No real number squares to a negative, so has no real solution. What if we allow a number whose square is ?

Algebra 2 · N.CN.A.1 · N.CN.C.7

Imaginary and complex numbers

No real number squares to a negative, so x^2 + 9 = 0 has no real solution. What if we allow a number whose square is -1?

You need

Square rootsMath 8 Multiplying and dividing powers with the same baseMath 8

Unlocks

Adding, subtracting and multiplying complex numbersAlgebra 2

Where you are 5 in Complex numbers

Number9

Complex numbers5

Adding, subtracting and multiplying complex numbers Conjugates and dividing complex numbers

Imaginary and complex numbers

Complex roots come in pairs The complex plane, modulus and argument

Why this exists

Try to solve x^2 + 9 = 0. Subtracting 9 gives x^2 = -9, and there the search stops. A positive number times itself is positive, a negative number times itself is positive too, and 0^2 = 0. No real number has a negative square. Figure 1 shows it: the curve y = x^2 meets the level y = 4 twice, above x = -2 and x = 2, and never comes down to the level y = -1.

Figure 1Every square is at least 0, so the curve y = x^2 meets the level 4 twice and never reaches the level -1.

For a long time that settled it: such equations had no solution. The trouble was that square roots of negative numbers kept turning up in the middle of problems with perfectly ordinary answers. In 1545 Gerolamo Cardano asked for two numbers that add to 10 and multiply to 40, and his method gave \displaystyle 5 + \sqrt{-15} and \displaystyle 5 - \sqrt{-15}. He did not know what to make of them, yet if \displaystyle \sqrt{-15} is treated as a number whose square is -15, they do add to 10, and they multiply to 25 - (-15) = 40. Later in the same century Rafael Bombelli took the cubic x^3 = 15x + 4, whose solution x = 4 anyone can check, and found that the known formula for cubics passes through \displaystyle \sqrt{-121} on the way to it. Refusing those roots meant refusing a route to real answers.

The way forward was the one taken before. Negative numbers were added so that 5 - 8 has an answer, and fractions so that 3 \div 4 has one. Here one new number is added, a number whose square is -1, the rules of arithmetic stay as they were, and we see what that forces. The result is the complex numbers, and in them every quadratic equation has solutions.

Definition

The imaginary unit i is a number whose square is -1:

i^2 = -1

and the square root of -1 is taken to be i, so \displaystyle \sqrt{-1} = i.

For a positive real number a, the square root of -a is

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

So \displaystyle \sqrt{-25} = i\sqrt{25} = 5i. The i is written in front of a radical, as in \displaystyle i\sqrt{3}, so that it cannot be read as sitting under the root sign.

A complex number is a number that can be written

a + bi \qquad (a \text{ and } b \text{ real})

and this is its standard form. For z = a + bi, the real number a is the real part, \operatorname{Re}(z) = a, and the real number b is the imaginary part, \operatorname{Im}(z) = b. The imaginary part is b, not bi: it is the real number that multiplies i.

If b = 0, then a + 0i = a is a real number. Every real number is a complex number.
If b \ne 0, then a + bi is an imaginary number.
If a = 0 and b \ne 0, the number bi is pure imaginary, such as 7i or -i.

Two complex numbers in standard form are equal exactly when their real parts are equal and their imaginary parts are equal.

Intuition

A complex number carries two real numbers at once: how much of the ordinary unit 1, and how much of the new unit i. Neither can be traded for the other, since no real number squares to -1. That is why 3 - 5i cannot be squeezed into a single real number, and why two complex numbers are equal only when both parts match. The complex plane later draws exactly this, the real part across and the imaginary part up.

With i in hand, every negative number has square roots. Since (5i)^2 = 25i^2 = -25, the number 5i squares to -25, and so does -5i, because (-5i)^2 = 25i^2 as well. That matches what happens for positive numbers: 25 has two square roots, 5 and -5, and the radical sign \displaystyle \sqrt{25} names one of them, the positive one (square roots). In the same way \displaystyle \sqrt{-25} names 5i, the root whose imaginary part is positive.

