1. The Square Root of Minus One!

If we want to calculate the square root of a negative number, it rapidly becomes clear that neither a positive or a negative number can do it.

For example, √−1 ≠ ±1, since 1² = (−1)² = +1.

To find √−1 we must introduce a new quantity, i, defined to be such that i² = −1. (Note that engineers often use the notation j.)

Example 1:

(a)

√−25	=	5i
Since (5i)²	=	5² × i²
	=	25 × (−1)
	=	−25

(b)

√− 16/9	=	4/3i
Since (4/3i)²	=	16/9 × (i)²
	=	−16/9

2. Real, Imaginary and Complex Numbers

Real numbers are the usual positive and negative numbers.

If we multiply a real number by i, we call the result an imaginary number. Examples of imaginary numbers are: i, 3i and −i⁄2. If we add or subtract a real number and an imaginary number, the result is a complex number. We write a complex number as:

z = a + ib

- where a and b are real numbers.

3. Adding and Subtracting Complex Numbers

If we want to add or subtract two complex numbers, z₁ = a + ib and z₂ = c + id, the rule is to add the real and imaginary parts separately:

z₁ + z₂	=	a + ib + c + id	=	a + c + i(b + d)
z₁ − z₂	=	a + ib − c − id	=	a − c + i(b − d)

Example 2:

(a)

(1 + i) + (3 + i)	=	1 + 3 + i(1 + 1)	=	4 + 2i

(b)

(2 + 5i) − (1 − 4i)	=	2 + 5i − 1 + 4i	=	1 + 9i

Exercise 1: Add or subtract the following complex numbers:

(a)(3 + 2i) + (3 + i)

Solution:

(3 + 2i) + (3 + i)	=	3 + 2i + 3 + i
	=	3 + 3 + 2i + 2i
	=	6 + 3i

(b)(4 − 2i) − (3 − 2i)

Solution: Here we need to be careful with the signs!

(4 − 2i) − (3 − 2i)	=	4 − 2i − 3 + 2i
	=	4 − 3 − 2i + 2i
	=	1

- a purely real result.

(c)(−1 + 3i) + ½(2 + 2i)

Solution: The factor of 1 ⁄ 2 multiplies both terms in the complex number:

(−1 + 3i) + ½(2 + 2i)	=	−1 + 3i + 1 + i
	=	4i

- a purely imaginary result.

(d) 1/3 (2 − 5i) − 1/6 (8 − 2i)

Solution:

1/3 (2 − 5i) − 1/6 (8 − 2i)	=	2/3 − 5/3 i − 8/6 + 2/6 i
	=	2/3 − 5/3 i − 4/3 + 1/3 i
	=	2/3 − 4/3 − 5/3 i + 1/3 i
	=	− 2/3 − 4/3 i

- which we may also write as −2/3(1 + 2i).

Quiz 1: To which of the following does the expression (4 − 3i) + (2 + 5i) simplify?

(a)6 − 8i Incorrect - please try again!

(b)6 + 2iCorrect - well done!

(d)9 − i Incorrect - please try again!

Quiz 2: To which of the following does the expression (3 − i) − (2 − 6i) simplify?

(a)3 − 9i Incorrect - please try again!

(b)1 − 7i Incorrect - please try again!

(d)1 − 5i Incorrect - please try again!

4. Multiplying Complex Numbers

We multiply two complex numbers just as we would multiply expressions of the form (x + y) together (see the module on Brackets):

(a + ib) (c + id)	=	ac + a(id) + (ib)c + (ib)(id)
	=	ac + iad + ibc − bd
	=	ac − bd + i(ad + bc)

Example 3:

(2 + 3i) (3 + 2i)	=	2 × 3 + 2 × 2i + 3i × 3 + 3i × 2i
	=	6 + 4i + 9i − 6
	=	13i

Exercise 2: Multiply the following complex numbers:

(a)(3 + 2i) (3 + i)

Solution:

(3 + 2i) (3 + i)	=	3 × 3 + 3 × i + 2i × 3 + 2i × i
	=	9 + 3i + 6i − 2
	=	9 − 2 + 3i + 6i
	=	7 + 9i

(b)(4 − 2i) (3 − 2i)

Solution:

(4 − 2i) (3 − 2i)	=	4 × 3 + 4 × (−2i) − 2i × 3 − 2i × (−2i)
	=	12 − 8i − 6i − 4
	=	12 − 4 −8i − 6i
	=	8 −14i

(c)(−1 + 3i) (2 + 2i)

Solution:

(−1 + 3i) (2 + 2i)	=	−1 × 2 − 1 × 2i + 3i × 2 + 3i × 2i
	=	−2 − 2i + 6i − 6
	=	−2 − 6 − 2i + 6i
	=	−8 + 4i

(d)(4 − 5i) (8 − 2i)

Solution:

(2 - 5i) (8 - 3i)	=	2 × 8 + 2 × (−3i) − 5i × 8 − 5i × (−3i)
	=	16 − 6i − 40i − 15
	=	16 − 15 −6i − 40i
	=	1 − 46i

Quiz 3: To which of the following does the expression (2 − i) (3 + 4i) simplify?

(a)5 + 4i Incorrect - please try again!

(b)6 + 11i Incorrect - please try again!

(d)6 + i Incorrect - please try again!

