Complex Numbers on the SAT

Perform operations with complex numbers.

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Complex Numbers on the SAT

What Is a Complex Number?

A complex number has the form: a+bia + bi

Where:

  • aa = real part
  • bb = imaginary part
  • i=−1i = \sqrt{-1} (the imaginary unit)

The Imaginary Unit ii

i=−1i = \sqrt{-1} i2=−1i^2 = -1 i3=i2⋅i=−ii^3 = i^2 \cdot i = -i i4=(i2)2=1i^4 = (i^2)^2 = 1

The powers of ii cycle every 4: i1=i,i2=−1,i3=−i,i4=1,i5=i,…i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1, \quad i^5 = i, \quad \ldots

To find ini^n: Divide nn by 4 and use the 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

Operations with Complex Numbers

Addition and Subtraction

Combine real parts and imaginary parts separately: (3+2i)+(5−4i)=8−2i(3 + 2i) + (5 - 4i) = 8 - 2i (3+2i)−(5−4i)=−2+6i(3 + 2i) - (5 - 4i) = -2 + 6i

Multiplication

Use FOIL and remember i2=−1i^2 = -1: (2+3i)(4−i)=8−2i+12i−3i2=8+10i−3(−1)=11+10i(2 + 3i)(4 - i) = 8 - 2i + 12i - 3i^2 = 8 + 10i - 3(-1) = 11 + 10i

Division (Conjugates)

Multiply numerator and denominator by the conjugate of the denominator:

3+2i1−i×1+i1+i=3+3i+2i+2i21−i2=3+5i−21+1=1+5i2=12+52i\frac{3 + 2i}{1 - i} \times \frac{1 + i}{1 + i} = \frac{3 + 3i + 2i + 2i^2}{1 - i^2} = \frac{3 + 5i - 2}{1 + 1} = \frac{1 + 5i}{2} = \frac{1}{2} + \frac{5}{2}i

Complex Conjugates

The 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 (always a real number!)

This is why we multiply by the conjugate to divide — it eliminates ii from the denominator.

Complex Numbers and Quadratics

When the discriminant b2−4ac<0b^2 - 4ac < 0, the quadratic has complex (non-real) solutions:

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Example: x2+4=0x^2 + 4 = 0 x2=−4x^2 = -4 x=±−4=±2ix = \pm \sqrt{-4} = \pm 2i

SAT Question Types

Type 1: Simplify Powers of ii

"What is i23i^{23}?" → 23÷4=523 \div 4 = 5 remainder 33 → i23=i3=−ii^{23} = i^3 = -i

Type 2: Add/Subtract Complex Numbers

Combine real and imaginary parts separately.

Type 3: Multiply Complex Numbers

Use FOIL, replace i2i^2 with −1-1.

Type 4: Divide Complex Numbers

Multiply by the conjugate.

Type 5: Solve Quadratics with Complex Solutions

Use the quadratic formula when discriminant < 0.

Common SAT Mistakes

  1. Forgetting that i2=−1i^2 = -1 (not −i-i or 11)
  2. Not multiplying by the conjugate when dividing
  3. Treating ii like a variable instead of replacing i2i^2 with −1-1
  4. Errors in the cycle of ii — remember the 4-cycle: i,−1,−i,1i, -1, -i, 1
  5. Panicking — complex number questions look harder than they are!

📚 Practice Problems

1Problem 1easy

❓ Question:

Simplify: (4+3i)+(2−7i)(4 + 3i) + (2 - 7i)

💡 Show Solution

Add real parts and imaginary parts separately:

Real: 4+2=64 + 2 = 6 Imaginary: 3i+(−7i)=−4i3i + (-7i) = -4i

Answer: 6−4i6 - 4i

2Problem 2easy

❓ Question:

Simplify: (4+3i)+(2−7i)(4 + 3i) + (2 - 7i)

💡 Show Solution

Add real parts and imaginary parts separately:

Real: 4+2=64 + 2 = 6 Imaginary: 3i+(−7i)=−4i3i + (-7i) = -4i

Answer: 6−4i6 - 4i

3Problem 3medium

❓ Question:

Simplify: (3+2i)(4−5i)(3 + 2i)(4 - 5i)

💡 Show Solution

Use FOIL:

(3)(4)+(3)(−5i)+(2i)(4)+(2i)(−5i)(3)(4) + (3)(-5i) + (2i)(4) + (2i)(-5i) =12−15i+8i−10i2= 12 - 15i + 8i - 10i^2

Replace i2i^2 with −1-1: =12−7i−10(−1)= 12 - 7i - 10(-1) =12−7i+10= 12 - 7i + 10 =22−7i= 22 - 7i

Answer: 22−7i22 - 7i

4Problem 4medium

❓ Question:

Simplify: (3+2i)(4−5i)(3 + 2i)(4 - 5i)

💡 Show Solution

Use FOIL:

(3)(4)+(3)(−5i)+(2i)(4)+(2i)(−5i)(3)(4) + (3)(-5i) + (2i)(4) + (2i)(-5i) =12−15i+8i−10i2= 12 - 15i + 8i - 10i^2

Replace i2i^2 with −1-1: =12−7i−10(−1)= 12 - 7i - 10(-1) =12−7i+10= 12 - 7i + 10 =22−7i= 22 - 7i

Answer: 22−7i22 - 7i

5Problem 5medium

❓ Question:

What is the value of i50i^{50}?