Powers of i go round in a loop. Each power is the one before it times i, and once i^2 is replaced by -1 the values come back after four steps:

n	i^n	n	i^n
1	i	5	i
2	-1	6	-1
3	-i	7	-i
4	1	8	1

For instance i^3 = i^2 \cdot i = -i and i^4 = i^3 \cdot i = -i^2 = 1. Figure 2 sets the four values in a ring; each factor of i moves one step round it.

Figure 2Each factor of i moves one step round the ring, so the powers of i repeat every four.

Results

The imaginary unit. i^2 = -1 and \displaystyle \sqrt{-1} = i.

Square roots of negative numbers. For every real number a > 0,

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

and the equation x^2 = -a has exactly two solutions, \displaystyle x = i\sqrt{a} and \displaystyle x = -i\sqrt{a}.

Powers of i. For every whole number k,

\begin{aligned} i^{4k} &= 1, & i^{4k+1} &= i, \\ i^{4k+2} &= -1, & i^{4k+3} &= -i \end{aligned}

so i^n depends only on the remainder when n is divided by 4.

Equality. For real numbers a, b, c and d, a + bi = c + di exactly when a = c and b = d.

Why it is true

No real number squares to a negative. If x > 0, then x^2 = x \cdot x is a product of two positives, so it is positive. If x < 0, then x^2 = (-x)(-x), again two positives. And 0^2 = 0. So x^2 \ge 0 for every real x, and a solution of x^2 = -1 has to be a new number.

The square root rule. The number i obeys the ordinary rules of multiplication, so \displaystyle (i\sqrt{a})^2 = i^2(\sqrt{a})^2 = (-1)a = -a. So \displaystyle i\sqrt{a} squares to -a, and so does \displaystyle -i\sqrt{a}, since (-1)^2 = 1. There are no others: if x^2 = -a, then \displaystyle x^2 - (i\sqrt{a})^2 = 0, which factors as a difference of two squares, \displaystyle (x - i\sqrt{a})(x + i\sqrt{a}) = 0, and a product of complex numbers is 0 only when a factor is 0, just as for real numbers.

The loop. By Multiplying and dividing powers with the same base, the exponents add, so i^4 = i^2 \cdot i^2 = (-1)(-1) = 1. Any whole number n is 4k + r with r one of 0, 1, 2, 3, and then

i^n = i^{4k} \cdot i^r = (i^4)^k \cdot i^r = 1^k \cdot i^r = i^r

Equality. Suppose a + bi = c + di. Then a - c = (d - b)i. If d \ne b, dividing by d - b gives \displaystyle i = \frac{a - c}{d - b}, a real number, but no real number squares to -1. So d = b, and then a - c = 0, so a = c.

Worked examples

Example 1

Square roots of negative numbers

Write \displaystyle \sqrt{-49} and \displaystyle \sqrt{-18} as multiples of i.

Solution

Each has the shape \displaystyle \sqrt{-a} with a positive, so \displaystyle \sqrt{-a} = i\sqrt{a} applies. Since 49 is a perfect square, \displaystyle \sqrt{-49} = i\sqrt{49} = 7i.

18 is not a perfect square, but it has the square factor 9, and simplifying the root gives \displaystyle \sqrt{18} = \sqrt{9}\sqrt{2} = 3\sqrt{2}. So \displaystyle \sqrt{-18} = i\sqrt{18} = 3i\sqrt{2}.

Check. (7i)^2 = 49i^2 = -49, and \displaystyle (3i\sqrt{2})^2 = 9 \cdot i^2 \cdot 2 = -18.

Example 2

Reading the parts of a complex number

Give the real part and the imaginary part of 3 - 5i, and say whether it is real, imaginary or pure imaginary.

Solution

Standard form is a + bi, with a plus sign. Subtracting 5i is the same as adding (-5)i, so 3 - 5i = 3 + (-5)i. So the real part is 3 and the imaginary part is -5. The minus sign belongs to the imaginary part, and the part is -5, not -5i, because it is the real number that multiplies i.

