Complex Numbers on the ACT: i, Powers, Operations, Conjugates, and Equations

Complex numbers look advanced, but the ACT usually tests a small set of repeatable rules.

The foundation is:

$$\boxed{i=\sqrt{-1}}$$

which means:

$$\boxed{i^2=-1}$$

A complex number has the form:

$$a+bi$$

where:

• (a) is the real part;
• (b) is the coefficient of the imaginary part.

ACT's current enhanced Math framework explicitly lists Complex Numbers within Number & Quantity and describes the skill as performing operations and solving equations involving complex numbers.

So this is an official current ACT topic, not an obsolete legacy detail.

Why imaginary numbers exist

In the real numbers:

$$x^2=-1$$

has no solution.

Complex numbers extend the number system by defining:

$$i^2=-1$$

Therefore:

$$x=\pm i$$

solves:

$$x^2=-1.$$

You do not need a deep theory of complex analysis for the ACT. You need the algebra.

Simplifying square roots of negative numbers

Use:

$$\sqrt{-a}=isqrt a$$

for positive (a).

Example:

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

Example:

$$\sqrt{-20} = isqrt{20} = 2isqrt5$$

Do not leave:

$$\sqrt{-20}$$

if the question expects standard complex form.

Important distinction from ordinary square roots

$$\sqrt{25}=5$$

The radical symbol gives the principal square root.

But:

$$x^2=25$$

has:

$$x=pm5$$

Likewise:

$$\sqrt{-25}=5i$$

while:

$$x^2=-25$$

has:

$$x=pm5i$$

The (pm) comes from solving the equation, not from the radical symbol itself.

Powers of i cycle every four

Start:

$$i^1=i$$ $$i^2=-1$$ $$i^3=-i$$ $$i^4=1$$

Then repeat:

$$i^5=i$$ $$i^6=-1$$

and so on.

Cycle:

$$\boxed{i, -1, -i, 1}$$

Reduce large powers using remainder mod 4

Example:

$$i^{37}$$

Divide 37 by 4:

$$37=4(9)+1$$

Remainder:

1

Therefore:

$$i^{37}=i^1=i$$

Example:

$$i^{58}$$ $$58=4(14)+2$$

So:

$$i^{58}=i^2=-1$$

This is much faster than multiplying (i) 58 times.

What if the exponent is divisible by 4?

Example:

$$i^{100}$$

Remainder:

0

So use:

$$i^4=1$$

Therefore:

$$i^{100}=1$$

A remainder of 0 corresponds to the fourth position in the cycle.

Adding complex numbers

Add real parts together and imaginary parts together.

$$(3+2i)+(5-7i)$$

Real:

$$3+5=8$$

Imaginary:

$$2i-7i=-5i$$

So:

$$\boxed{8-5i}$$

This is just combining like terms.

Subtracting complex numbers

Distribute the subtraction to the entire second expression.

$$(4+3i)-(2-5i)$$ $$4+3i-2+5i$$ $$\boxed{2+8i}$$

A common trap is failing to change both signs inside the second parentheses.

Multiplying complex numbers

Use ordinary distribution/FOIL, then replace:

$$i^2=-1$$

Example:

$$(2+3i)(4-i)$$

Expand:

$$8-2i+12i-3i^2$$

Since:

$$i^2=-1$$

then:

$$-3i^2=3$$

So:

$$8+10i+3 = \boxed{11+10i}$$

The most common multiplication mistake

Students expand correctly but forget to simplify:

$$i^2$$

to:

$$-1.$$

Never leave $i^2$ in final standard form.

Squaring a complex binomial

$$(3+2i)^2$$

Use:

$$(a+b)^2=a^2+2ab+b^2$$ $$9+12i+4i^2$$ $$9+12i-4$$ $$\boxed{5+12i}$$

Do not incorrectly square term-by-term as:

$$9+4i^2$$

and forget the middle term.

Conjugates

The conjugate of:

$$a+bi$$

is:

$$a-bi$$

Examples:

$$3+4i \rightarrow 3-4i$$ $$7-2i \rightarrow 7+2i$$

Why are conjugates useful?

