🎯⭐ INTERACTIVE LESSON

Complex Numbers on the SAT

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Complex Numbers on the SAT - Complete Interactive Lesson

Part 1: Imaginary Unit

🔢 The Imaginary Unit $i$

Part 1 of 7 — Definition, Powers of $i$ , and the Cycle

The imaginary unit $i$ was invented to give meaning to the square root of negative numbers:

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

Expression	Value
$\sqrt{-4}$	$2i$
$\sqrt{- 9}$

General rule: $\sqrt{-a} = i\sqrt{a}$ for any positive number .

A complex number has the form $a + bi$ , where $a$ is the real part and $b$ is the imaginary part.

Powers of $i$ — The 4-Step Cycle

The powers of $i$ repeat every 4:

Power	Value	Why
$i^1$	$i$	Definition

Practice — Powers of $i$ 🔍

Simplifying Square Roots of Negatives

Always extract $i$ first, then simplify the radical:

Example 1: $\sqrt{-48}$ $\sqrt{- 48} = i \sqrt{48} = i \sqrt{16 \cdot 3} = 4$

Simplify these expressions. 🧮

Write answers using $i$ (e.g. type "5i" or "-3i" or "1").

$i^{36} =$
$i^{75} =$

Match each expression to its simplified value. 🔍

SAT-Style Questions 📋

Part 2: Complex Arithmetic

➕ Adding & Subtracting Complex Numbers

Part 2 of 7 — Combine Real with Real, Imaginary with Imaginary

Complex numbers have the form $a + bi$ . When adding or subtracting, treat $i$ like a variable — combine like terms.

(a + bi) + (c + di) = (a + c) + (b + d)i

Part 3: Complex Conjugates

✖️ Multiplying Complex Numbers

Part 3 of 7 — FOIL Method, $i^2 = -1$ , and Special Products

To multiply two complex numbers, use FOIL just like with binomials, then replace $i^2$ with $-1$ :

Part 4: Quadratics & Complex Roots

➗ Complex Conjugates & Division

Part 4 of 7 — Conjugates, Rationalizing, Division

The complex conjugate of $a + bi$ is $a - bi$ . The product of a number and its conjugate is always real:

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

Part 5: Powers of i

🔍 Solving Equations with Complex Solutions

Part 5 of 7 — Negative Discriminants, $x^2 + k = 0$ , Quadratic Formula

Not every quadratic equation has real solutions. When the discriminant $b^2-4ac < 0$ , the solutions are .

Part 6: Problem-Solving Workshop

🎯 Complex Numbers on the SAT

Part 6 of 7 — Pattern Recognition, Common Traps, $i^n$ Cycles

Complex number questions appear in the Calculator and No Calculator sections. They test three main skills:

Skill	Frequency
Powers of $i$	★★★ Very common
Add/subtract/multiply	★★★ Very common
Division (conjugates)	★★ Common
Discriminant / non-real solutions	★★ Common
Writing in form

Part 7: Review & Applications

📋 Review & Mixed Practice

Part 7 of 7 — Cheat Sheet, Mixed Problems, Strategies

Complex Numbers Cheat Sheet

Concept	Formula / Rule
Imaginary unit	$i = \sqrt{-1},\; i^2 = -1$

Shortcut: To find $i^n$ , divide $n$ by $4$ and look at the remainder:

Remainder	$i^n$ equals
$0$	$1$
$1$	$i$
$2$	$-1$
$3$	$-i$

Example: $i^{23}$ . Divide $23 \div 4 = 5$ remainder $3$ . So $i^{23} = -i$ .

Example: $i^{100}$ . Divide $100 \div 4 = 25$ remainder $0$ . So $i^{100} = 1$ .

- 48 = i 48 = i 16 \cdot 3 = 4 i 3

Example 2: $\sqrt{-72}$ $\sqrt{-72} = i\sqrt{72} = i\sqrt{36 \cdot 2} = 6i\sqrt{2}$

Example 3: $3\sqrt{-50}$ $3\sqrt{-50} = 3 \cdot i\sqrt{50} = 3i\sqrt{25 \cdot 2} = 15i\sqrt{2}$

⚠️ Common mistake: $\sqrt{-4} \cdot \sqrt{-9} \neq \sqrt{36} = 6$ . You must convert to $i$ -form first: $(2i)(3i) = 6i^2 = -6$ .

