Complex Numbers and Operations

Complex Numbers in Secondary Certification Math

Not every state high school course spends weeks on complex numbers, but Praxis 5165 includes them in the official Number & Quantity and Algebra outline. Expect straightforward arithmetic, simplification of i^n, and occasional context linking to quadratic equations with negative discriminants.

A complex number has the form a + bi, where a and b are real and i^2 = −1. The real part is a; the imaginary part is b (some texts write bi as the imaginary part—ETS stems usually make the intended part clear).

Powers of i

i cycles every four powers:

Worked Example 1

Simplify i^23. Since 23 = 4·5 + 3, i^23 = i^3 = −i.

Worked Example 2

Simplify i^100 + i^101 + i^102 + i^103 = 1 + i + (−1) + (−i) = 0. Any block of four consecutive powers sums to zero.

Addition and Subtraction

Combine like parts: (3 + 2i) + (5 − 7i) = 8 − 5i.

Subtraction distributes the sign: (4 − i) − (−2 + 3i) = 4 − i + 2 − 3i = 6 − 4i.

Geometrically on the complex plane, addition is vector addition of components.

Multiplication

Use distribution and i^2 = −1.

Worked Example 3 — (2 + i)(3 − 4i)

(2)(3) + (2)(−4i) + (i)(3) + (i)(−4i) = 6 − 8i + 3i − 4i^2 = 6 − 5i + 4 = 10 − 5i

Worked Example 4 — (1 − 2i)^2

(1 − 2i)^2 = 1 − 4i + 4i^2 = 1 − 4i − 4 = −3 − 4i.

Watch the middle term: (a − bi)^2 is not a^2 − b^2.

Conjugates and Division

The complex conjugate of a + bi is a − bi. Product: (a + bi)(a − bi) = a^2 + b^2.

To divide, multiply numerator and denominator by the conjugate of the denominator.

Worked Example 5 — (1 + 2i) / (3 − i)

Multiply top and bottom by (3 + i):

Numerator: (1 + 2i)(3 + i) = 3 + i + 6i + 2i^2 = 1 + 7i.

Denominator: (3 − i)(3 + i) = 9 + 1 = 10.

Result: (1 + 7i)/10 or 1/10 + (7/10)i.

Worked Example 6 — 5 / (2i)

Multiply by i/i: 5i/(2i^2) = 5i/(−2) = −(5/2)i.

The Complex Plane (Brief)

Plot a + bi at (a, b). Modulus |a + bi| = sqrt(a^2 + b^2). Praxis rarely asks for argument/angle, but distance from the origin supports checking multiplication magnitude informally.

Quadratic Connection

When b^2 − 4ac < 0, real solutions do not exist, but complex conjugate solutions do. For x^2 + 4 = 0, x = ±2i. For x^2 − 2x + 5 = 0, quadratic formula gives x = 1 ± 2i.

Linking this to graphing (parabola not crossing the x-axis) is a common teaching-scenario angle.

Traps to Diagnose

• Treating i^2 as +1.
• Dropping the imaginary unit when combining terms (writing 3 + 2i + 5 = 8 instead of 8 + 2i).
• Forgetting to multiply both parts of a fraction by the conjugate when dividing.
• Writing (a + bi)(a + bi) = a^2 + b^2 without the middle term 2abi.

Worked Example 7 — Student check

A student claims (1 + i)^2 = 1 + i^2 = 0. The error is failing to expand the binomial: (1 + i)^2 = 1 + 2i + i^2 = 2i, not 0.

Exam Tips

Memorize the i-cycle table. For division, always rationalize with the conjugate. Expand to check equivalence when two answer choices look similar. Complex items are usually one-step or two-step—do not overcomplicate with polar form unless a stem explicitly requires it.

Combining Operations in One Expression

Worked Example 8

Simplify (1 + i)^3.

(1 + i)^2 = 2i, then (2i)(1 + i) = 2i + 2i^2 = −2 + 2i.

Order matters—cube after squaring, not by adding exponents.

Solving with Complex Solutions

For 2x^2 + 8 = 0, divide by 2: x^2 = −4, so x = ±2i. Check: 2(2i)^2 + 8 = 2(−4) + 8 = 0.

Teaching stems may show a student rejecting i as "not a real answer." The correct response is that the solution set in complex numbers is {2i, −2i}.

Modulus as Distance

|3 − 4i| = sqrt(9 + 16) = 5. Modulus answers "how far from the origin" and appears when checking division results—if |z| = 1, then 1/z equals the conjugate of z.

Standard Form Requirement

Always write a + bi with real part first. 3i − 7 should be rewritten −7 + 3i if the stem asks for standard form. Small formatting slips can make two equivalent answers look different in multiple choice.