Imaginary Numbers Advanced. This lesson goes beyond core Algebra I. You can skip it.

Learning goals

Define $i$ by $i^2 = -1$ , and know that $-i$ squares to $-1$ too
Reduce $i^n$ using the remainder of $n$ divided by four
Simplify $\sqrt{-n}$ as $i\sqrt{n}$ , and multiply negative square roots safely
Solve $x^2 = -k$ as $x = \pm\sqrt{k}\,i$

Why the real numbers are not enough

You can solve $x^2 = 9$ because $3^2 = 9$ , and $(-3)^2 = 9$ as well, so it has the two answers $x = 3$ and $x = -3$ . You can even solve $x^2 = 7$ , whose answers $\pm\sqrt{7}$ are irrational but still perfectly real. Try $x^2 = -1$ , though, and the search fails. No real number works, and the reason is not that we have not looked hard enough. It is built into how signs behave under multiplication.

Why no real number squares to a negative number#

Take any real number $a$ and look at its square $a^2 = a \times a$ . There are only three possibilities for the sign of $a$ . If $a$ is positive, then $a \times a$ is a positive times a positive, which is positive. If $a$ is negative, then $a \times a$ is a negative times a negative, and two negatives make a positive, so $a^2$ is again positive. If $a$ is zero, then $a^2 = 0$ .

In every case $a^2 \ge 0$ , so a real square is never negative. That rules out any real solution of $a^2 = -1$ , or $a^2 = -9$ , or any equation that asks for a negative square. The gap is genuine and permanent, and closing it needs a number that does not sit on the real number line at all.

Defining the imaginary unit

Define the imaginary unit $i$ to be a number whose square is $-1$ :

i^2 = -1.

That equation is the entire definition, and everything else in this lesson follows from it. The symbol $\sqrt{-1}$ is just another name for $i$ . That follows the same convention you already know from real numbers: $\sqrt{9}$ means the positive root $3$ , even though $-3$ also squares to $9$ . In the same way, $\sqrt{-1}$ points to $i$ , even though (as the next paragraph shows) $-i$ squares to $-1$ too.

With $i$ available, $x^2 = -1$ finally has answers. One is $x = i$ , because $i^2 = -1$ by definition. The other is $x = -i$ , because

(-i)^2 = (-1)^2\, i^2 = (1)(-1) = -1

as well. So $x^2 = -1$ has the two solutions $x = i$ and $x = -i$ , written together as $x = \pm i$ , matching the same two-answer pattern you already know from $x^2 = 9$ .

A number like $2 + 3i$ , combining a real number and a real multiple of $i$ , is called a complex number. This lesson stays with $i$ itself and its multiples; the arithmetic that combines the two parts of a complex number is the subject of the next lesson.

The powers of i

Since $i$ is a number, you can raise it to powers, and something useful happens: the powers repeat in a short cycle. Work them out one at a time, each from the one before, leaning on $i^2 = -1$ at every step.

i^1 = i, \qquad i^2 = -1, \qquad i^3 = -i, \qquad i^4 = 1.

Multiplying by i steps forward through a four-value cycle (i, then -1, then -i, then 1) and returns to i. Because the pattern repeats every four steps, a power of i is fixed by the remainder of its exponent divided by 4.

Why the powers of $i$ repeat every four steps#

Start from the definition and climb. The first power is $i$ itself. The second is $i^2 = -1$ , the definition. Multiply by one more factor of $i$ for the third power, then again for the fourth:

i^3 = i^2 \cdot i = -i, \qquad i^4 = i^2 \cdot i^2 = 1.

Reaching $i^4 = 1$ is the key event: four factors of $i$ multiply to $1$ . One factor beyond that returns to the start, since $i^5 = i^4 \cdot i = 1 \cdot i = i$ . From there the same four values $i, -1, -i, 1$ repeat forever, exactly as the diagram shows.

That fact turns any power into a small one. Take $i^{11}$ . Eleven factors of $i$ make two complete groups of four, worth $1$ each, with three factors left over:

i^{11} = (i^4)^2 \cdot i^3 = 1^2 \cdot i^3 = i^3 = -i.

Only the leftover factors matter. The same idea works for any exponent $n$ : write $n = 4q + r$ , where $r$ is the remainder of $n$ divided by $4$ , one of $0, 1, 2, 3$ . Then

i^n = i^{4q + r} = (i^4)^q \cdot i^r = 1^q \cdot i^r = i^r.

So $i^n$ depends only on $r$ , the remainder of $n$ divided by $4$ . When that remainder is $0$ , the power is $i^0 = 1$ .