Because:

$$(a+bi)(a-bi)$$

is real.

Product of conjugates

$$(a+bi)(a-bi)$$

Difference of squares:

$$a^2-(bi)^2$$

Since:

$$i^2=-1$$ $$a^2+b^2$$

Example:

$$(3+4i)(3-4i) = 9+16 = \boxed{25}$$

This pattern is worth recognizing immediately.

Dividing complex numbers

To divide:

$$\frac{a+bi}{c+di}$$

multiply numerator and denominator by the conjugate of the denominator:

$$c-di$$

Example:

$$\frac{1+i}{2+i}$$

Multiply by:

$$\frac{2-i}{2-i}$$

Numerator:

$$(1+i)(2-i)$$ $$2-i+2i-i^2$$ $$3+i$$

Denominator:

$$(2+i)(2-i) = 4+1 = 5$$

So:

$$\boxed{\frac35+\frac15i}$$

or:

$$\boxed{\frac{3+i}{5}}$$

depending on answer format.

Why the conjugate works

The denominator becomes:

$$c^2+d^2$$

which is real.

The goal is the same as rationalizing a radical denominator:

remove the awkward form from the denominator.

Solving simple equations with i

Example:

$$x+3i=5-2i$$

Subtract (3i):

$$x=5-5i$$

If (x) is allowed to be complex, that is the solution.

But ACT questions may instead ask you to solve for real and imaginary coefficients separately.

Equality of complex numbers

If:

$$a+bi=c+di$$

then:

$$a=c$$

and:

$$b=d.$$

Example:

$$(2x+1)+(y-3)i=7+5i$$

Match real parts:

$$2x+1=7$$ $$x=3$$

Match imaginary coefficients:

$$y-3=5$$ $$y=8$$

So:

$$\boxed{(x,y)=(3,8)}$$

Quadratics with complex roots

Example:

$$x^2+9=0$$ $$x^2=-9$$ $$x=pm3i$$

These roots occur because the discriminant is negative.

For:

$$ax^2+bx+c=0$$

discriminant:

$$D=b^2-4ac$$

If:

$$D<0$$

the quadratic has two nonreal complex roots.

See ACT Quadratics.

Quadratic formula with complex roots

Solve:

$$x^2+2x+5=0$$ $$x=\frac{-2pmsqrt{4-20}}2$$ $$x=\frac{-2pmsqrt{-16}}2$$ $$x=\frac{-2pm4i}2$$ $$\boxed{x=-1pm2i}$$

Complex roots come in conjugate pairs for real-coefficient polynomials

If a polynomial with real coefficients has:

$$3+2i$$

as a root, then:

$$3-2i$$

is also a root.

This can help on harder ACT polynomial questions.

Build a quadratic from complex roots

Roots:

$$2+i,quad2-i$$

Factors:

$$(x-(2+i))(x-(2-i))$$

Rewrite:

$$((x-2)-i)((x-2)+i)$$

Conjugate product:

$$(x-2)^2+1$$ $$x^2-4x+5$$

So a monic quadratic with those roots is:

$$\boxed{x^2-4x+5}$$

Real and imaginary parts

For:

$$z=7-3i$$

real part:

$$\operatorname{Re}(z)=7$$

imaginary part is commonly described as:

$$\operatorname{Im}(z)=-3$$

The imaginary component is:

(-3i).

Be careful about how a question phrases it.

Complex number standard form

Final answers are usually expressed as:

$$\boxed{a+bi}$$

Examples:

• (4+3i)
• (-2-i)
• (7) is also complex with (b=0)
• (5i) is complex with (a=0)

Every real number is also a complex number.

Common ACT complex-number traps

Trap 1: i²=1

Wrong.

$$i^2=-1$$

Trap 2: power cycle forgotten

Reduce exponent modulo 4.

Trap 3: subtraction sign not distributed

$$(a+bi)-(c+di)$$

requires subtracting both terms.

Trap 4: leaving i² unsimplified

Always replace with (-1).

Trap 5: adding real and imaginary terms together

$$3+2i$$

cannot become:

(5i) or 5.