(a + bi) - (c + di) = (a - c) + (b - d)i

Operation	Example	Result
Addition	$(3 + 2i) + (5 + 4i)$	$8 + 6i$
Subtraction	$(7 + 3i) - (2 + 5i)$	$5 - 2i$
Mixed	$(4 - i) + (-1 + 6i)$	$3 + 5i$

Worked Examples

Example 1: $(6 + 3i) + (-2 + 7i)$

Real parts: $6 + (-2) = 4$
Imaginary parts: $3i + 7i = 10i$
Result: $4 + 10i$

Example 2: $(9 - 4i) - (3 - 8i)$

Distribute the minus sign: $9 - 4i - 3 + 8i$
Real parts: $9 - 3 = 6$
Imaginary parts: $- 4 i + 8 i =$

Example 3: $(-5 + 2i) - (-5 + 2i)$

$-5 + 2i + 5 - 2i = 0 + 0i = 0$

⚠️ Watch the signs! The most common error is forgetting to distribute the negative sign when subtracting. $(a+bi)-(c+di) = a+bi-c-di$ , NOT .

Practice — Addition & Subtraction 🔍

Multi-Step Problems

Example 4: If $z_1 = 2 + 3i$ and $z_2 = -1 + 4i$ , find $2z_1 - z_2$ .

$2z_1 = 2(2+3i) = 4 + 6i$
$2 z_{}$

Example 5: Find the value of $a$ and $b$ such that $(a + bi) + (3 - 2i) = 7 + i$ .

$(a+3) + (b-2)i = 7 + i$
Real parts: $a+3=7 \Rightarrow a=4$

SAT Tip: Some problems ask for just the real part or just the imaginary part. Read the question carefully!

Compute each result. 🧮

Write answers in $a + bi$ form (e.g. "3 + 2i" or "-1 - 4i"). For real answers, just write the number.

$(4 + 9i) + (-4 + i) =$
$(10 - 6i) - (3 - 6i) =$
$3(2 + i) - (1 + 5i) =$

Identify the real and imaginary parts. 🔍

SAT-Style Questions 📋

(a+bi)(c+di) = ac + adi + bci + bdi^2

= ac + adi + bci + bd(-1)

= (ac - bd) + (ad + bc)i

(a+bi)(c+di) = (ac-bd) + (ad+bc)i

Don't memorize the formula — just FOIL and replace $i^2 = -1$ .

Worked Examples — FOIL

Example 1: $(3 + 2i)(4 + 5i)$

Step	Calculation
First	$3 \cdot 4 = 12$
Outer	$3 \cdot 5i = 15i$
Inner	$2i \cdot 4 = 8i$
Last	$2i \cdot 5i = 10i^2 = -10$

$12 + 15i + 8i - 10 = 2 + 23i$

Example 2: $(1 - 3i)(2 + i)$

$= 2 + i - 6i - 3i^2 = 2 - 5i - 3(-1) = 2 - 5i + 3 = 5 - 5i$

Example 3: $(4i)(3 - 2i)$

$= 12i - 8i^2 = 12i - 8(-1) = 8 + 12i$

Practice — Multiplication 🔍

Special Products

Conjugate pairs give real results — this is crucial for division (Part 4):

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

Example	Conjugate Pair	Product
$(3+4i)(3-4i)$	Yes	$9+16=25$

Squaring a complex number:

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

Example: $(3+i)^2 = 9 + 6i + i^2 = 9 + 6i - 1 = 8 + 6i$

Multiply and simplify. 🧮

Write answers in $a + bi$ form.

$(5 + i)(2 + 3i) =$
$(4 - 2i)^2 =$
$(6 + i)(6 - i) =$

Classify each product. 🔍

SAT-Style Questions 📋

Number	Conjugate	Product
$3 + 4i$	$3 - 4i$	$25$
$2 - i$	$2 + i$	$5$
$7i$	$-7i$	$49$

This property is the key to dividing complex numbers.

Division — Multiply by the Conjugate

To divide $\frac{a + bi}{c + di}$ , multiply top and bottom by the conjugate of the denominator:

\frac{a+bi}{c+di} = \frac{(a+bi)(c-di)}{(c+di)(c-di)} = \frac{(a+bi)(c-di)}{c^2+d^2}

Example 1: $\frac{3+i}{1-2i}$

$= \frac{(3+i)(1+2i)}{(1-2i)(1+2i)} = \frac{3+6i+i+2i^2}{1+4} = \frac{3+7i-2}{5} = \frac{1+7i}{5} = \frac{1}{5}+\frac{7}{5}i$

