Introduction to Complex Numbers |
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PPLATO / Basic Mathematics |
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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 |
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Since (5i)2 | = | 52 × i2 |
= | 25 × (−1) | |
= | −25 |
(b)
√− 169 | = | 43i |
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Since (43i)2 | = | 169 × (i)2 |
= | −169 |
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:
(1 + i) + (3 + i) | = | 1 + 3 + i(1 + 1) | = | 4 + 2i |
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(2 + 5i) − (1 − 4i) | = | 2 + 5i − 1 + 4i | = | 1 + 9i |
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Exercise 1: Add or subtract the following complex numbers:
Solution:
(3 + 2i) + (3 + i) | = | 3 + 2i + 3 + i |
= | 3 + 3 + 2i + 2i | |
= | 6 + 3i |
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.
Solution: The factor of 1 ⁄ 2 multiplies both terms in the complex number:
(−1 + 3i) + ½(2 + 2i) | = | −1 + 3i + 1 + i |
= | 4i |
Solution:
13 (2 − 5i) − 16 (8 − 2i) | = | 23 − 53 i − 86 + 26 i |
= | 23 − 53 i − 43 + 13 i | |
= | 23 − 43 − 53 i + 13 i | |
= | − 23 − 43 i |
- which we may also write as −23(1 + 2i).
Quiz 1: To which of the following does the expression (4 − 3i) + (2 + 5i) simplify?
Quiz 2: To which of the following does the expression (3 − i) − (2 − 6i) simplify?
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:
Solution:
(3 + 2i) (3 + i) | = | 3 × 3 + 3 × i + 2i × 3 + 2i × i |
= | 9 + 3i + 6i − 2 | |
= | 9 − 2 + 3i + 6i | |
= | 7 + 9i |
Solution:
(4 − 2i) (3 − 2i) | = | 4 × 3 + 4 × (−2i) − 2i × 3 − 2i × (−2i) |
= | 12 − 8i − 6i − 4 | |
= | 12 − 4 −8i − 6i | |
= | 8 −14i |
Solution:
(−1 + 3i) (2 + 2i) | = | −1 × 2 − 1 × 2i + 3i × 2 + 3i × 2i |
= | −2 − 2i + 6i − 6 | |
= | −2 − 6 − 2i + 6i | |
= | −8 + 4i |
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?
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 | ||
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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:
Solution:
(3 + 2i) + (3 + 2i)∗ | = | (3 + 2i) + (3 − 2i) |
= | 3 + 2i + 3 − 2i | |
= | 3 + 3 + 2i − 2i | |
= | 6 |
Solution:
(3 - 2i) (3 - 2i)∗ | = | (3 - 2i) (3 + 2i) |
= | 9 + 6i − 6i − 2i × (2i) | |
= | 9 −4i2 | |
= | 9 + 4 = 13 |
Solution:
(−1 + 3i) (−1 + 3i)∗ | = | (−1 + 3i) (−1 − 3i) |
= | (−1) × (−1) + (−1)(−3i) + 3i(−1) + 3i(−3i) | |
= | 1 + 3i − 3i − 9i2 | |
= | 1 + 9 = 10 |
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!
Explanation:
The 'trick' for dividing two complex numbers is to multiply top and bottom by the complex conjugate of the denominator:
The denominator, z2z∗2, is now a real number.
Example 4:
1i | = | 1i × −i−i |
= | −ii × (−i) | |
= | −i1 | |
= | −i |
Example 5:
(2 + 3i)(1 + 2i) | = | (2 + 3i) (1 + 2i) × (1 − 2i)(1 − 2i) |
= | (2 + 3i) (1 − 2i)(1 + 4) | |
= | 15(2 + 3i) (1 − 2i) | |
= | 15(2 − 4i + 3 i + 6) | |
= | 15(8 − i) |
Exercise 4: Perform the following divisions:
Solution:
(2 + 4i)i | = | (2 + 4i)i × −i−i |
= | (2 + 4i) × (−i)+1 | |
= | (2 + 4i) (−i) | |
= | −2i −i2 | |
= | 4 − 2i |
Solution:
(−2 + 6i) (1 + 2i) | = | (−2 + 6i) (1 + 2i) × (1 − 2i)(1 − 2i) |
= | (−2 + 6i) (1 − 2i) (1 + 4) | |
= | 15 (−2 + 6i) (1 − 2i) | |
= | 15 (−2 + 4i + 6i − 12i2) | |
= | 15 (−2 + 10i + 12) | |
= | 15 (−2 + 12 + 10i) | |
= | 15 (10 + 10i) | |
= | 2 + 2i |
Solution:
(1 + 3i)(2 + i) | = | (1 + 3i)(2 + i) × (2 − i)(2 − i) |
= | (1 + 3i) (2 − i) (4 + 1) | |
= | 15 (2 − i + 6i − 3i2) | |
= | 15 (2 + 3 + 5i) | |
= | 15 (5 + 5i) | |
= | 1 + i |
Solution:
(3 + 2i)(3 + i) | = | (3 + 2i)(3 + i) × (3 − i)(3 − i) |
= | (3 + 2i) (3 − i) (9 + 1) | |
= | 110 (3 + 2i) (3 − i) | |
= | 110 (9 − 3i + 6i - 2i2) | |
= | 110 (9 + 2 + 3i) | |
= | 110 (11 + 3i) |
Quiz 5: To which of the following does the expression: 8 − i2 + i simplify?
Explanation:
8 − i2 + i | = | 8 − i2 + i 2 − i2 − 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 + i2 + i simplify?
Explanation:
−2 + i2 + i | = | −2 + i2 + i 2 − i2 − 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.