Trap 6: forgetting conjugate in division

Multiply by denominator's conjugate.

Trap 7: ± placed on a radical itself

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

while solving:

$$x^2=-9$$

gives:

$$x=pm3i.$$

Worked example 1: large power

$$i^{73}$$ $$73=4(18)+1$$ $$\boxed{i}$$

Worked example 2: addition

$$(6-4i)+(-2+7i)$$ $$\boxed{4+3i}$$

Worked example 3: multiplication

$$(1+2i)(3+i)$$ $$3+i+6i+2i^2$$ $$3+7i-2$$ $$\boxed{1+7i}$$

Worked example 4: conjugates

$$(5+2i)(5-2i)$$ $$25+4$$ $$\boxed{29}$$

Worked example 5: negative square root

$$\sqrt{-72}$$ $$isqrt{72} = 6isqrt2$$ $$\boxed{6isqrt2}$$

Worked example 6: match components

$$(3x-2)+(2y+1)i=10+9i$$

Real:

$$3x-2=10$$ $$x=4$$

Imaginary:

$$2y+1=9$$ $$y=4$$ $$\boxed{x=y=4}$$

Powers of i: fastest memory pattern

Write:

$$i, -1, -i, 1$$

Then cycle.

Or remember:

• remainder 1 → (i)
• remainder 2 → (-1)
• remainder 3 → (-i)
• remainder 0 → (1)

This is one of the most efficient ACT Math patterns to memorize.

Calculator caution

Some calculators can evaluate complex-number expressions.

That can be useful.

But if you do not understand the form:

$$a+bi$$

or the $i^2=-1$ rule, calculator syntax can create more problems than it solves.

For powers of (i), mod-4 reasoning is often faster than calculator entry.

Current ACT Math blueprint

The enhanced ACT Math section has 45 questions in 50 minutes.

ACT's 2026 design framework explicitly lists Complex Numbers under Number & Quantity and describes the skill as:

• performing operations;
• solving equations involving complex numbers.

Number & Quantity currently accounts for roughly 10–12% of Math reporting content and also includes exponents, vectors, matrices, and real-number reasoning.

That makes complex numbers a small but legitimate current ACT skill worth learning efficiently.

Practice set

1

$$i^{18}$$

Remainder 2:

$$\boxed{-1}$$

2

$$(2+5i)+(3-2i)$$ $$\boxed{5+3i}$$

3

$$(2+i)(2-i)$$ $$4+1 = \boxed{5}$$

4

$$\sqrt{-49}=7i$$

5

$$x^2+16=0$$ $$\boxed{x=pm4i}$$

6

If:

$$a+bi=8-3i$$

then:

$$\boxed{a=8, b=-3}$$

How to review complex-number mistakes

Bad note:

imaginary-number mistake.

Better:

I forgot that the power-of-i cycle repeats every four.

Or:

I multiplied correctly but failed to replace $i^2$ with (-1).

Or:

I wrote ± in the value of the radical instead of only when solving the equation.

Solve equations by separating real and imaginary parts

Suppose:

$$(2x-1)+(3y+4)i=7-2i$$

Two complex numbers are equal only if their real parts match and their imaginary coefficients match.

Real:

$$2x-1=7$$ $$x=4$$

Imaginary:

$$3y+4=-2$$ $$3y=-6$$ $$y=-2$$

So:

$$\boxed{(x,y)=(4,-2)}$$

This turns one complex equation into two ordinary real equations.

Use conjugates strategically, not ceremonially

If a denominator is already real, you do not need a conjugate.

Example:

$$\frac{3+4i}{5}$$

is already acceptable.

But:

$$\frac{3+4i}{2-i}$$

has a complex denominator.

Multiply by:

$$\frac{2+i}{2+i}$$

because $2+i$ is the conjugate of $2-i$.

The goal is not "always use conjugates."

The goal is:

make the denominator real when division requires it.

Recognize quadratic structure before using the formula

Example:

$$x^2+6x+13=0$$

Complete the square:

$$x^2+6x+9=-4$$ $$(x+3)^2=-4$$ $$x+3=\pm2i$$ $$\boxed{x=-3\pm2i}$$

The quadratic formula also works, but completing the square is cleaner here.

