The Imaginary Unit $i$

$i = \sqrt{-1}$ $i^2 = -1$ $i^3 = i^2 \cdot i = -i$ $i^4 = (i^2)^2 = 1$

The powers of $i$ cycle every 4: $i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1, \quad i^5 = i, \quad \ldots$

To find $i^n$: Divide $n$ by 4 and use the remainder.

• Remainder 0: $i^n = 1$
• Remainder 1: $i^n = i$
• Remainder 2: $i^n = -1$
• Remainder 3: $i^n = -i$

Operations with Complex Numbers

Addition and Subtraction

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

Multiplication

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

Division (Conjugates)

Multiply numerator and denominator by the conjugate of the denominator:

$\frac{3 + 2i}{1 - i} \times \frac{1 + i}{1 + i} = \frac{3 + 3i + 2i + 2i^2}{1 - i^2} = \frac{3 + 5i - 2}{1 + 1} = \frac{1 + 5i}{2} = \frac{1}{2} + \frac{5}{2}i$

Complex Conjugates

The conjugate of $a + bi$ is $a - bi$.

Key property: $(a + bi)(a - bi) = a^2 + b^2$ (always a real number!)

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

Complex Numbers and Quadratics

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

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

Example: $x^2 + 4 = 0$ $x^2 = -4$ $x = \pm \sqrt{-4} = \pm 2i$

SAT Question Types

Type 1: Simplify Powers of $i$

"What is $i^{23}$?" → $23 \div 4 = 5$ remainder $3$ → $i^{23} = i^3 = -i$

Type 2: Add/Subtract Complex Numbers

Combine real and imaginary parts separately.

Type 3: Multiply Complex Numbers

Use FOIL, replace $i^2$ with $-1$.

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

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

Remainder 2 → $i^2 = -1$

Answer: $-1$

SAT Tip: Find remainder when dividing by 4:

• Remainder 0 → 1
• Remainder 1 → $i$
• Remainder 2 → $-1$
• Remainder 3 → $-i$

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

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

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

Step 4: Divide: $\frac{7 + 11i}{5} = \frac{7}{5} + \frac{11}{5}i$

Answer: $\frac{7}{5} + \frac{11}{5}i$

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

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

Step 4: Divide: $\frac{7 + 11i}{5} = \frac{7}{5} + \frac{11}{5}i$

Answer: $\frac{7}{5} + \frac{11}{5}i$

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

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

Step 4: Divide: $\frac{7 + 11i}{5} = \frac{7}{5} + \frac{11}{5}i$

Answer: $\frac{7}{5} + \frac{11}{5}i$

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

Answer: $x = -1 + 2i$ and $x = -1 - 2i$

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

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

Answer: $x = -1 + 2i$ and $x = -1 - 2i$

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

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

Answer: $x = -1 + 2i$ and $x = -1 - 2i$

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