Since b = -5 \ne 0, the number is imaginary. Since a = 3 \ne 0, it is not pure imaginary.

Example 3

Powers of i

Simplify i^7 and i^{22}.

Solution

The powers repeat every four, so take out as many factors i^4 = 1 as possible. Since 7 = 4 + 3, the exponents add by Multiplying and dividing powers with the same base, and

i^7 = i^4 \cdot i^3 = 1 \cdot (-i) = -i

Since 22 = 4 \cdot 5 + 2,

i^{22} = (i^4)^5 \cdot i^2 = 1^5 \cdot (-1) = -1

Check. Group the factors in pairs instead: i^{22} = (i^2)^{11} = (-1)^{11} = -1.

Example 4

An equation with no real solution

Solve x^2 + 9 = 0.

Solution

Subtracting 9 from both sides gives x^2 = -9. There is no real solution, since every real square is at least 0. Over the complex numbers, -9 is -a with a = 9, and the results give exactly two solutions:

x = \pm i\sqrt{9}, \qquad x = \pm 3i

Check. (3i)^2 = 9i^2 = -9 and (-3i)^2 = 9i^2 = -9, so x^2 + 9 = 0 both times. The graph of y = x^2 + 9 never meets the x-axis, which matches its negative discriminant, 0^2 - 4 \cdot 1 \cdot 9 = -36. The solutions \pm 3i are not points on that graph.

Example 5

Putting a root into standard form

Write \displaystyle \dfrac{8 - \sqrt{-12}}{4} in standard form, and give its real and imaginary parts.

Solution

First the root: \displaystyle \sqrt{-12} = i\sqrt{12} = i\sqrt{4}\sqrt{3} = 2i\sqrt{3}, so the numerator is \displaystyle 8 - 2i\sqrt{3}. Dividing a difference by 4 divides each term by 4:

\frac{8 - 2i\sqrt{3}}{4} = \frac{8}{4} - \frac{2\sqrt{3}}{4}i = 2 - \frac{\sqrt{3}}{2}i

The real part is 2, and the imaginary part \displaystyle -\frac{\sqrt{3}}{2}. Expressions of this shape come out of the quadratic formula whenever the discriminant is negative. Dividing by a real number only splits the fraction; dividing by a number that is not real is another job, done in Conjugates and dividing complex numbers.

Example 6

Two negatives under two roots

Evaluate \displaystyle \sqrt{-4} \cdot \sqrt{-9}.

Solution

Write each root with i first: \displaystyle \sqrt{-4} = 2i and \displaystyle \sqrt{-9} = 3i. Reordering the factors (the commutative property) and regrouping them (the associative property),

2i \cdot 3i = (2 \cdot 3)(i \cdot i) = 6i^2 = -6

The rule \displaystyle \sqrt{a}\sqrt{b} = \sqrt{ab} would have given \displaystyle \sqrt{(-4)(-9)} = \sqrt{36} = 6, which has the wrong sign. That rule does not hold when both numbers are negative; Boundaries says why.

Example 7

Matching parts

Find the real numbers x and y for which (2x - 6) + (y + 4)i = 10 - i.

Solution

Since x and y are real, 2x - 6 and y + 4 are real, so the left side is in standard form. The right side is 10 + (-1)i. Equal complex numbers have equal real parts and equal imaginary parts:

2x - 6 = 10 \qquad\text{and}\qquad y + 4 = -1

So x = 8 and y = -5. One equation between complex numbers carries two equations between real numbers.

Check. 2(8) - 6 = 10 and -5 + 4 = -1.