In short, divide the exponent by $4$ and keep only the remainder. A remainder of $1$ gives $i$ , a remainder of $2$ gives $-1$ , a remainder of $3$ gives $-i$ , and a remainder of $0$ gives $1$ .

Worked example 1 Evaluate $i^{50}$ , $i^{23}$ , and $i^{100}$

Divide each exponent by $4$ and keep only the remainder.

For $i^{50}$ , since $50 = 4 \times 12 + 2$ , the remainder is $2$ :

i^{50} = i^2 = -1.

For $i^{23}$ , since $23 = 4 \times 5 + 3$ , the remainder is $3$ :

i^{23} = i^3 = -i.

For $i^{100}$ , since $100 = 4 \times 25$ exactly, the remainder is $0$ :

i^{100} = i^0 = 1.

No matter how large the exponent, only its remainder after dividing by $4$ ever matters.

Check your understanding

Simplify $i^{27}$ .

i

-i

1

-1

Square roots of negative numbers

The imaginary unit lets you take the square root of any negative number, not just $-1$ . The rule is short:

\sqrt{-n} = i\sqrt{n} \quad \text{for } n > 0.

Here $\sqrt{n}$ is the ordinary real square root you already know, and the factor of $i$ carries the negative sign. A real number multiplied by $i$ , such as the $3i$ you are about to compute, is called a pure imaginary number. These really are new numbers: $3i$ is not equal to any real number, because its square is $(3i)^2 = 9i^2 = -9$ , and no real number squares to $-9$ .

Why $\sqrt{-n} = i\sqrt{n}$ #

To call something the square root of $-n$ , it has to square to $-n$ . Test the candidate $i\sqrt{n}$ by squaring it:

\left(i\sqrt{n}\right)^2 = i^2 \left(\sqrt{n}\right)^2 = (-1)(n) = -n.

It works: $i\sqrt{n}$ squares to $-n$ , so it is a square root of $-n$ . The step $\left(\sqrt{n}\right)^2 = n$ is just the meaning of a real square root, and $i^2 = -1$ supplies the sign. Splitting the radical as $\sqrt{-n} = \sqrt{-1}\cdot\sqrt{n} = i\sqrt{n}$ is allowed here because only one of the two factors under the roots is negative. The next section shows why that restriction is not optional.

The other square root of $-n$ is $-i\sqrt{n}$ , but $\sqrt{-n}$ always names $i\sqrt{n}$ , the same way $\sqrt{9}$ always names $3$ and not $-3$ . Both roots matter when you solve an equation instead of just simplifying a radical, which is exactly what the last section of this lesson does.

Worked example 2 Simplify $\sqrt{-9}$ , $\sqrt{-12}$ , and $\sqrt{-72}$

Pull out the factor of $i$ first, then simplify the real square root that is left.

The root $\sqrt{-9}$ has a perfect square inside:

\sqrt{-9} = i\sqrt{9} = 3i.

The root $\sqrt{-12}$ needs the radical simplified, using $12 = 4 \times 3$ and $\sqrt{4} = 2$ :

\sqrt{-12} = i\sqrt{12} = i\sqrt{4 \times 3} = 2i\sqrt{3}.

The root $\sqrt{-72}$ works the same way, with $72 = 36 \times 2$ :

\sqrt{-72} = i\sqrt{72} = i\sqrt{36 \times 2} = 6i\sqrt{2}.

In each case the answer is a real number (or a simplified radical) times $i$ , that is, a pure imaginary number.

Check your understanding

Simplify $\sqrt{-50}$ .

5\sqrt{2}

25i\sqrt{2}

5i\sqrt{2}

5i

The trap: two negative radicands

There is one place where square roots of negatives bite back, and nearly everyone is caught by it once. You have used the product rule for square roots, $\sqrt{a}\cdot\sqrt{b} = \sqrt{ab}$ , to simplify radicals. That rule is valid only when $a$ and $b$ , the numbers under the root signs (the radicands), are not both negative. Apply it blindly when both radicands are negative and you get the wrong answer, including the wrong sign.

Watch it fail on $\sqrt{-4}\cdot\sqrt{-9}$ . The correct method rewrites each square root using $i$ first, and only then multiplies:

\sqrt{-4}\cdot\sqrt{-9} = (2i)(3i) = 6i^2 = -6.

The tempting shortcut multiplies the radicands first, and it disagrees:

\sqrt{-4}\cdot\sqrt{-9} \ne \sqrt{(-4)(-9)} = \sqrt{36} = 6.

The two results, $-6$ and $6$ , are not equal, so the product rule genuinely breaks for two negatives. The fix is a firm habit: convert every square root of a negative into the form $i\sqrt{\;}$ before doing anything else. Once the factors of $i$ are out in the open, they multiply like any other factors, and $i \cdot i = i^2 = -1$ supplies the correct sign on its own.

