Complex numbers and the unit i

A square that comes out negative

Your book starts with a question: is there a real number whose square is negative? Try x² − 1 = 0 first. It gives x² = 1, so x = 1 or x = −1, and its graph crosses the x-axis twice (Fig. 1.1).

Now try x² + 1 = 0. It needs x² = −1, but squaring a real number never gives a negative value. The graph never touches the x-axis (Fig. 1.2). So we need a new kind of number.

i² = −1. Every complex number is x + iy: a real part x and an imaginary part y, both real numbers.

Key terms

Imaginary unit (i)

The number whose square is −1, read 'iota'.

Complex number

A real part plus a real multiple of i.

Real part, imaginary part

The two real numbers that make up z. The imaginary part does not include i.

Pure imaginary number

A complex number with no real part. With y = 0 instead, the number is real.

Short questions with model answers

1. Q1. Example 1(iii) (p. 3): simplify i¹⁷.

   • i¹⁷ = i¹⁶ × i
   • = (i²)⁸ × i
   • = (−1)⁸ × i

   i¹⁷ = i, because (−1)⁸ = 1.

2. Q2. Example 1(iv) (p. 3): simplify i⁻²⁵.

   • i⁻²⁵ = 1 ÷ i²⁵ = 1 ÷ ((i²)¹² × i)
   • = 1 ÷ ((−1)¹² × i) = 1 ÷ i
   • = i ÷ i² = i ÷ (−1)

   i⁻²⁵ = −i

Common mistakes

• ✗ Im(5 − 7i) = −7i.

  ✓ The imaginary part is the real number −7, without i (p. 3).

• ✗ i² = 1, because a square is never negative.

  ✓ That rule is for real numbers. i is not real, and i² = −1 by definition.

• ✗ If y = 0, the number is pure imaginary.

  ✓ y = 0 leaves only x, a real number. Pure imaginary needs x = 0.

MCQs

1. 1. i² + i⁴ =

   1. (a) −1
   2. (b) 0
   3. (c) 1
   4. (d) 2
   Show answer

   (b) Review Q1(i) (p. 20): −1 + 1 = 0.

2. 2. x + iy is a pure imaginary number when:

   1. (a) y = 0
   2. (b) i = 0
   3. (c) x = 0, y = 0
   4. (d) x = 0
   Show answer

   (d) Review Q1(iv) (p. 20): no real part means x = 0.

3. 3. Im(5 − 7i) is:

   1. (a) −7
   2. (b) 5
   3. (c) −7i
   4. (d) 7
   Show answer

   (a) The imaginary part is the real coefficient of i (p. 3).

4. 4. i¹⁶ simplifies to (Exercise 1.1 Q1(ii)):

   1. (a) −1
   2. (b) i
   3. (c) 1
   4. (d) −i
   Show answer

   (c) (i²)⁸ = (−1)⁸ = 1.

5. 5. Which equation has no real solution?

   1. (a) x² − 1 = 0
   2. (b) x² + 1 = 0
   3. (c) x − 1 = 0
   4. (d) x² = 1
   Show answer

   (b) Its graph never crosses the x-axis (Fig. 1.2, p. 2).

Quick revision

• x² + 1 = 0 has no real root, so we use i, with i² = −1.
• Simplify a power of i by writing it with i² and using i² = −1.
• z = x + iy has Re(z) = x and Im(z) = y, both real.