Example 2: $\frac{4}{2+i}$

$= \frac{4(2-i)}{(2+i)(2-i)} = \frac{8-4i}{4+1} = \frac{8-4i}{5} = \frac{8}{5}-\frac{4}{5}i$

Practice — Division 🔍

Step-by-Step Division Checklist

When dividing complex numbers on the SAT:

Identify the denominator (e.g., $c + di$ )
Write its conjugate ( $c - di$ )
Multiply numerator and denominator by the conjugate
FOIL the numerator
Use $i^2 = -1$ to simplify
Denominator becomes $c^2 + d^2$ (always a real number)
Split into real and imaginary parts if needed

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

$= \frac{(2-3i)(4-i)}{(4+i)(4-i)} = \frac{8-2i-12i+3i^2}{16+1} = \frac{8-14i-3}{17} = \frac{5-14i}{17}$

$= \frac{5}{17} - \frac{14}{17}i$

Divide and simplify. 🧮

Give answers as simplified fractions or whole numbers.

$\frac{10}{1+3i}$ — what is the real part?
$\frac{10}{1+3i}$ — what is the coefficient of $i$ ?
$(3+4i)(3-4i) =$

Match each expression to the correct conjugate. 🔍

SAT-Style Questions 📋

x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}

Discriminant $\Delta = b^2-4ac$	Solutions
$\Delta > 0$	Two real solutions
$\Delta = 0$	One repeated real solution
$\Delta < 0$	Two complex conjugate solutions

Key fact: Complex solutions always come in conjugate pairs: if $a+bi$ is a solution, then $a-bi$ is also a solution.

Simple Cases: $x^2 + k = 0$

Example 1: Solve $x^2 + 4 = 0$ .

$x^2 = -4 \Rightarrow x = \pm\sqrt{-4} = \pm 2i$

Example 2: Solve $x^2 + 9 = 0$ .

$x^2 = -9 \Rightarrow x = \pm 3i$

Example 3: Solve $2x^2 + 50 = 0$ .

$2x^2 = -50 \Rightarrow x^2 = -25 \Rightarrow x = \pm 5i$

Pattern: $x^2 + k = 0$ (with $k > 0$ ) always gives $x = \pm i\sqrt{k}$ .

Practice — Simple Equations 🔍

Using the Quadratic Formula

Example: Solve $x^2 - 4x + 13 = 0$ .

Here $a=1$ , $b=-4$ , $c=13$ .

$\Delta = (-4)^2 - 4(1)(13) = 16 - 52 = -36$

$x = \frac{4 \pm \sqrt{-36}}{2} = \frac{4 \pm 6i}{2} = 2 \pm 3i$

Solutions: $x = 2 + 3i$ and $x = 2 - 3i$ (conjugate pair!).

Example: Solve $x^2 + 2x + 5 = 0$ .

$\Delta = 4 - 20 = -16$

$x = \frac{-2 \pm \sqrt{-16}}{2} = \frac{-2 \pm 4i}{2} = -1 \pm 2i$

SAT Tip: If a problem says "non-real solutions," it means the discriminant is negative.

Solve each equation. 🧮

$x^2 + 36 = 0$ — the positive imaginary solution is $x =$
$x^2 - 6x + 13 = 0$ — the discriminant is $\Delta =$
For the equation in (2), the solutions are $x = 3 \pm \_\_ i$ . Fill in the blank.

Classify each equation's solutions. 🔍

SAT-Style Questions 📋

\frac{a + bi}{c + d i}

Key pattern: Most SAT complex-number problems take 30–60 seconds if you know the rules. They reward memorization of the $i$ -cycle and fluency with FOIL.

Common SAT Traps

Trap 1: Forgetting $i^2 = -1$ when multiplying

Wrong: $(2+i)(3+i) = 6 + 2i + 3i + i^2 = 6 + 5i + i^2 = ?$

Students who leave $i^2$ or write $+1$ instead of $-1$ get the wrong answer.

Correct: $6 + 5i + (-1) = 5 + 5i$

Trap 2: Sign errors when subtracting

Wrong: $(4+3i)-(2+5i) = 4+3i-2+5i = 2+8i$ ❌

Correct: $(4+3i)-(2+5i) = 4+3i-2-5i = 2-2i$ ✅

Trap 3: Confusing $r^2$ in the discriminant

When $\Delta < 0$ , students sometimes forget to put $\sqrt{|\Delta|}$ over $2a$ .

