^{403} + i^{404} + i^{405} + i^{406}$

Four consecutive powers of i. The cycle has period 4, so consecutive powers always form a predictable pattern.

Step 1. Find the remainder of 403 \div 4.

403 = 4 \times 100 + 3

So the remainder is 3, and i^{403} = i^3 = -i.

Why: the cycle i, -1, -i, 1 repeats every four powers. Only the remainder matters.

Step 2. The next three powers follow from the cycle.

i^{404} = i^{403} \cdot i = (-i)(i) = -i^2 = 1
i^{405} = i^{404} \cdot i = 1 \cdot i = i
i^{406} = i^{405} \cdot i = i \cdot i = i^2 = -1

Why: each power is the previous one times i. Four consecutive powers of i always cycle through all four values in the set \{i, -1, -i, 1\}.

Step 3. Add them up.

(-i) + 1 + i + (-1) = (-i + i) + (1 - 1) = 0 + 0 = 0

Why: the four values i, -1, -i, 1 sum to zero — they cancel in pairs. This is always true for any four consecutive powers of i.

Result: i^{403} + i^{404} + i^{405} + i^{406} = 0.

Four consecutive powers of i cancelling to zero on the cycle diagramA circle with the four values of powers of i marked at compass points: 1 on the right, i at the top, negative 1 on the left, negative i at the bottom. Each of the four points is highlighted with a red dot and labelled with its corresponding power. Arrows from each pair that cancels (i and negative i, 1 and negative 1) meet at the centre, where a large zero is written, showing the sum is zero.1 = i⁴⁰⁴i = i⁴⁰⁵−1 = i⁴⁰⁶−i = i⁴⁰³0opposite pairs cancel: i + (−i) = 0 and 1 + (−1) = 0
The four consecutive powers $i^{403}$ through $i^{406}$ land on all four points of the cycle. The dashed lines connect the pairs that cancel: $i$ and $-i$ sum to zero, and $1$ and $-1$ sum to zero. The total is always zero for any run of four consecutive powers.

The fact that four consecutive powers of i always sum to zero is a useful shortcut. If you are asked for i^1 + i^2 + i^3 + \cdots + i^{100}, just count: 100 terms form 25 complete groups of four, each summing to zero. The answer is 0.

Example 2: Find the complex roots of $x^2 + 6x + 13 = 0$

A quadratic with real coefficients and a negative discriminant. The roots will be complex conjugates.

Step 1. Compute the discriminant.

D = b^2 - 4ac = 36 - 52 = -16

Why: D < 0 confirms no real roots, and the value of D tells you exactly what goes under the square root in the formula.

Step 2. Apply the quadratic formula.

x = \frac{-6 \pm \sqrt{-16}}{2}

Why: the formula works for all quadratics, including those with negative discriminants. The new ingredient is expressing \sqrt{-16} using i.

Step 3. Simplify \sqrt{-16}.

\sqrt{-16} = \sqrt{16} \cdot \sqrt{-1} = 4i

So the roots are:

x = \frac{-6 \pm 4i}{2} = -3 \pm 2i

Why: separate the negative sign from the 16, convert to i, then simplify the fraction by dividing numerator and denominator by 2.

Step 4. Verify by substitution. Take x = -3 + 2i.

(-3 + 2i)^2 + 6(-3 + 2i) + 13
= (9 - 12i + 4i^2) + (-18 + 12i) + 13
= (9 - 12i - 4) + (-18 + 12i) + 13
= 5 - 12i - 18 + 12i + 13
= (5 - 18 + 13) + (-12i + 12i) = 0 + 0i = 0. \checkmark

Why: substitution is the definitive check. The real terms sum to zero and the imaginary terms sum to zero independently — both conditions are needed for the complex number to equal zero.

Result: The roots are x = -3 + 2i and x = -3 - 2i.

Complex roots of x squared plus 6x plus 13 shown on the complex planeA coordinate cross representing the complex plane. The horizontal axis is the real axis, the vertical is the imaginary axis. Two red dots are plotted at (negative 3, 2) and (negative 3, negative 2), representing the conjugate roots negative 3 plus 2i and negative 3 minus 2i. A dashed vertical line connects them. The midpoint of the pair is marked at (negative 3, 0) on the real axis.ReIm−3−2−1012i−2i−3 + 2i−3 − 2imidpoint = −3= −b/2a
The two roots $-3 + 2i$ and $-3 - 2i$ are conjugates: same real part, opposite imaginary parts. They sit at mirror-image positions above and below the real axis. The midpoint on the real axis is $-3$, which equals $-b/2a$ — the $x$-coordinate of the parabola's vertex.

The conjugate symmetry is visible in the picture: the two roots are reflections of each other across the real axis. The midpoint of the pair lies on the real axis at -3, which is -b/(2a) — the same formula that gives the vertex of the parabola. The real part of the complex roots remembers where the vertex is; the imaginary part measures how far above the axis the vertex sits.

Common confusions

Going deeper

If you now know that i^2 = -1, that every complex number is a + bi, that the powers of i cycle with period four, and that quadratics with negative discriminants have complex conjugate roots, you have the core of the introduction. The rest of this section is for readers who want to see where the idea came from and where it leads.

The historical nudge

Complex numbers were not invented to solve x^2 = -1. That equation was easy to dismiss — just say "no solution" and move on. The real pressure came from cubic equations, where a formula discovered in the 16th century sometimes required computing the square root of a negative number even when all three roots were real. The formula forced mathematicians to pass through complex arithmetic to reach a real answer. They could not just say "no solution" — the solution was right there, real and correct, but the path to it went through \sqrt{-1}. That is what made complex numbers impossible to ignore.

Brahmagupta's 7th-century rules for negative numbers had a similar flavour — he wrote that "the square of a negative number is positive," which implicitly rules out real square roots of negatives. The recognition that you could define a useful number whose square is negative came much later, but the arithmetic foundation Brahmagupta laid for negative numbers was part of what made the eventual leap possible.

\mathbb{C} is algebraically closed

Once you have the complex numbers, no polynomial equation — not just quadratics, but cubics, quartics, and polynomials of any degree — can escape the system. Every polynomial with complex coefficients has all its roots in \mathbb{C}. This is the Fundamental Theorem of Algebra: a polynomial of degree n has exactly n roots in \mathbb{C} (counting multiplicity).

This means \mathbb{C} is the last extension you will ever need for the purpose of solving polynomial equations. There is no equation that forces a further patch. The number system that started with counting mangoes, and grew through negatives and fractions and irrationals, finally closes at \mathbb{C}. That is a remarkable place to stop.

The complex plane — a preview

A real number lives on a line. A complex number a + bi has two independent pieces of information — a and b — so it lives on a plane. The horizontal axis represents the real part; the vertical axis represents the imaginary part. Every complex number is a point in this plane, and every point is a complex number.

Multiplying by i rotates a point 90° counterclockwise around the origin. Multiplying by -1 rotates by 180°. The four-cycle 1 \to i \to -1 \to -i \to 1 is a quarter-turn rotation repeated four times. This geometric picture — that multiplication by a complex number is a rotation-and-scaling — is the subject of Complex Numbers — Modulus and Argument and Complex Numbers — Argand Plane.

Where this leads next

The imaginary unit i is the first step. The articles below build the full picture.