What is the value of i squared on the ACT?

By definition, i = √(-1), so i² = -1. This single rule drives every complex-number simplification: whenever i² appears, replace it with -1 before finishing the problem.

How do you simplify a high power of i?

Powers of i cycle every four: i¹ = i, i² = -1, i³ = -i, i⁴ = 1. Divide the exponent by 4 and use the remainder (0, 1, 2, or 3) to pick the matching value from the cycle.

How do you divide complex numbers?

Multiply the numerator and denominator by the conjugate of the denominator. Since (a + bi)(a - bi) = a² + b², this makes the denominator a real number, and you can then split the result into a + bi form.

What is the conjugate of a complex number?

The conjugate of a + bi is a - bi: keep the real part and flip the sign of the imaginary part. Conjugates are used to divide complex numbers because a number times its conjugate is always real.

CramClub • Complex Numbers — ACT® Guide & Practice

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

$i^1 = i,\; i^2 = -1,\; i^3 = -i,\; i^4 = 1$

$(a + bi)(a - bi) = a^2 + b^2$

$\overline{a + bi} = a - bi$

Beyond the Real Numbers

The real numbers cannot solve $x^2 = -1$ , because no real number squares to a negative. Mathematicians resolved this by defining a new number, the imaginary unit $i$ , with the single defining property:

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

A complex number has the form $a + bi$ , where $a$ is the real part and $b$ is the imaginary part. On the ACT, complex-number questions test whether you can simplify powers of $i$ , add and subtract complex numbers, multiply them, and rationalize a fraction using the conjugate. None of it requires deep theory — it requires treating $i$ like a variable while remembering the one rule that $i^2 = -1$ .

The Powers of $i$ Cycle

Because $i^2 = -1$ , higher powers of $i$ repeat in a cycle of length four:

$i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1$

After $i^4 = 1$ the pattern restarts: $i^5 = i$ , $i^6 = -1$ , and so on. To simplify any power $i^n$ , divide the exponent by 4 and keep only the remainder. For example, $i^{27}$ : since $27 = 4 \cdot 6 + 3$ , the remainder is $3$ , so $i^{27} = i^3 = -i$ .

Adding and Subtracting

Combine complex numbers by matching like parts — real with real, imaginary with imaginary — exactly as you would combine like terms:

$(3 + 5i) + (2 - 8i) = (3 + 2) + (5 - 8)i = 5 - 3i$

Multiplying

Multiply complex numbers with FOIL, just like binomials, then replace $i^2$ with $-1$ :

(2 + 3i)(4 - i) &= 8 - 2i + 12i - 3i^2 \\ &= 8 + 10i - 3(-1) \\ &= 11 + 10i \end{aligned}$$ The crucial final move is turning the $i^2$ term into a real number, which shifts value into the real part. ## Conjugates and Division The **conjugate** of $a + bi$ is $a - bi$ — same real part, opposite imaginary sign. Conjugates are powerful because multiplying a complex number by its conjugate always produces a **real** number: $$(a + bi)(a - bi) = a^2 - (bi)^2 = a^2 + b^2$$ That fact lets you divide complex numbers: to simplify $\dfrac{1}{a + bi}$ or any complex fraction, multiply the top and bottom by the conjugate of the denominator, which clears $i$ out of the denominator. $$\frac{3 + i}{2 - i} \cdot \frac{2 + i}{2 + i} = \frac{(3 + i)(2 + i)}{2^2 + 1^2} = \frac{6 + 3i + 2i + i^2}{5} = \frac{5 + 5i}{5} = 1 + i$$ ## Quick Reference | Task | Move | |---|---| | Simplify $i^n$ | Remainder of $n \div 4$ picks $i^0, i^1, i^2, i^3$ | | Add / subtract | Combine real parts and imaginary parts separately | | Multiply | FOIL, then replace $i^2$ with $-1$ | | Divide | Multiply top and bottom by the conjugate of the denominator | | Conjugate of $a + bi$ | $a - bi$ (flip the sign of the imaginary part) | ## A Reliable Game Plan - **Treat $i$ like a variable** through every operation, but the instant you see $i^2$, replace it with $-1$. - **For powers,** reduce the exponent by 4 and read off the cycle $i, -1, -i, 1$. - **For a quotient,** multiply by the conjugate of the denominator to make it real. - **Report the answer as $a + bi$** — a single real number, then the imaginary term. Complex numbers reward one habit above all: never leave an $i^2$ (or a higher power) unsimplified. Convert it immediately and the arithmetic falls into place.

Compute $(4 + 7i) + (2 - 3i)$ .

Compute $(3 + 2i)(1 - 4i)$ .

Compute $(2 - 3i)^2$ .

Write $\dfrac{4 + 2i}{1 - i}$ in the form $a + bi$ .

Simplify $i^{23} + (2 + i)(3 - 2i)$ and write the result as $a + bi$ .

Treating $i^2$ as $+1$ or leaving it in the answer.

By definition $i^2 = -1$ . Substitute it immediately every time it appears; never leave $i^2$ in a final answer.

Reducing a power of $i$ with the quotient instead of the remainder — e.g. saying $i^{10} = i^{2.5}$ .

Divide the exponent by 4 and use only the whole-number REMAINDER (0, 1, 2, or 3) to pick from the cycle $1, i, -1, -i$ .

Taking the conjugate by flipping the sign of the real part instead of the imaginary part.

The conjugate of $a + bi$ is $a - bi$ : the real part stays, only the imaginary sign flips.

Adding real parts to imaginary parts, e.g. writing $(3 + 2i) + 4$ as $9i$ .

Real and imaginary parts never merge. Keep them in separate columns and report the sum as $a + bi$ .

What is the value of $i^{15}$ ?

