Complex Numbers

Operations with imaginary and complex numbers

Complex Numbers (SAT)

Imaginary Unit

Definition of ii

i=−1i = \sqrt{-1} i2=−1i^2 = -1

Powers of ii

Pattern repeats every 4:

i1=ii^1 = i i2=−1i^2 = -1 i3=i2⋅i=−1⋅i=−ii^3 = i^2 \cdot i = -1 \cdot i = -i i4=i2⋅i2=(−1)(−1)=1i^4 = i^2 \cdot i^2 = (-1)(-1) = 1 i5=i4⋅i=1⋅i=ii^5 = i^4 \cdot i = 1 \cdot i = i (pattern repeats)

To find ini^n: Divide nn by 4, use remainder

  • Remainder 0 → in=1i^n = 1
  • Remainder 1 → in=ii^n = i
  • Remainder 2 → in=−1i^n = -1
  • Remainder 3 → in=−ii^n = -i

Complex Numbers

Standard Form

a+bia + bi

Where:

  • aa = real part
  • bb = imaginary part (coefficient of ii)

Examples:

  • 3+4i3 + 4i → real: 3, imaginary: 4
  • 5−2i5 - 2i → real: 5, imaginary: -2
  • 7i7i → real: 0, imaginary: 7
  • −3-3 → real: -3, imaginary: 0

Operations with Complex Numbers

Addition and Subtraction

Combine real parts and imaginary parts separately

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

Example: (3+2i)+(1−4i)=4−2i(3 + 2i) + (1 - 4i) = 4 - 2i

(5+3i)−(2+i)=3+2i(5 + 3i) - (2 + i) = 3 + 2i

Multiplication

Use distributive property (FOIL)

Remember: i2=−1i^2 = -1

(a+bi)(c+di)=ac+adi+bci+bdi2(a + bi)(c + di) = ac + adi + bci + bdi^2 =ac+(ad+bc)i+bd(−1)= ac + (ad + bc)i + bd(-1) =(ac−bd)+(ad+bc)i= (ac - bd) + (ad + bc)i

Example: (2+3i)(1+4i)(2 + 3i)(1 + 4i) =2+8i+3i+12i2= 2 + 8i + 3i + 12i^2 =2+11i+12(−1)= 2 + 11i + 12(-1) =2+11i−12= 2 + 11i - 12 =−10+11i= -10 + 11i

Division

Multiply by conjugate to eliminate ii from denominator

Conjugate: Change sign of imaginary part

  • Conjugate of a+bia + bi is a−bia - bi
  • Conjugate of a−bia - bi is a+bia + bi

Key property: (a+bi)(a−bi)=a2+b2(a + bi)(a - bi) = a^2 + b^2 (real number!)

Example: 3+2i1−i\frac{3 + 2i}{1 - i}

Multiply by conjugate of denominator: 3+2i1−i⋅1+i1+i\frac{3 + 2i}{1 - i} \cdot \frac{1 + i}{1 + i}

=(3+2i)(1+i)(1−i)(1+i)= \frac{(3 + 2i)(1 + i)}{(1 - i)(1 + i)}

Numerator: 3+3i+2i+2i2=3+5i−2=1+5i3 + 3i + 2i + 2i^2 = 3 + 5i - 2 = 1 + 5i

Denominator: 1−i2=1−(−1)=21 - i^2 = 1 - (-1) = 2

=1+5i2=12+52i= \frac{1 + 5i}{2} = \frac{1}{2} + \frac{5}{2}i

Simplifying Square Roots of Negatives

−a=ia\sqrt{-a} = i\sqrt{a}

Examples: −9=i9=3i\sqrt{-9} = i\sqrt{9} = 3i −16=4i\sqrt{-16} = 4i −5=i5\sqrt{-5} = i\sqrt{5}

Conjugate Pairs

Property: If a+bia + bi is a solution to polynomial equation with real coefficients, then a−bia - bi is also a solution

Example: If 3+2i3 + 2i is a solution, then 3−2i3 - 2i must also be a solution

SAT Complex Number Questions

Type 1: Simplify Powers of ii

Find i27i^{27}

Divide: 27÷4=627 ÷ 4 = 6 remainder 33 Answer: i3=−ii^3 = -i

Type 2: Add/Subtract Complex Numbers

(5+3i)−(2−i)=?(5 + 3i) - (2 - i) = ?

