Know the number i, write complex numbers in the form a + bi, and simplify powers of i

Learning goal: Know there is a complex number i such that i² = −1, know that every complex number can be written in the form a + bi with a and b real, write square roots of negative numbers in that form, and simplify any whole-number power of i.

Why a new number is needed. Inside the real numbers, the equation x² = −1 has no solution, because the square of a positive number is positive, the square of a negative number is positive, and 0² = 0. No real number squares to a negative. For centuries that was the end of the matter: an equation such as x² + 1 = 0 was simply "impossible." The complex numbers remove that limitation by adding one new number and then keeping the rules of addition and multiplication we already trust (commutative, associative, distributive); a few other rules, such as the product rule for square roots, keep only under their stated conditions. The new number is called i, and it is defined by one property: i² = −1. That is all we assume about it. Everything else in this unit follows from that one relation together with the ordinary commutative, associative and distributive properties of addition and multiplication.

i is a number, not a variable. Students sometimes treat i like the x in an equation, something waiting to be solved for. It is not. The letter i names one specific number, in the same way that π names one specific number. Its defining fact is i² = −1, so also (−i)² = (−1)²·i² = 1·(−1) = −1, which means −i is a second number whose square is −1. When we write √(−1) we mean i by convention, just as √4 means 2 rather than −2.

The form a + bi. A complex number is any number that can be written as a + bi, where a and b are real numbers. The number a is the real part and the number b is the imaginary part; note that the imaginary part is the real number b, not bi. Examples: 3 + 2i has real part 3 and imaginary part 2. The number −5i equals 0 + (−5)i, so its real part is 0 and its imaginary part is −5. The number 7 equals 7 + 0i, so every real number is also a complex number, with imaginary part 0. A complex number whose real part is 0 and whose imaginary part is not 0, such as −5i or i itself, is called pure imaginary. Two complex numbers a + bi and c + di are equal exactly when a = c and b = d: the real parts must match and the imaginary parts must match. This is the rule that later lets us compare and solve.

Square roots of negative numbers. Once i exists, every negative real number has a square root written with i. For a positive real number p, √(−p) = i√p, because (i√p)² = i²·(√p)² = (−1)·p = −p. So √(−9) = 3i, since (3i)² = 9i² = −9. And √(−12) = i√12 = 2i√3, since (2√3 i)² = 4·3·i² = −12. Always simplify the real square root as usual, then attach the i. A common misconception is to apply the product rule for square roots to two negative numbers: a student writes √(−4)·√(−9) = √36 = 6. That is wrong. The rule √a·√b = √(ab) is guaranteed only when at least one of a and b is nonnegative; when both are negative it fails, as it does here. Convert first: √(−4)·√(−9) = (2i)(3i) = 6i² = −6. The correct product is −6, not 6. The reliable habit is to rewrite every square root of a negative number in the form bi before doing any other operation.

Powers of i cycle with period four. Starting from i² = −1 and multiplying by i each time: i¹ = i; i² = −1; i³ = i²·i = −i; i⁴ = i³·i = −i·i = −i² = 1. Then i⁵ = i⁴·i = i, and the pattern repeats: i, −1, −i, 1, i, −1, −i, 1, … forever. Because i⁴ = 1, any power of i can be reduced by removing multiples of four from the exponent. To simplify iⁿ, divide n by 4 and keep the remainder r; then iⁿ = iʳ, where i⁰ = 1, i¹ = i, i² = −1 and i³ = −i.

A worked model inside the lesson. Simplify i²⁷. Divide 27 by 4: 27 = 4·6 + 3, so the remainder is 3 and i²⁷ = (i⁴)⁶·i³ = 1⁶·(−i) = −i. Check the reasoning, not just the answer: (i⁴)⁶ is 1 because i⁴ = 1, and multiplying by i³ = −i leaves −i. Now simplify i⁵⁰: 50 = 4·12 + 2, so i⁵⁰ = i² = −1. And i¹⁰¹: 101 = 4·25 + 1, so i¹⁰¹ = i¹ = i. The exponent 2026 gives 2026 = 4·506 + 2, so i²⁰²⁶ = i² = −1.

