Size and reflection: two quantities that turn algebra into geometry

What to assume they know

• Multiplying, then inverting: how division becomes possible — multiplication, the multiplicative inverse, and division defined as multiplication by an inverse
• Which school algebra identities survive the enlargement, and why — the difference-of-two-squares identity for complex numbers
• Splitting a number into two parts, and when two such numbers agree — real part and imaginary part
• That a sum of two real squares is zero only when both terms are zero, and is positive otherwise

What they should be able to do

• Compute the modulus of a given complex number and state that the value is a non-negative real number
• Write down the conjugate of a given complex number, including cases where one of the two parts is zero
• Show that a number multiplied by its conjugate gives the square of the modulus, and identify where each cancellation happens
• Recover the inverse formula of §4.3.3 by dividing the conjugate by the squared modulus
• Turn a quotient into standard form by multiplying above and below by the conjugate of the divisor
• State the five results the chapter lists for moduli and conjugates of products and quotients, including the conditions attached to two of them
• Use the multiplicative property of the modulus to shortcut a problem that would otherwise need full expansion

Where it usually goes wrong

• "The modulus of a complex number is a complex number." It is always a real number, and never negative. That is what makes it usable as a size.
• "The conjugate is the additive inverse." The additive inverse flips both slots; the conjugate flips only the second. Confusing them is the single most common slip in this section.
• "Conjugating changes the size." It cannot: squaring −b gives the same value as squaring b, so a number and its conjugate have the same modulus.
• "A number times its conjugate is complex, like any other product." It is real, because the two cross terms are equal and opposite. This is the identity the whole section rests on.
• "Rationalising the denominator is a trick with no justification." It is multiplication by a fraction whose numerator and denominator are the same conjugate, so it multiplies by 1 and changes nothing except the shape.
• "The modulus could be zero for some non-zero number." A sum of two real squares is zero only when both parts are zero. This is exactly the guarantee that makes the inverse formula of §4.3.3 safe.
• "The chapter proves the five listed results." It states them and says they can be derived. A student should know which claims in this section come with an argument and which do not.

Questions to check understanding

• Find the modulus of a given complex number, including quotients that must be simplified first
• Find the conjugate of a product or a quotient of given complex numbers
• Find a multiplicative inverse using the conjugate over the squared modulus
• Express a quotient in standard form by clearing the denominator
• Prove that a stated sum of two squares equals a given expression, using the squared-modulus identity
• Items where recognising a number times its conjugate saves an entire expansion, as in Miscellaneous item 7
• Find real unknowns from a condition involving a conjugate, combining this topic with the matching step of §4.2

Examples worth working on the board

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

• The two definitions (§4.4, p. 81). For a number with parts a and b, the modulus is the square root of a² + b², taken as the non-negative root, and the conjugate has a in the first slot and −b in the second.
• The chapter's printed moduli (§4.4, p. 81): for 3 + i the value is the square root of 10, and for 2 − 5i it is the square root of 29. Verified: 9 + 1 = 10, and 4 + 25 = 29 — note that the minus sign in the second number makes no difference, because it is squared.
• The chapter's printed conjugates (§4.4, p. 81): 3 + i pairs with 3 − i; 2 − 5i pairs with 2 + 5i; and −5 − 3i pairs with −5 + 3i, which the page writes with the terms in the other order. Verified by applying the definition: only the sign attached to i changes.
• The central identity (§4.4, p. 81). A number times its conjugate equals the squared modulus. Verified by expanding with the multiplication rule of §4.3.3: the first slot receives a·a − b·(−b), which is a² + b²; the second slot receives a·(−b) + b·a, which is 0. The two cross terms cancel because they are the same product with opposite signs, and the i² term flips from −b² to +b². Run this expansion in full — it is the topic.
• The inverse rewritten (§4.4, p. 81). Dividing the conjugate by the squared modulus gives the inverse. Verified: this is the same object as the formula printed in §4.3.3 on p. 78, with a²+b² recognised as the squared modulus and a − ib recognised as the conjugate.
• Example 5 (p. 82). Input: the multiplicative inverse of 2 − 3i, worked two ways on the page. Verified: the conjugate is 2 + 3i and the squared modulus is 4 + 9 = 13, so the inverse is 2/13 + (3/13)i. The second route on the page multiplies above and below by 2 + 3i and gets there through a difference of squares; both routes are the same identity.
• Example 6, first part (p. 82). Input exactly as printed: the quotient of (5 + √2·i) by (1 − √2·i). Verified: multiply above and below by 1 + √2·i. The denominator becomes 1 − (√2·i)², which is 1 + 2 = 3. The numerator expands to 5 + 5√2·i + √2·i + 2i², which is 3 + 6√2·i. Dividing by 3 gives 1 + 2√2·i. The chapter reaches the same value.
• Example 7 (p. 85). Inputs: the conjugate is wanted of a quotient whose numerator is the product of (3 − 2i) with (2 + 3i) and whose denominator is the product of (1 + 2i) with (2 − i). Verified: the numerator is 12 + 5i and the denominator is 4 + 3i; multiplying above and below by 4 − 3i turns the denominator into 25 and the numerator into 63 − 16i, so the quotient is 63/25 − (16/25)i and its conjugate is 63/25 + (16/25)i. The chapter reaches the same pair of values.
• The five stated results (§4.4, p. 81): the modulus of a product is the product of the moduli; the modulus of a quotient is the quotient of the moduli, provided the divisor is not zero; the conjugate of a product is the product of the conjugates; the conjugate of a sum or a difference splits the same way; and the conjugate of a quotient splits, again provided the divisor is not zero. The chapter says these can be derived and does not derive them.
• A numerical test to run. Verified: take 3 + i and 2 − 5i. Their product is 6 − 15i + 2i − 5i², which is 11 − 13i, whose squared modulus is 121 + 169 = 290; and 10 × 29 is also 290. The multiplicative property holds on the chapter's own two sample numbers. These two numbers are the chapter's; the test is added here.
• Miscellaneous items that this topic unlocks (p. 86), as inputs only. Item 5: with z₁ = 2 − i and z₂ = 1 + i, a modulus of a quotient is wanted. Verified: the numerator combination is 4 and the denominator combination is 2 − 2i, whose modulus is 2√2, so the value is √2. Item 7, second part: with z₁ = 2 − i, the imaginary part is wanted of the reciprocal of z₁ times its conjugate. Verified: that product is the squared modulus, which is 5, so the reciprocal is 1/5 and its imaginary part is 0 — a one-line answer for anyone who has this identity and a long one for anyone who has not. Item 9: the modulus is wanted of the difference of two reciprocal-looking fractions. Verified: each fraction reduces to i and −i respectively, the difference is 2i, and the modulus is 2. Item 13: a product of four brackets is given as A + iB, and the product of the four sums of squares is to be shown equal to A² + B². Verified as an application of the multiplicative property: each sum of squares is one squared modulus, and the squared modulus of the product is the product of them. Item 11: two different complex numbers are given with the modulus of one equal to 1, and a modulus of a quotient is wanted. Item 6 asks for a sum of two squares to be shown equal to a stated expression, which is the squared-modulus identity in disguise. Item 4 asks for the square of such a sum, so the identity has to be applied twice there — once for the square root the item starts from and once for the sum itself. Do not describe the two as the same task.

All values marked verified are worked out here on the printed items; the chapter prints no answers on these pages.

Figures to have open

• A four-cell expansion grid for a number times its conjugate, with the two cross cells struck out and the i² cell flipping sign. Standard schematic, and the central image of this topic; §4.4 has no figure.
• A right-angled triangle with legs of lengths given by the two parts and the modulus on the hypotenuse. Standard schematic, and worth drawing here even though the geometric reading formally arrives in §4.5 on p. 84 — it makes the sum of two squares mean something.
• A matched pair of panels putting the §4.3.3 inverse formula and its §4.4 rewriting side by side with connecting lines. Standard schematic.
• A conditions strip listing the five results with the two provisos attached. Standard schematic.

Where this sits in the book

• NCERT Class XI Mathematics, Chapter 4, whose printed title is Complex Numbers and Quadratic Equations; §4.4, pp. 81–82, for both definitions, the identity, the rewritten inverse and the five listed results.
• Example 5 and Example 6, p. 82.
• Example 7, p. 85, among the Miscellaneous Examples.
• Miscellaneous Exercise on Chapter 4, items 3 to 13, p. 86.
• Summary, p. 87, which restates the conjugate but not the modulus.
• Backward pointer inside the same chapter: the inverse formula this section rewrites is printed in §4.3.3 on p. 78.