What is $(5 - 2i) - (3 + 4i)$ ?

What is $(2 + 3i)(4 + i)$ written in the form $a + bi$ ?

What is $(3 + i)^2$ in the form $a + bi$ ?

When $\dfrac{2 + 4i}{1 + i}$ is written in the form $a + bi$ , what is it?

What is the value of $i^{18} + i^{33}$ ?

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

$i^1 = i,\; i^2 = -1,\; i^3 = -i,\; i^4 = 1$

$(a + bi)(a - bi) = a^2 + b^2$

$\overline{a + bi} = a - bi$

Beyond the Real Numbers

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

The Powers of $i$ Cycle

Because $i^2 = -1$ , higher powers of $i$ repeat in a cycle of length four:

$i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1$

Adding and Subtracting

Combine complex numbers by matching like parts — real with real, imaginary with imaginary — exactly as you would combine like terms:

$(3 + 5i) + (2 - 8i) = (3 + 2) + (5 - 8)i = 5 - 3i$

Multiplying

Multiply complex numbers with FOIL, just like binomials, then replace $i^2$ with $-1$ :

(2 + 3i)(4 - i) &= 8 - 2i + 12i - 3i^2 \\ &= 8 + 10i - 3(-1) \\ &= 11 + 10i \end{aligned}$$ The crucial final move is turning the $i^2$ term into a real number, which shifts value into the real part. ## Conjugates and Division The **conjugate** of $a + bi$ is $a - bi$ — same real part, opposite imaginary sign. Conjugates are powerful because multiplying a complex number by its conjugate always produces a **real** number: $$(a + bi)(a - bi) = a^2 - (bi)^2 = a^2 + b^2$$ That fact lets you divide complex numbers: to simplify $\dfrac{1}{a + bi}$ or any complex fraction, multiply the top and bottom by the conjugate of the denominator, which clears $i$ out of the denominator. $$\frac{3 + i}{2 - i} \cdot \frac{2 + i}{2 + i} = \frac{(3 + i)(2 + i)}{2^2 + 1^2} = \frac{6 + 3i + 2i + i^2}{5} = \frac{5 + 5i}{5} = 1 + i$$ ## Quick Reference | Task | Move | |---|---| | Simplify $i^n$ | Remainder of $n \div 4$ picks $i^0, i^1, i^2, i^3$ | | Add / subtract | Combine real parts and imaginary parts separately | | Multiply | FOIL, then replace $i^2$ with $-1$ | | Divide | Multiply top and bottom by the conjugate of the denominator | | Conjugate of $a + bi$ | $a - bi$ (flip the sign of the imaginary part) | ## A Reliable Game Plan - **Treat $i$ like a variable** through every operation, but the instant you see $i^2$, replace it with $-1$. - **For powers,** reduce the exponent by 4 and read off the cycle $i, -1, -i, 1$. - **For a quotient,** multiply by the conjugate of the denominator to make it real. - **Report the answer as $a + bi$** — a single real number, then the imaginary term. Complex numbers reward one habit above all: never leave an $i^2$ (or a higher power) unsimplified. Convert it immediately and the arithmetic falls into place.

Compute $(4 + 7i) + (2 - 3i)$ .

Compute $(3 + 2i)(1 - 4i)$ .

Compute $(2 - 3i)^2$ .

Write $\dfrac{4 + 2i}{1 - i}$ in the form $a + bi$ .

Simplify $i^{23} + (2 + i)(3 - 2i)$ and write the result as $a + bi$ .

Treating $i^2$ as $+1$ or leaving it in the answer.

By definition $i^2 = -1$ . Substitute it immediately every time it appears; never leave $i^2$ in a final answer.

Reducing a power of $i$ with the quotient instead of the remainder — e.g. saying $i^{10} = i^{2.5}$ .

Divide the exponent by 4 and use only the whole-number REMAINDER (0, 1, 2, or 3) to pick from the cycle $1, i, -1, -i$ .

Taking the conjugate by flipping the sign of the real part instead of the imaginary part.

The conjugate of $a + bi$ is $a - bi$ : the real part stays, only the imaginary sign flips.

Adding real parts to imaginary parts, e.g. writing $(3 + 2i) + 4$ as $9i$ .

Real and imaginary parts never merge. Keep them in separate columns and report the sum as $a + bi$ .

What is the value of $i^{15}$ ?

What is $(5 - 2i) - (3 + 4i)$ ?

What is $(2 + 3i)(4 + i)$ written in the form $a + bi$ ?

What is $(3 + i)^2$ in the form $a + bi$ ?

When $\dfrac{2 + 4i}{1 + i}$ is written in the form $a + bi$ , what is it?

What is the value of $i^{18} + i^{33}$ ?

Complex Numbers

Key Formulas

Video Lesson

Explanation

Beyond the Real Numbers

The Powers of iii Cycle

Adding and Subtracting

Multiplying

Worked Examples

Simplify a power of i

Add complex numbers

Multiply two complex numbers

Square a complex number

Divide using the conjugate

Combine a high power and arithmetic

Watch Out For

Practice Problems

Frequently Asked Questions

Keep Going

Quadratics and the Parabola

Powers and Roots

Matrices

Ready to test yourself?

Complex Numbers

Key Formulas

Video Lesson

Explanation

Beyond the Real Numbers

The Powers of iii Cycle

Adding and Subtracting

Multiplying

Worked Examples

Simplify a power of i

Add complex numbers

Multiply two complex numbers

Square a complex number

Divide using the conjugate

Combine a high power and arithmetic

Watch Out For

Practice Problems

Frequently Asked Questions

Keep Going

Quadratics and the Parabola

Powers and Roots

Matrices

Ready to test yourself?

The Powers of $i$ Cycle

The Powers of $i$ Cycle