Boundaries

The product rule fails when both numbers are negative. \displaystyle \sqrt{a}\sqrt{b} = \sqrt{ab} holds for a \ge 0 and b \ge 0, and still holds when just one is negative: \displaystyle \sqrt{-4}\sqrt{9} = 2i \cdot 3 = 6i = \sqrt{-36}. With both negative it is off by a sign. For a > 0 and b > 0,

\begin{aligned} \sqrt{-a}\sqrt{-b} &= i\sqrt{a} \cdot i\sqrt{b} = i^2\sqrt{ab} = -\sqrt{ab}, \\ \sqrt{(-a)(-b)} &= \sqrt{ab} \end{aligned}

So write every root of a negative number as a multiple of i before multiplying.

Complex numbers are not ordered. The signs < and > compare real numbers only. If i > 0 were true, multiplying both sides by the positive number i would give i^2 > 0, that is -1 > 0. If i < 0 were true, then -i > 0, and the same step gives (-i)^2 > 0, again -1 > 0. So neither holds. An inequality such as x^2 + 9 > 0, solved in Quadratic inequalities, is a statement about real x.

The parts are read from standard form. In \displaystyle 5 + \sqrt{-4} the imaginary part is 2, not \displaystyle \sqrt{-4}, because \displaystyle \sqrt{-4} = 2i; and in 4i + 2 the real part is 2. The numbers a and b in a + bi must both be real.

The loop is about whole-number exponents. The rule i^n = i^r is proved here for n = 0, 1, 2, \dots. It extends to negative integer exponents once dividing by i is in hand, but powers such as i^{1/2} are a different question, not settled by the loop.

Where this goes next

With i^2 = -1 in place, complex numbers can be added, subtracted and multiplied by treating i like a variable and then replacing i^2 with -1, which is Adding, subtracting and multiplying complex numbers; the powers of i from this lesson turn up there constantly. Dividing by a complex number uses the conjugate, a - bi, in Conjugates and dividing complex numbers. In the complex plane multiplying by i turns out to be a quarter turn, which the ring in Figure 2 already hints at.

Every quadratic equation now has solutions: the quadratic formula gives complex ones whenever the discriminant is negative, and for real coefficients they come as a pair p \pm qi, the subject of Complex roots come in pairs. And no further invention is ever needed: by the The fundamental theorem of algebra, every polynomial equation has its solutions among the complex numbers.

Exercises

Square roots of negative numbers

Write each as a multiple of i, with any root simplified.

\displaystyle \sqrt{-100}
\displaystyle \sqrt{-7}
\displaystyle \sqrt{-50}
\displaystyle -\sqrt{-64}
\displaystyle \sqrt{-\frac{9}{25}}
\displaystyle \sqrt{-45}

Real and imaginary parts

Give the real part and the imaginary part of -2 + 9i.
Give the real part and the imaginary part of 11i, and say whether it is real, imaginary or pure imaginary.
Give the real part and the imaginary part of -4, and say whether it is real, imaginary or pure imaginary.
Write -6i + 4 in standard form and give its real part and imaginary part.
Write \displaystyle \dfrac{10 + \sqrt{-8}}{2} in standard form.
Write \displaystyle \dfrac{-6 - \sqrt{-27}}{3} in standard form.

Powers of i

Simplify.

i^{11}
i^{40}
i^{33}
i^{102}
i^3 \cdot i^9
(-i)^6

Equations

Solve x^2 + 16 = 0 over the complex numbers.
Solve x^2 = -20 over the complex numbers.
Solve 4x^2 + 81 = 0 over the complex numbers.
Find the real numbers x and y for which (3x + 1) + (y - 2)i = 13 + 6i.

Reasoning

Mahala writes \displaystyle \sqrt{-2}\cdot\sqrt{-8} = \sqrt{16} = 4. Find the step that breaks, and give the correct value.
Rhona notices that (-i)^2 = -1 as well. Is -i a square root of -1? If it is, why is \displaystyle \sqrt{-1} defined to be i?
Svetlana says the last two digits of a whole number n are enough to find i^n. Explain why she is right, and use it to find i^{2026}.

