Multiplying, then inverting: how division becomes possible

What to assume they know

• An equation with no real solution, and the symbol invented to solve it — the form a + ib and the relation i² = −1
• Adding and subtracting componentwise, and undoing an addition — addition, the additive inverse, and the way the two slots stay separate
• Expanding a product of two binomials and collecting like terms
• That a sum of two real squares is zero only when both squares are zero

What they should be able to do

• Derive the multiplication rule by expanding a product of two numbers in the form a + ib and substituting for i²
• Apply the rule to numerical cases and present the answer in the form a + ib
• Name the multiplicative identity of the system
• Write down the multiplicative inverse of a given non-zero complex number and verify it by multiplying back
• Explain why the denominator in the inverse formula can never be zero for a number that is not itself zero
• Define division as multiplication by an inverse, and carry out a quotient in that form
• Identify the exact condition under which a quotient is defined, and say why the condition printed alongside the inverse formula is stricter than the chapter's own exercises require

Where it usually goes wrong

• "(a + ib)(c + id) = ac + i(bd)." The commonest error in the chapter. Expand and there are four terms, not two; the two cross terms are what make the second slot, and the fourth term is what puts the minus sign in the first.
• "The product rule is arbitrary and must be memorised." It is one bracket expansion plus one substitution. A student who can expand (p + q)(r + s) can reconstruct it in ten seconds and never needs to recall it.
• "You cannot divide by a complex number." You can divide by any complex number except zero, and the reason is arithmetic on the reals: a²+b² is a positive real whenever a and b are not both zero.
• "A purely imaginary number has no inverse, because the printed condition asks for a ≠ 0." The chapter's own Q13 asks for the inverse of −i. The working condition is simply that the number is not zero.
• "Multiplying by the inverse is a different operation from dividing." In this chapter it is the definition of dividing. There is nothing else to do.
• "i² = −1 only applies to the symbol on its own." It applies wherever i² appears, including inside an expanded product — which is where the rule's minus sign comes from.

Questions to check understanding

• Express a product or a quotient of given complex numbers in the form a + ib
• Find the multiplicative inverse of a given number, including cases where one of the two parts is zero
• Verify a claimed inverse by multiplying back and reaching 1 + i0
• Simplify an expression combining a product with an addition, as in Exercise 4.1 item 4
• State the condition on the divisor for a quotient to be defined, and justify it
• Powers of a bracketed complex number, where repeated multiplication is faster than expanding directly

Examples worth working on the board

Items marked verified are worked out here from the chapter's stated data.

• The derivation to run (§4.3.3, p. 78). Start from the product of a + ib with c + id, expand into four terms — ac, i·ad, i·bc and i²·bd — and substitute −1 for i². Verified: the fourth term becomes −bd and joins the first slot, while the two middle terms combine into the second slot. That reproduces the printed rule, which places ac − bd in the first slot and ad + bc in the second.
• The printed product (§4.3.3, p. 78): (3 + i5)(2 + i6) gives −24 + i28. Verified: 3×2 − 5×6 = 6 − 30 = −24, and 3×6 + 5×2 = 18 + 10 = 28.
• The six properties (§4.3.3, p. 78), stated on the page without proof: closure, commutativity, associativity, the identity 1 + i0, the inverse for a non-zero number, and the distributive law in both its left and right forms.
• The inverse formula (§4.3.3, p. 78). For z = a + ib the inverse has a/(a²+b²) in the first slot and −b/(a²+b²) in the second. Verified by multiplying back with the product rule: the first slot of the product is a·a/(a²+b²) − b·(−b)/(a²+b²), which is (a²+b²)/(a²+b²) = 1; the second slot is a·(−b)/(a²+b²) + b·a/(a²+b²), which is 0. So the product is 1 + i0, the identity. This one-line check is the section's whole argument.
• The condition, and where it bites. The denominator a²+b² is a sum of two real squares, so it is never negative, and it is zero only when a and b are both zero. Verified: that is the case z = 0, the one number excluded.
• The chapter's stricter printed condition, and its own counter-item. The parenthesis beside the inverse formula on p. 78, repeated in the Summary on p. 87, asks for a ≠ 0 and b ≠ 0. Exercise 4.1 Q13 on p. 83 asks for the inverse of −i, whose first part is 0. Verified: −i has a = 0 and b = −1, so a²+b² = 1, and the formula returns 0 + i(1) = i; multiplying back, (−i)(i) = −i², which is 1. The inverse exists and the exercise expects it. Flag this as a place where the printed wording is narrower than the mathematics and than the chapter's own §4.3.4, which requires only that the divisor is not zero.
• Division defined (§4.3.4, p. 78). The quotient of z₁ by a non-zero z₂ is defined as z₁ multiplied by the inverse of z₂ — no separate division rule appears.
• The chapter's worked quotient (§4.3.4, pp. 78–79). Inputs: z₁ = 6 + 3i and z₂ = 2 − i. Verified: the inverse of 2 − i has 2/5 in the first slot and 1/5 in the second, so it is (2 + i)/5; multiplying, (6 + 3i)(2 + i) expands to 12 + 6i + 6i + 3i², which is 9 + 12i, and the quotient is (9 + 12i)/5, that is 9/5 + (12/5)i. The chapter carries the working to the same point.
• Exercise 4.1 items for this topic (pp. 82–83). Q1: (5i)(−3i/5). Q4: 3(7 + i7) + i(7 + i7). Q8: (1 − i)⁴. Q11: the inverse of 4 − 3i. Q12: the inverse of √5 + 3i — note the radical, which the plain text layer of this page loses. Q13: the inverse of −i. Verified: Q1 gives 3 + i0, since −3i² = 3. Q4 gives 14 + 28i, since 21 + 21i + 7i + 7i² = 21 − 7 + 28i. Q8 gives −4 + i0, since (1 − i)² = −2i and (−2i)² = −4. Q11 gives 4/25 + (3/25)i, since 16 + 9 = 25. Q12 gives √5/14 − (3/14)i, since 5 + 9 = 14. Q13 gives i. All of this is worked out here on the printed items; the chapter prints no answers on these pages.
• The conjugate shortcut, cross-referenced forward. Example 6(i) on p. 82 clears a denominator by multiplying above and below by a number that differs from the divisor only in the sign of its second slot. That device is §4.4's and belongs to Size and reflection: two quantities that turn algebra into geometry; here it should be shown as the same computation the inverse formula performs, arranged differently.

Figures to have open

• A four-cell expansion grid for (a + ib)(c + id), with the bottom-right cell carrying i²·bd and flipping sign as it is substituted. Standard schematic, and the single most valuable image in this topic; the chapter has no figure in §4.3.
• A slot-labelled panel for the inverse formula with a²+b² highlighted as the common denominator of both slots. Standard schematic.
• A two-column comparison of the condition as printed on p. 78 and the condition the mathematics needs. Standard schematic; the comparison is added here.
• No textbook figure is required for this topic.

Where this sits in the book

• NCERT Class XI Mathematics, Chapter 4, whose printed title is Complex Numbers and Quadratic Equations; §4.3.3 and §4.3.4, pp. 78–79.
• Exercise 4.1, items 1, 4, 8, 11, 12 and 13, pp. 82–83.
• Summary, p. 87, which restates the multiplication rule and the inverse formula, carrying the same narrow parenthetical condition as p. 78.
• Forward pointer inside the same chapter: §4.4 on p. 81 gives the conjugate and restates the inverse as the conjugate over the squared modulus; Example 5 on p. 82 works one inverse both ways.