Method choice matters even in complex-number questions.

Negative discriminant tells you the form before calculation

For:

$$ax^2+bx+c=0$$

if:

$$b^2-4ac<0$$

you know before simplifying that the roots will contain $i$.

Example:

$$x^2-4x+8=0$$

Discriminant:

$$16-32=-16$$

So the roots are nonreal.

Then:

$$x=\frac{4\pm4i}{2}$$ $$\boxed{x=2\pm2i}$$

This lets you reject real-only answer choices immediately.

Conjugate roots can save a full calculation

If a real-coefficient quadratic has one root:

$$5-3i$$

the other nonreal root must be:

$$5+3i$$

That is often faster than rebuilding the polynomial or using the quadratic formula.

Conversely, if a proposed pair is:

$$5-3i,\quad4+3i$$

they are not conjugates, so they cannot be the only two roots of a real-coefficient quadratic.

Sum and product of conjugate roots

For roots:

$$a+bi,\quad a-bi$$

sum:

$$2a$$

product:

$$a^2+b^2$$

Both are real.

That explains why multiplying the corresponding factors produces a polynomial with real coefficients.

Do not confuse the imaginary coefficient with the imaginary term

For:

$$z=4-7i$$

the coefficient of $i$ is:

$$-7$$

The imaginary term is:

$$-7i$$

If a question asks for $b$ in:

$$a+bi$$

the answer is the coefficient, not the entire term.

Complex numbers and exponent/radical skills connect

To simplify:

$$\sqrt{-48}$$

first separate the negative:

$$i\sqrt{48}$$

then simplify the positive radical:

$$i(4\sqrt3)$$

so:

$$\boxed{4i\sqrt3}$$

The complex-number step is small; most of the work is ordinary radical simplification.

See ACT Exponents and Radicals.

A compact method-selection guide

If you see:

Large power of i

Use remainder mod 4.

Addition/subtraction

Combine real with real and imaginary with imaginary.

Product

Distribute, then replace $i^2$ with $-1$.

Division

Use a conjugate only if the denominator is complex.

Quadratic with negative discriminant

Simplify the square root of the negative and write $a+bi$ form.

Equality of two complex expressions

Match real parts and imaginary coefficients separately.

FAQ: complex numbers on the ACT

Are complex numbers still on the enhanced ACT?

Yes. ACT's current 2026 design framework explicitly includes complex numbers in Number & Quantity.

What is the most important rule?

$$i^2=-1$$

How do I simplify large powers of i?

Reduce the exponent modulo 4.

What is a conjugate?

For (a+bi), the conjugate is (a-bi).

Why use conjugates?

They turn products such as ((a+bi)(a-bi)) into real numbers and help simplify complex division.

Can quadratics have complex answers on ACT Math?

Yes. A negative discriminant produces nonreal complex roots.

Related MastaPrep guides

• ACT Algebra
• ACT Exponents and Radicals
• ACT Quadratics
• ACT Math Formulas
• Enhanced ACT 2026–27 Guide

Final takeaway

Complex numbers are a compact ACT topic.

Master these patterns:

$$\boxed{i^2=-1}$$ $$\boxed{i, -1, -i, 1\text{ repeats}}$$ $$\boxed{(a+bi)+(c+di)=(a+c)+(b+d)i}$$ $$\boxed{(a+bi)(a-bi)=a^2+b^2}$$

Then connect complex numbers to negative square roots and quadratic equations.

That covers most of the ACT value without turning the topic into a college-level course.

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Official ACT references:
https://www.act.org/content/act/en/products-and-services/the-act/test-preparation/description-of-math-test.html
https://www.act.org/content/act/en/products-and-services/the-act/test-preparation/act-exam-sections-and-structure.html
https://www.act.org/content/dam/act/unsecured/documents/R2519-Design-Framework-for-the-ACT-Enhancements-2026-02.pdf

Trademark note: ACT® is a registered trademark of ACT Education Corp. MastaPrep is independent and is not affiliated with or endorsed by ACT Education Corp.