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Introduction to Complex Numbers

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Presentation on theme: "Introduction to Complex Numbers"— Presentation transcript:

1 Introduction to Complex Numbers

2 Imaginary Numbers Consider x2 = –1. x2 = –1 or 1 – = x
∵ and are not real numbers. ∴ The equation x2 = –1 has no real roots. x2 = –1 or 1 = x They are called imaginary numbers. The square roots of negative numbers are called imaginary numbers. e.g , , , 2 - 3

3 (c) For any positive real number p,
(a) is denoted by i. 1 -  i.e. 1 - = i (b) 1 - =  i.e. i 2 = –1 (c) For any positive real number p, 1 - = p  i.e. i p = - e.g. 4 - = 1 4 - i 2 = = 9 - 1 9 - 3i =

4 Complex Numbers A complex number is a number that can be written in the form a + bi, where a and b are real numbers, and 1 - i = a + b i Complex Number: Real part Imaginary part

5 Complex Numbers A complex number is a number that can be written in the form a + bi, where a and b are real numbers, and 1 - i = e.g. (i) 2 – i (ii) –3 (iii) 4i Real part: 2, imaginary part: –1 Real part: –3, imaginary part: 0 Real part: 0, imaginary part: 4

6 Complex Numbers A complex number is a number that can be written in the form a + bi, where a and b are real numbers, and 1 - i = e.g. (i) 2 – i (ii) –3 (iii) 4i Real part: 2, imaginary part: –1 Real part: –3, imaginary part: 0 Real part: 0, imaginary part: 4 For a complex number a + bi, if b = 0, then a + bi is a real number.

7 Complex Numbers A complex number is a number that can be written in the form a + bi, where a and b are real numbers, and 1 - i = e.g. (i) 2 – i (ii) –3 (iii) 4i Real part: 2, imaginary part: –1 Real part: –3, imaginary part: 0 Real part: 0, imaginary part: 4 For a complex number a + bi, if a = 0 and b ≠ 0, then a + bi is an imaginary number.

8 Complex Number System Complex numbers Real numbers
e.g. –3, 0 (i) Imaginary numbers e.g. 4i, –2i (ii) Sum of a non-zero real number and an imaginary number e.g. 2 – i, 1 + 5i

9 Follow-up question It is given that z = (k – 3) + (k + 1)i. If the imaginary part of z is 4, find the value of k, is z an imaginary number? (a) ∵ Imaginary part of z = 4 ∴ k + 1 = 4 k = 3

10 Follow-up question It is given that z = (k – 3) + (k + 1)i. If the imaginary part of z is 4, find the value of k, is z an imaginary number? (b) ∵ Real part = 3 – 3 = 0 Imaginary part ≠ 0 ∴ z is an imaginary number.

11 & a = c a + bi = c + di b = d Equality of Complex Numbers
Two complex numbers (a + bi and c + di) are equal when both their and imaginary real parts parts are equal. a = c & a + bi = c + di Equality b = d

12 If x – 3i = yi, find the values of the real numbers x and y.
∵ The real parts are equal. ∴ x = 0 ∵ The imaginary parts are equal. ∴ y = –3

13 Follow-up question Find the values of the real numbers x and y if 2x + 4i = –8 + (y + 1)i. 2x + 4i = –8 + (y + 1)i By comparing the real parts, we have 8 2 = x 4 = x By comparing the imaginary parts, we have 1 4 + = y 3 = y

14 Operations of Complex Numbers

15 Let a + bi and c + di be two complex numbers.
Addition (a + bi) + (c + di) = (a + c) + (b + d)i e.g. (1 + 2i) + (2 – i) = 1 + 2i + 2 – i = (1 + 2) + (2 – 1)i = 3 + i

16 Subtraction (a + bi) – (c + di) = (a – c) + (b – d)i e.g. (1 + 2i) – (2 – i) = 1 + 2i – 2 + i = (1 – 2) + (2 + 1)i = –1 + 3i

17 Follow-up question Simplify and express each of the following in the form a + bi. (a) (4 – 2i) + (3 + i) (b) (–5 + 3i) – (1 + 2i) (a) (4 – 2i) + (3 + i) = 4 – 2i i = (4 + 3) + (–2 + 1)i = 7 – i (b) (–5 + 3i) – (1 + 2i) = –5 + 3i – 1 – 2i = (–5 – 1) + (3 – 2)i = –6 + i

18 Multiplication (a + bi)(c + di) = = ac + bci + adi + bdi 2 = ac + bci + adi + bd(–1) = (ac – bd) + (bc + ad)i (a + bi)(c) + (a + bi)(di) e.g. (1 + 2i)(2 – i) = (1 + 2i)(2) + (1 + 2i)(–i) = 2 + 4i – i – 2i2 = 2 + 3i – 2(–1) = 4 + 3i

19 Division i + - = 2 1 di c bi a + = di c - e.g. i - + = 2 4 ) ( )( di c
2 1 di c bi a + = di c - e.g. i - + = 2 4 2 ) ( )( di c bi a - + = i - + = ) 1 ( 4 2 5 d c i ad bc bd ac 2 ) ( + - = i = 5 i d c ad bc bd ac 2 + - = i =

20 Follow-up question Simplify and express each of the following in the form a + bi. (a) (1 + 3i)(–2 + 2i) (b) (a) (1 + 3i)(–2 + 2i) = (1 + 3i)(–2) + (1 + 3i)(2i) = –2 – 6i + 2i + 6i2 = –2 – 4i + 6(–1) = –8 – 4i

21 (b) i - + = 3 1 2 4 i - + = ) 3 ( 1 6 12 2 4 i - + = ) 1 ( 9 6 10 4 i - = 10 i - = 1


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