💡 Show Solution

The powers of ii cycle every 4: i,−1,−i,1,i,−1,…i, -1, -i, 1, i, -1, \ldots

Divide the exponent by 4: 50÷4=1250 \div 4 = 12 remainder 22

Use the remainder: i50=i2=−1i^{50} = i^2 = -1

Answer: −1-1

Quick reference:

  • Remainder 0 → 11
  • Remainder 1 → ii
  • Remainder 2 → −1-1
  • Remainder 3 → −i-i

6Problem 6medium

❓ Question:

What is the value of i50i^{50}?

💡 Show Solution

The powers of ii cycle every 4: i,−1,−i,1,i,−1,…i, -1, -i, 1, i, -1, \ldots

Divide the exponent by 4: 50÷4=1250 \div 4 = 12 remainder 22

Use the remainder: i50=i2=−1i^{50} = i^2 = -1

Answer: −1-1

Quick reference:

  • Remainder 0 → 11
  • Remainder 1 → ii
  • Remainder 2 → −1-1
  • Remainder 3 → −i-i

7Problem 7hard

❓ Question:

Write 5+3i2−i\frac{5 + 3i}{2 - i} in the form a+bia + bi.

💡 Show Solution

Step 1: Multiply by the conjugate of the denominator: 5+3i2−i×2+i2+i\frac{5 + 3i}{2 - i} \times \frac{2 + i}{2 + i}

Step 2: Multiply the numerator: (5+3i)(2+i)=10+5i+6i+3i2=10+11i+3(−1)=7+11i(5 + 3i)(2 + i) = 10 + 5i + 6i + 3i^2 = 10 + 11i + 3(-1) = 7 + 11i

Step 3: Multiply the denominator: (2−i)(2+i)=4−i2=4−(−1)=5(2 - i)(2 + i) = 4 - i^2 = 4 - (-1) = 5

Step 4: Divide: 7+11i5=75+115i\frac{7 + 11i}{5} = \frac{7}{5} + \frac{11}{5}i

Answer: 75+115i\frac{7}{5} + \frac{11}{5}i

8Problem 8hard

❓ Question:

Write 5+3i2−i\frac{5 + 3i}{2 - i} in the form a+bia + bi.

💡 Show Solution

Step 1: Multiply by the conjugate of the denominator: 5+3i2−i×2+i2+i\frac{5 + 3i}{2 - i} \times \frac{2 + i}{2 + i}

Step 2: Multiply the numerator: (5+3i)(2+i)=10+5i+6i+3i2=10+11i+3(−1)=7+11i(5 + 3i)(2 + i) = 10 + 5i + 6i + 3i^2 = 10 + 11i + 3(-1) = 7 + 11i

Step 3: Multiply the denominator: (2−i)(2+i)=4−i2=4−(−1)=5(2 - i)(2 + i) = 4 - i^2 = 4 - (-1) = 5

Step 4: Divide: 7+11i5=75+115i\frac{7 + 11i}{5} = \frac{7}{5} + \frac{11}{5}i

Answer: 75+115i\frac{7}{5} + \frac{11}{5}i

9Problem 9expert

❓ Question:

Find both solutions to x2+2x+5=0x^2 + 2x + 5 = 0.

💡 Show Solution

Step 1: Check the discriminant: b2−4ac=4−20=−16b^2 - 4ac = 4 - 20 = -16

Since Δ<0\Delta < 0, the solutions are complex.

Step 2: Apply the quadratic formula: x=−2±−162x = \frac{-2 \pm \sqrt{-16}}{2} =−2±4i2= \frac{-2 \pm 4i}{2} =−1±2i= -1 \pm 2i

Answer: x=−1+2ix = -1 + 2i and x=−1−2ix = -1 - 2i

Notice: The solutions are complex conjugates of each other. This is always the case for quadratics with real coefficients and complex roots.

10Problem 10expert

❓ Question:

Find both solutions to x2+2x+5=0x^2 + 2x + 5 = 0.

💡 Show Solution

Step 1: Check the discriminant: b2−4ac=4−20=−16b^2 - 4ac = 4 - 20 = -16

Since Δ<0\Delta < 0, the solutions are complex.

Step 2: Apply the quadratic formula: x=−2±−162x = \frac{-2 \pm \sqrt{-16}}{2} =−2±4i2= \frac{-2 \pm 4i}{2} =−1±2i= -1 \pm 2i

Answer: x=−1+2ix = -1 + 2i and x=−1−2ix = -1 - 2i

Notice: The solutions are complex conjugates of each other. This is always the case for quadratics with real coefficients and complex roots.

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