Combine like terms: 3+4i3 + 4i

Type 3: Multiply Complex Numbers

(2+i)(3−i)=?(2 + i)(3 - i) = ?

FOIL and simplify

Type 4: Simplify −n\sqrt{-n}

−25=?\sqrt{-25} = ?

Answer: 5i5i

SAT Strategies

Remember i2=−1i^2 = -1

Always substitute when you see i2i^2

Pattern of Powers

i,−1,−i,1i, -1, -i, 1 repeats

Combine Like Terms

Real with real, imaginary with imaginary

Rationalize Denominators

Multiply by conjugate

Common SAT Traps

Trap 1: Forgetting i2=−1i^2 = -1

3i2=3(−1)=−33i^2 = 3(-1) = -3 (not 3i23i^2 left alone!)

Trap 2: Sign Errors with Conjugates

Conjugate of 3−4i3 - 4i is 3+4i3 + 4i (flip the sign!)

Trap 3: Not Simplifying Completely

(1+i)2=1+2i+i2=1+2i−1=2i(1 + i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i (Don't stop at 1+2i+i21 + 2i + i^2!)

Trap 4: Dividing Incorrectly

Must multiply by conjugate, not just cross-multiply

SAT Tips

  • i2=−1i^2 = -1 (most important rule!)
  • Powers of ii repeat: i,−1,−i,1,i,−1,−i,1...i, -1, -i, 1, i, -1, -i, 1...
  • Add/subtract: Combine real and imaginary parts separately
  • Multiply: FOIL and replace i2i^2 with −1-1
  • Divide: Multiply by conjugate to eliminate ii from denominator
  • −a=ia\sqrt{-a} = i\sqrt{a}
  • Conjugate: Flip the sign of imaginary part

📚 Practice Problems

1Problem 1easy

❓ Question:

What is i18i^{18}?

💡 Show Solution

Solution:

Powers of ii repeat every 4: i,−1,−i,1i, -1, -i, 1

Divide exponent by 4: 18÷4=4 remainder 218 ÷ 4 = 4 \text{ remainder } 2

Remainder 2 → i2=−1i^2 = -1

Answer: −1-1

SAT Tip: Find remainder when dividing by 4:

  • Remainder 0 → 1
  • Remainder 1 → ii
  • Remainder 2 → −1-1
  • Remainder 3 → −i-i

2Problem 2medium

❓ Question:

Simplify: (3+2i)+(1−5i)(3 + 2i) + (1 - 5i)

💡 Show Solution

Solution:

Combine real parts: 3+1=43 + 1 = 4 Combine imaginary parts: 2i−5i=−3i2i - 5i = -3i

Answer: 4−3i4 - 3i

SAT Tip: Treat real and imaginary parts separately, like combining like terms!

3Problem 3hard

❓ Question:

Simplify: (2+i)(3−2i)(2 + i)(3 - 2i)

💡 Show Solution

Solution:

FOIL: F:2×3=6F: 2 \times 3 = 6 O:2×(−2i)=−4iO: 2 \times (-2i) = -4i I:i×3=3iI: i \times 3 = 3i L:i×(−2i)=−2i2L: i \times (-2i) = -2i^2

Combine: 6−4i+3i−2i26 - 4i + 3i - 2i^2

Substitute i2=−1i^2 = -1: 6−i−2(−1)6 - i - 2(-1) =6−i+2= 6 - i + 2 =8−i= 8 - i

Answer: 8−i8 - i

SAT Tip: Don't forget to replace i2i^2 with −1-1 at the end!

🎴

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