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, since12 = (−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 |
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.
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!
(c)6 + 8i Incorrect - please try again!
(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!
(c)1 + 5iCorrect - well done!
(d)1 − 5i Incorrect - please try again!
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!
(c)10 + 5iCorrect - well done!
(d)6 + i Incorrect - please try again!
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!
(c)10 + 5i Incorrect - please try again!
(d)6 + i Incorrect - please try again!
Explanation:
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!
(c)4 − ½i 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 |
In each of the following, simplify the expression and choose the solution from the options given.
1.(3 + 4i) − (2 − 3i)
(a)3 −i
(b)5 + 7i
(c)1 + 7i
(d)1 − i
2.(3 + 3i) (2 − 3i)
(a)6 − 8i
(b)6 + 8i
(c)−3 + 3i
(d)15 − 3i
3.|12 − 5i|
(a)13
(b)√7
(c)√119
(d)−12.5
4. (7 − 17i)/(5 − i)
(a)13⁄5 + 17i
(b)3 + i
(c)3 + 2i
(d)2 − 3i