Answers

10i. \displaystyle \sqrt{-100} = i\sqrt{100} = 10i.
\displaystyle i\sqrt{7}. The only square factor of 7 is 1, so the root stays as it is.
\displaystyle 5i\sqrt{2}. \displaystyle \sqrt{-50} = i\sqrt{25}\sqrt{2} = 5i\sqrt{2}.
-8i. \displaystyle \sqrt{-64} = 8i, and the minus sign in front makes it -8i.
\displaystyle \frac{3}{5}i. \displaystyle \sqrt{-\frac{9}{25}} = i\sqrt{\frac{9}{25}} = \frac{3}{5}i.
\displaystyle 3i\sqrt{5}. \displaystyle \sqrt{45} = \sqrt{9}\sqrt{5} = 3\sqrt{5}.
Real part -2, imaginary part 9.
Real part 0, imaginary part 11. 11i = 0 + 11i, so it is pure imaginary.
Real part -4, imaginary part 0. -4 = -4 + 0i, so it is real, and not imaginary.
4 - 6i, with real part 4, imaginary part -6. Addition is commutative, so the real term can move to the front.
\displaystyle 5 + i\sqrt{2}. \displaystyle \sqrt{-8} = 2i\sqrt{2}, then \displaystyle \frac{10}{2} + \frac{2i\sqrt{2}}{2}.
\displaystyle -2 - i\sqrt{3}. \displaystyle \sqrt{-27} = 3i\sqrt{3}, then \displaystyle \frac{-6}{3} - \frac{3i\sqrt{3}}{3}.
-i. 11 = 8 + 3, so i^{11} = (i^4)^2 \cdot i^3 = -i.
1. 40 = 4 \cdot 10, so i^{40} = (i^4)^{10} = 1.
i. 33 = 32 + 1, so i^{33} = (i^4)^8 \cdot i = i.
-1. 102 = 100 + 2 and 100 = 4 \cdot 25, so i^{102} = i^2 = -1.
1. The exponents add, so i^3 \cdot i^9 = i^{12} = (i^4)^3 = 1.
-1. (-i)^2 = i^2 = -1, so (-i)^6 = \left((-i)^2\right)^3 = (-1)^3 = -1.
x = \pm 4i. x^2 = -16, so \displaystyle x = \pm i\sqrt{16}.
\displaystyle x = \pm 2i\sqrt{5}. \displaystyle x = \pm i\sqrt{20} and \displaystyle \sqrt{20} = \sqrt{4}\sqrt{5} = 2\sqrt{5}.
\displaystyle x = \pm \frac{9}{2}i. 4x^2 = -81, so \displaystyle x^2 = -\frac{81}{4} and \displaystyle x = \pm i\sqrt{\frac{81}{4}}.
x = 4, y = 8. Real parts: 3x + 1 = 13. Imaginary parts: y - 2 = 6.
-4. The step \displaystyle \sqrt{-2}\cdot\sqrt{-8} = \sqrt{16} uses \displaystyle \sqrt{a}\sqrt{b} = \sqrt{ab} with both numbers negative, where it fails. Correctly, \displaystyle \sqrt{-2}\cdot\sqrt{-8} = i\sqrt{2} \cdot 2i\sqrt{2} = 2i^2(\sqrt{2})^2 = -4.
Yes: (-i)^2 = (-1)^2 i^2 = -1. So -1 has two square roots, i and -i, just as 25 has 5 and -5. The radical sign has to name one of them, and which one is an agreement, just as \displaystyle \sqrt{25} is agreed to name 5, the root that is not negative. The agreement is that \displaystyle \sqrt{-1} is i, the root whose imaginary part is positive. Nothing in the arithmetic itself can tell i and -i apart, since both square to -1.
i^{2026} = -1. The value of i^n depends only on the remainder when n is divided by 4. A whole number is 100q + r, where r is the number its last two digits make, and 100 = 4 \cdot 25 is a multiple of 4, so n and r leave the same remainder. For 2026, r = 26 = 4 \cdot 6 + 2, so i^{2026} = i^{26} = i^2 = -1.