5. Complex Conjugation

For any complex number, z = a + ib, we define the complex conjugate to be: z^* = a − ib. It is very useful since:

z + z^*	=	a + ib + (a − ib)	=	2a
zz^*	=	(a + ib)(c − id)	=	a² + iab − iab − a² − (ib)²	=	a² + b²

The modulus of a complex number is defined as: |z| = √zz^∗

Exercise 3: Combine the following complex numbers and their conjugates:

(a)If z = (3 + 2i), find z + z^∗

Solution:

(3 + 2i) + (3 + 2i)^∗	=	(3 + 2i) + (3 − 2i)
	=	3 + 2i + 3 − 2i
	=	3 + 3 + 2i − 2i
	=	6

(b)If z = (3 − 2i), find zz^∗

Solution:

(3 - 2i) (3 - 2i)^∗	=	(3 - 2i) (3 + 2i)
	=	9 + 6i − 6i − 2i × (2i)
	=	9 −4i²
	=	9 + 4 = 13

(c)If z = (−1 + 3i), find zz^∗

Solution:

(−1 + 3i) (−1 + 3i)^∗	=	(−1 + 3i) (−1 − 3i)
	=	(−1) × (−1) + (−1)(−3i) + 3i(−1) + 3i(−3i)
	=	1 + 3i − 3i − 9i²
	=	1 + 9 = 10

(d)If z = (4 − 3i), find |z|

Solution:

√(4 − 3i) (4 + 3i)	=	√4² + 4 × 3i − 3i × 4 − 3i × 3i
	=	√16 + 12i − 12i − 9i²
	=	√16 + 9
	=	√25 = 5

Quiz 4: Which of the following is the modulus of 4 − 2i? ANS: NONE OF THEM!

(a)5 + 4i Incorrect - please try again!

(b)6 + 11i Correct - well done!

(d)6 + i Incorrect - please try again!

Explanation:

6. Dividing Complex Numbers

The 'trick' for dividing two complex numbers is to multiply top and bottom by the complex conjugate of the denominator:

z₁/z₂ = z₁/z₂ × z^*₂/z^*₂ = z₁z^*₂/z₂z^*₂

The denominator, z₂z^∗₂, is now a real number.

Example 4:

1/i	=	1/i × −i/−i
	=	−i/i × (−i)
	=	−i/1
	=	−i

Example 5:

(2 + 3i)/(1 + 2i)	=	(2 + 3i) /(1 + 2i) × (1 − 2i)/(1 − 2i)
	=	(2 + 3i) (1 − 2i)/(1 + 4)
	=	1/5(2 + 3i) (1 − 2i)
	=	1/5(2 − 4i + 3 i + 6)
	=	1/5(8 − i)

Exercise 4: Perform the following divisions:

(a)(2 + 4i)/i

Solution:

(2 + 4i)/i	=	(2 + 4i)/i × −i/−i
	=	(2 + 4i) × (−i)/+1
	=	(2 + 4i) (−i)
	=	−2i −i²
	=	4 − 2i

(b)(−2 + 6i)/ (1 + 2i)

Solution:

(−2 + 6i)/ (1 + 2i)	=	(−2 + 6i)/ (1 + 2i) × (1 − 2i)/(1 − 2i)
	=	(−2 + 6i) (1 − 2i)/ (1 + 4)
	=	1/5 (−2 + 6i) (1 − 2i)
	=	1/5 (−2 + 4i + 6i − 12i²)
	=	1/5 (−2 + 10i + 12)
	=	1/5 (−2 + 12 + 10i)
	=	1/5 (10 + 10i)
	=	2 + 2i

(c)(1 + 3i)/ (2 + i)

Solution:

(1 + 3i)/(2 + i)	=	(1 + 3i)/(2 + i) × (2 − i)/(2 − i)
	=	(1 + 3i) (2 − i)/ (4 + 1)
	=	1/5 (2 − i + 6i − 3i²)
	=	1/5 (2 + 3 + 5i)
	=	1/5 (5 + 5i)
	=	1 + i

(d)(3 + 2i)/ (3 + i)

Solution:

(3 + 2i)/(3 + i)	=	(3 + 2i)/(3 + i) × (3 − i)/(3 − i)
	=	(3 + 2i) (3 − i)/ (9 + 1)
	=	1/10 (3 + 2i) (3 − i)
	=	1/10 (9 − 3i + 6i - 2i²)
	=	1/10 (9 + 2 + 3i)
	=	1/10 (11 + 3i)

Quiz 5: To which of the following does the expression: 8 − i/2 + i simplify?

(a)3 − 2i Correct - well done!

(b)2 + 3i Incorrect - please try again!

(d)4 Incorrect - please try again!

Explanation:

8 − i/2 + i	=	8 − i/2 + i 2 − i/2 − i
	=	(8 − i)(2 − i)/ 2² + 1¹
	=	(8 × 2 + 8 × (−i) − i × 2 − i × (−i))/ 5
	=	1/ 5 (16 − 8i − 2i − 1)
	=	1/ 5 (15 − 10i) = 3 − 2i

Quiz 6: To which of the following does the expression −2 + i/2 + i simplify?

(a)−1 Incorrect - please try again!

(b)¹/₅(−5 + 7i) Incorrect - please try again!

(c)−1 + ½i Incorrect - please try again!

(d) ¹/₅(−3 + 4i) Correct - well done!

Explanation:

−2 + i/2 + i	=	−2 + i/2 + i 2 − i/2 − i
	=	(−2 + i)(2 − i)/ 2² + 1¹
	=	1/ 5 (−2 × 2 − 2 × (−i) + i × 2 + i × (−i))
	=	1/ 5 (−4 + 2i + 2i + 1 = 1/ 5 (−3 + 4)i