Writing an expression in the form a + bi. The expression 2 − √(−25) becomes 2 − 5i, because √(−25) = 5i. Here a = 2 and b = −5. The expression (√(−16) + 3)/2 becomes (4i + 3)/2 = 3/2 + 2i, with a = 3/2 and b = 2. Notice that dividing by a real number divides both the real and imaginary parts. The expression 4 + i³ becomes 4 − i, because i³ = −i, so a = 4 and b = −1.

Solving the simplest equations. The equation x² + 16 = 0 has no real solution, but in the complex numbers x² = −16 gives x = ±√(−16) = ±4i. Check: (4i)² = 16i² = −16, so (4i)² + 16 = 0, and likewise for −4i. The equation x² = −7 gives x = ±i√7. Both solutions are pure imaginary and both are written in the form a + bi with a = 0.

Keeping the vocabulary straight. "Complex" is the name of the whole set; it contains the real numbers and the pure imaginary numbers and everything of the form a + bi. "Imaginary" is not an insult and does not mean "not really there"; it is a historical label. Within this course, the word "real" means the imaginary part is 0, "pure imaginary" means the real part is 0 and the imaginary part is not, and "complex" covers all of them. When a question asks for "the form a + bi," it wants two real numbers named: the real part a and the imaginary part b.

Worked square root: Write √(−50) in the form bi. Since 50 = 25·2, √50 = 5√2, so √(−50) = i√50 = 5√2 i. Check by squaring: (5√2 i)² = 25·2·i² = 50·(−1) = −50, which is the number we started with. In the form a + bi, a = 0 and b = 5√2.

Worked product of two negative square roots: Compute √(−3)·√(−27). Rewrite each first: √(−3) = i√3 and √(−27) = i√27 = 3√3 i. Multiply: (i√3)(3√3 i) = 3·(√3·√3)·i² = 3·3·(−1) = −9. A student who combined the radicands first would write √81 = 9, which has the wrong sign, because the product rule for square roots fails when both radicands are negative.

Worked power of i: Simplify i⁷³ and i⁴⁰⁰. For 73, divide by 4: 73 = 4·18 + 1, remainder 1, so i⁷³ = i¹ = i. For 400, 400 = 4·100 + 0, remainder 0, so i⁴⁰⁰ = i⁰ = 1. Check the first with the cycle: i⁷² = (i⁴)¹⁸ = 1, and one more factor of i gives i.

Worked equation: Solve 3x² + 75 = 0. Subtract 75 and divide by 3: x² = −25, so x = ±√(−25) = ±5i. Check x = 5i: 3(5i)² + 75 = 3(25i²) + 75 = 3(−25) + 75 = −75 + 75 = 0. The solutions are 5i and −5i, each in the form a + bi with a = 0 and b = ±5.

Form-a + bi record: Make a three-column table headed expression, real part a, imaginary part b. Fill it for these expressions: 6 − 2i, −i, 11, √(−36), 4 + √(−8), (10 − √(−4))/2, i⁶, and 2i³ + 5. For each row, show the one or two steps that produced a and b, and mark the row as real, pure imaginary or neither. Then write a sentence explaining why 11 belongs in the table at all.

Independent performance: Write a short guide for a classmate titled "Three things to know about i." Part one states the defining relation i² = −1 and uses it to show that (−i)² is also −1. Part two shows, with your own example of two negative radicands, why √(−a)·√(−b) must be rewritten as (i√a)(i√b) before multiplying, and what wrong answer the shortcut produces. Part three explains the four-step cycle of powers of i and uses it to simplify a power with an exponent above 100 that you choose yourself. Success criteria: every claim is checked by squaring or multiplying back, the powers-of-i example shows the division by 4 and its remainder, and every final answer is in the form a + bi with a and b identified.