$x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-b \pm i\sqrt{|\Delta|}}{2a}$

Trap-Spotting Practice 🔍

The Cycle Sum Trick

One complete cycle of consecutive powers of $i$ sums to $0$ :

$i^1 + i^2 + i^3 + i^4 = i + (-1) + (-i) + 1 = 0$

This means:

$i + i^2 + i^3 + \cdots + i^{100} = 0$ (25 complete cycles)

Shortcut for sums: Divide the number of terms by 4. Complete groups of 4 cancel to $0$ . Then add up the remaining terms.

Example: $i + i^2 + \cdots + i^{42}$

42 terms: $42 \div 4 = 10$ complete groups (40 terms) + 2 remaining.

Remaining: $i^{41} + i^{42} = i + (-1) = i - 1 = -1 + i$ .

Quick calculations. 🧮

$(2i)^3 =$
$i + i^2 + i^3 + i^4 + i^5 =$
$\frac{1}{i} =$ (Write in the form $ai$ or $-ai$ )

Identify the correct approach for each SAT problem type. 🔍

SAT-Style Questions 📋

SAT Test-Day Strategies

1. Time management: Complex number problems are usually quick (30–60 sec). Don't skip them!

2. Powers of $i$ : Always divide the exponent by 4 and use the remainder. This takes 5 seconds.

3. Multiplication: FOIL and replace $i^2 = -1$ . Double-check the sign on the last term.

4. Division: Multiply top and bottom by the conjugate. The denominator becomes $a^2+b^2$ .

5. "Which is equivalent to…": These problems usually test multiplication or division. Just compute carefully.

6. Discriminant questions: If they ask about "non-real" or "no real" solutions, compute $b^2-4ac$ and check if it's negative.

7. Verify with conjugate pairs: If one complex solution is $a+bi$ , the other is $a-bi$ . You can check by adding them ( $=2a$ ) or multiplying them ( $=a^2+b^2$ ).

Mixed Review — Set 1 🔍

Mixed computations. 🧮

$(3+i)(1-2i) + (2+3i) =$
The discriminant of $x^2 + 4x + 8 = 0$ is $\Delta =$
$\frac{(2+i)^2}{i} =$ (Write in $a+bi$ form)

Final review — true or false? 🔍

Final SAT-Style Questions 📋

🎉 You've Completed SAT Complex Numbers!

What you've mastered:

✅ The imaginary unit $i$ and its power cycle

✅ Adding, subtracting, multiplying complex numbers

✅ Complex conjugates and division

✅ Solving equations with complex solutions

✅ SAT-specific patterns and traps

Key takeaways for test day:

Powers of $i$ : divide exponent by 4, use remainder
Multiply: FOIL + replace $i^2 = -1$
Divide: multiply by the conjugate
Complex solutions: discriminant $< 0$
Solutions come in conjugate pairs

Keep practicing — complex numbers are some of the easiest points on the SAT once you know the rules!

Result:

6 + 4i

2 z_{1} - z_{2} = (4 + 6 i) - (- 1 + 4 i) = 4 + 6 i + 1 - 4 i = 5 + 2 i

Imaginary parts:

b-2=1 \Rightarrow b=3

16+18−2i−12i+3i2=

i + i^2 + i^3 + \cdots + i^{99} = i + i^2 + i^3 = i - 1 - i = -1

(24 complete cycles + 3 extra)

Complex Numbers on the SAT

Complex Numbers on the SAT - Complete Interactive Lesson

Part 1: Imaginary Unit

🔢 The Imaginary Unit iii

Powers of iii — The 4-Step Cycle

Simplifying Square Roots of Negatives

Part 2: Complex Arithmetic

➕ Adding & Subtracting Complex Numbers

Part 3: Complex Conjugates

✖️ Multiplying Complex Numbers

Part 4: Quadratics & Complex Roots

➗ Complex Conjugates & Division

Part 5: Powers of i

🔍 Solving Equations with Complex Solutions

Part 6: Problem-Solving Workshop

🎯 Complex Numbers on the SAT

Part 7: Review & Applications

📋 Review & Mixed Practice

Complex Numbers Cheat Sheet

Worked Examples

Multi-Step Problems

Worked Examples — FOIL

Special Products

Division — Multiply by the Conjugate

Step-by-Step Division Checklist

Simple Cases: x2+k=0x^2 + k = 0x2+k=0

Using the Quadratic Formula

Common SAT Traps

The Cycle Sum Trick

SAT Test-Day Strategies

🎉 You've Completed SAT Complex Numbers!

🔢 The Imaginary Unit $i$

Powers of $i$ — The 4-Step Cycle

Simple Cases: $x^2 + k = 0$