Worked example 3 Evaluate $\sqrt{-2}\cdot\sqrt{-8}$

Convert each root to the $i\sqrt{\;}$ form first. Here $\sqrt{-2} = i\sqrt{2}$ , and $\sqrt{-8} = i\sqrt{8} = 2i\sqrt{2}$ . Now multiply the two pure imaginary numbers:

\left(i\sqrt{2}\right)\left(2i\sqrt{2}\right) = 2i^2\left(\sqrt{2}\cdot\sqrt{2}\right) = 2(-1)(2) = -4.

The wrong route, $\sqrt{(-2)(-8)} = \sqrt{16} = 4$ , once again flips the sign. Convert first, and the answer $-4$ comes out right.

Check your understanding

Evaluate $\sqrt{-4}\cdot\sqrt{-25}$ .

10

-10

10i

-10i

Solving x squared equals a negative

Putting the pieces together, you can now solve any equation of the form $x^2 = -k$ with $k > 0$ , or equally $x^2 + k = 0$ . Take the square root of both sides, and keep both signs.

Worked example 4 Solve $x^2 + 9 = 0$ and $x^2 = -20$

For $x^2 + 9 = 0$ , move the constant across to isolate the square, then take the root of both sides:

x^2 = -9 \quad\Rightarrow\quad x = \pm\sqrt{-9} = \pm 3i.

Both $3i$ and $-3i$ check out, since $(\pm 3i)^2 = 9i^2 = -9$ . For $x^2 = -20$ , the root needs simplifying, with $20 = 4 \times 5$ :

x = \pm\sqrt{-20} = \pm i\sqrt{20} = \pm 2i\sqrt{5}.

Every equation of this shape has two pure imaginary solutions, each the negative of the other.

Check your understanding

Solve $x^2 + 12 = 0$ .

x = \pm 12i

x = \pm 2i\sqrt{3}

x = \pm 2\sqrt{3}

x = 2i\sqrt{3}

Common mistakes

Thinking $i^2 = 1$ . By definition $i^2 = -1$ . That single minus sign is the whole reason $i$ exists; a positive square would just be an ordinary real number.
The two-negatives radical trap. $\sqrt{-4}\cdot\sqrt{-9} = -6$ , not $6$ . Rewrite each root as $i\sqrt{\;}$ before multiplying, because $\sqrt{a}\cdot\sqrt{b} = \sqrt{ab}$ fails when both radicands are negative.
Writing $\sqrt{-9} = -3$ or $\pm 3$ . A square root of a negative is imaginary, not a negative real: $\sqrt{-9} = 3i$ . No real number, positive or negative, squares to $-9$ .
Multiplying the exponent instead of reducing it. $i^{20}$ is not $20i$ . Divide the exponent by $4$ and keep the remainder: since $20$ is a multiple of $4$ , the remainder is $0$ , so $i^{20} = 1$ .
Forgetting the second solution. Both $i$ and $-i$ square to $-1$ , since $(-i)^2 = (-1)^2 i^2 = -1$ too. So $x^2 = -25$ has both $x = 5i$ and $x = -5i$ : write $\pm 5i$ , not just one sign.

Key takeaways

The imaginary unit is defined by $i^2 = -1$ , equivalently $i = \sqrt{-1}$ . It fills the gap left by the real numbers, none of which square to a negative.
The powers of $i$ cycle every four: $i^1 = i$ , $i^2 = -1$ , $i^3 = -i$ , $i^4 = 1$ . Reduce $i^n$ using the remainder of $n$ divided by $4$ .
A square root of a negative simplifies as $\sqrt{-n} = i\sqrt{n}$ for $n > 0$ , for example $\sqrt{-9} = 3i$ and $\sqrt{-12} = 2i\sqrt{3}$ .
The product rule $\sqrt{a}\cdot\sqrt{b} = \sqrt{ab}$ fails for two negative radicands. Convert each root to $i\sqrt{\;}$ first: $\sqrt{-4}\cdot\sqrt{-9} = 6i^2 = -6$ , not $6$ .
A real multiple of $i$ , such as $3i$ , is a pure imaginary number, and $x^2 = -k$ with $k > 0$ has the two solutions $x = \pm\sqrt{k}\,i$ . Add a real number to a pure imaginary one, as in $2 + 3i$ , and the result is a complex number, the subject of the next lesson.

Imaginary Numbers Advanced. This lesson goes beyond core Algebra I. You can skip it.

Learning goals

Why the real numbers are not enough

Defining the imaginary unit

The powers of i

Square roots of negative numbers

The trap: two negative radicands

Solving x squared equals a negative

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice