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Topos Uranos
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The complex numbers: operations and equations

Summary

This lesson constructs the complex numbers as ordered pairs of real numbers, with an addition and a multiplication defined by explicit formulas, so that the equality i2=−1i^2 = -1 ceases to be a postulate and becomes a theorem. It introduces the rectangular form, the complex plane, the powers of the imaginary unit, the conjugate and the modulus, with their properties proved, among them the triangle inequality, and division by means of the conjugate. With these tools it solves the quadratic equations with negative discriminant, proves that the non-real roots of a polynomial with real coefficients occur in conjugate pairs, and solves equations such as z3=1z^3 = 1 or those that combine zz with its conjugate.

Learning objectives

  1. Construct the complex numbers as pairs of real numbers, deduce from the definition of the product that i2=−1i^2 = -1, and operate in rectangular form.
  2. Represent the complex numbers in the plane and interpret geometrically the sum, the conjugate, the modulus and the product by ii.
  3. Prove the properties of the conjugate and of the modulus, among them the triangle inequality, and divide by multiplying by the conjugate.
  4. Solve in C\mathbb{C} quadratic equations, simple polynomial equations and linear equations in zz and in its conjugate.

The complex numbers as pairs of real numbers

In the lesson on number systems, the imaginary unit ii was introduced as a new number whose square is −1-1, and the expressions a+bia + bi were handled by the same rules as the real numbers. This way of proceeding leaves two questions open: whether an object with that property really exists, and whether the rules of the real numbers can be applied to it without contradiction. The answer, which Hamilton published in 1837, consists in constructing the complex numbers from the real numbers, without postulating anything new: a complex number is an ordered pair of real numbers.

Definition (complex number). A complex number is an ordered pair (a,b)(a, b) of real numbers; their set is denoted by C\mathbb{C}. Two complex numbers are equal when their first components coincide and their second components coincide. The sum and the product are defined by the following formulas.
(a,b)+(c,d)=(a+c,b+d),(a,b)⋅(c,d)=(ac−bd,ad+bc)(a, b) + (c, d) = (a + c, b + d), \qquad (a, b) \cdot (c, d) = (ac - bd, ad + bc)

The sum operates component by component. The product, on the other hand, is not the component-by-component product, which would give (ac,bd)(ac, bd), and its form seems arbitrary; it is justified by its consequences. With the component-by-component product, the product of (1,0)(1, 0) by (0,1)(0, 1) would be (0,0)(0, 0) without either factor being zero, and there could be no division; with the product as defined, by contrast, the pair (0,1)(0, 1) has as its square the pair (−1,0)(-1, 0), and the pairs of the form (a,0)(a, 0) behave exactly like the real numbers.

The real numbers inside the complex numbers

The pairs whose second component is zero are added and multiplied like the real numbers:

(a,0)+(c,0)=(a+c,0),(a,0)⋅(c,0)=(ac,0)(a, 0) + (c, 0) = (a + c, 0), \qquad (a, 0) \cdot (c, 0) = (ac, 0)

Therefore the pair (a,0)(a, 0) can be identified with the real number aa without any operation changing its result, and R\mathbb{R} is contained in C\mathbb{C}, as the principle of extensions stated in the lesson on number systems requires. Multiplying a pair by a real number, moreover, multiplies both its components, because (r,0)⋅(a,b)=(ra,rb)(r, 0) \cdot (a, b) = (ra, rb). The pair i=(0,1)i = (0, 1) is called the imaginary unit.

TheoremThe imaginary unit and the rectangular form

The square of i=(0,1)i = (0, 1) is the real number −1-1, and every complex number (a,b)(a, b) can be written in exactly one way as a+bia + bi, with aa and bb real. The proof uses only the definition of the product and the identification of each real number aa with the pair (a,0)(a, 0). The figure represents each pair (a,b)(a, b) as the point of the plane with coordinates aa and bb, an idea developed in the following section.

Proof

  1. i2=(0,1)⋅(0,1)=(0⋅0−1⋅1,0⋅1+1⋅0)=(−1,0)i^2 = \resaltar{(0, 1) \cdot (0, 1)} = (0 \cdot 0 - 1 \cdot 1, 0 \cdot 1 + 1 \cdot 0) = \resaltar{(-1, 0)}

    We apply the definition of the product with a=c=0a = c = 0 and b=d=1b = d = 1.

  2. i2=−1\resaltar{i^2} = \resaltar{-1}

    The pair (−1,0)(-1, 0) is the real number −1-1: the imaginary unit is a square root of −1-1. The other one is its opposite, −i=(0,−1)-i = (0, -1), because (−i)2=i2(-i)^2 = i^2.

  3. b⋅i=(b,0)⋅(0,1)=(b⋅0−0⋅1,b⋅1+0⋅0)=(0,b)b \cdot i = \resaltar{(b, 0) \cdot (0, 1)} = (b \cdot 0 - 0 \cdot 1, b \cdot 1 + 0 \cdot 0) = \dato{b}{\resaltar{(0, b)}}

    A real number bb multiplied by ii gives the pair with first component zero and second component bb.

  4. (a,b)=(a,0)+(0,b)=a+bi(a, b) = \resaltar{(a, 0) + (0, b)} = \dato{f}{\resaltar{a + bi}}

    Every pair is the sum of a pair on the first axis and another on the second.

  5. a+bi=c+di↔(a=c∧b=d)a + bi = c + di \leftrightarrow \resaltar{(a = c \wedge b = d)}

    The representation is unique: if a+bi=c+dia + bi = c + di, the pairs (a,b)(a, b) and (c,d)(c, d) are equal, and by the definition of equality their components coincide.

The expression a+bia + bi is the rectangular form of the complex number z=(a,b)z = (a, b). The real number aa is the real part of zz, and the real number bb, its imaginary part; note that the imaginary part is the real number bb, and not the product bibi. The complex numbers with zero imaginary part are the real numbers, and those with zero real part, such as 3i3i, are called purely imaginary. By the last assertion of the theorem, an equality between complex numbers is equivalent to two equalities between real numbers; it follows that every complex equation can be converted into a system of two real equations, the main tool of the last section.

The properties of the operations

The rectangular form is useful only if one can operate with it as with the algebraic expressions of the real numbers, expanding products and replacing i2i^2 by −1-1. This requires the operations of C\mathbb{C} to have the same properties as those of R\mathbb{R}.

TheoremThe properties of the operations

The addition and the multiplication of complex numbers are commutative and associative, multiplication is distributive over addition, 0=(0,0)0 = (0, 0) and 1=(1,0)1 = (1, 0) are the identity elements of addition and of multiplication, and every complex number (a,b)(a, b) has as its opposite (−a,−b)(-a, -b). The properties of addition are inherited from those of the real numbers, component by component. Those of multiplication are checked by expanding both sides with the properties of the real numbers. The existence of the multiplicative inverse is proved later, with the conjugate. The figure follows the proof with (a,b)=2+i(a, b) = 2 + i, (c,d)=1+i(c, d) = 1 + i and (e,f)=1−i(e, f) = 1 - i, which are called zz, ww and uu there: the three products involved in associativity and in distributivity lead to the same point, 4+2i4 + 2i.

Proof

  1. (c,d)⋅(a,b)=(ca−db,cb+da)=(a,b)⋅(c,d)(c, d) \cdot (a, b) = (\resaltar{ca - db}, \resaltar{cb + da}) = (a, b) \cdot (c, d)

    The commutativity of the product can be read off the formula: interchanging (a,b)(a, b) with (c,d)(c, d) alters no component, because ca=acca = ac, db=bddb = bd and cb+da=ad+bccb + da = ad + bc.

  2. [(a,b)⋅(c,d)]⋅(e,f)=(ac−bd,ad+bc)⋅(e,f)=(ace−bde−adf−bcf,acf−bdf+ade+bce)\left[ (a, b) \cdot (c, d) \right] \cdot (e, f) = \resaltar{(ac - bd, ad + bc)} \cdot (e, f) = \dato{p}{(ace - bde - adf - bcf, acf - bdf + ade + bce)}

    For associativity, we first multiply the first two factors and expand.

  3. (a,b)⋅[(c,d)⋅(e,f)]=(a,b)⋅(ce−df,cf+de)=(ace−adf−bcf−bde,acf+ade+bce−bdf)(a, b) \cdot \left[ (c, d) \cdot (e, f) \right] = (a, b) \cdot \resaltar{(ce - df, cf + de)} = \resaltar{(ace - adf - bcf - bde, acf + ade + bce - bdf)}

    We now first multiply the last two factors: the same eight terms result, in another order.

  4. (a,b)⋅(c+e,d+f)=(ac−bd+ae−bf,ad+bc+af+be)(a, b) \cdot (c + e, d + f) = (\resaltar{ac - bd} + \resaltar{ae - bf}, \resaltar{ad + bc} + \resaltar{af + be})

    For distributivity, we expand each component and regroup the terms: the result is the sum of (a,b)⋅(c,d)(a, b) \cdot (c, d) and (a,b)⋅(e,f)(a, b) \cdot (e, f).

  5. (a,b)⋅(1,0)=(a⋅1−b⋅0,a⋅0+b⋅1)=(a,b)(a, b) \cdot (1, 0) = (a \cdot 1 - b \cdot 0, a \cdot 0 + b \cdot 1) = \resaltar{(a, b)}

    Finally, (1,0)(1, 0) is the identity element of multiplication, and (−a,−b)(-a, -b) added to (a,b)(a, b) gives (0,0)(0, 0).

Consequently, one operates with rectangular forms as with algebraic expressions in the letter ii, and each i2i^2 is replaced by −1-1. In particular, the product recovers the formula that defined it:

(a+bi)(c+di)=ac+adi+bci+bdi2=(ac−bd)+(ad+bc)i(a + bi)(c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i

For example, (2+3i)(4−i)=8−2i+12i−3i2=11+10i(2 + 3i)(4 - i) = 8 - 2i + 12i - 3i^2 = 11 + 10i. The special products, such as the square of a binomial or the difference of squares, and the binomial theorem from the lesson on combinatorics hold in C\mathbb{C} without change, because their proofs use only these properties. What is not preserved is the order: by solved problem 5 of the lesson on number systems, there is no order on C\mathbb{C} compatible with the operations, so an inequality between non-real complex numbers is meaningless. The size of a complex number will be measured by its modulus.