Complex Numbers Introduction. Number Systems. Natural Numbers ℵ Integer Z Rational Q Real Complex C

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1 Number Systems Natural Numbers ℵ Integer Z Rational Q R Real Complex C

2 Number Systems Natural Numbers ℵ Integer Z Rational Q R Real Complex C The Natural Number System All whole numbers greater then zero Some Mathematicians think this should also include 0

3 Number Systems Natural Numbers ℵ Integer Z Rational Q R Real Complex C The Integer Number System Zero, positive and negative integers

4 Number Systems Natural Numbers ℵ Integer Z Rational Q R Real Complex C The Rational Number System All Integer numbers Plus fractions

5 Number Systems Natural Numbers ℵ Integer Z Rational Q R Real Complex C The Real Number System All Rational Numbers Plus irrational numbers such as π and It does not contain the square root of negative numbers

6 Number Systems Natural Numbers ℵ Integer Z Rational Q R Real Complex C The Complex Number System All Real Numbers Plus Complex Numbers This allows for the square root of a negative number

7 Definition of a Complex Number In the Complex Number System, the square root of a negative number exists Are also called imaginary numbers Are represented by i for imaginary i = 1 noting i 2 = 1 A Complex Number can be represented by z = x + iy Where: The real part is Re z = x The imaginary part is Im z = y

8 Definition of a Complex Number k = ki Example 1: 64 = 64 1 = 8i Raising i to a power goes through cycles of 4 i 0 = 1 i 1 = i i 2 = 1 i 3 = i i 4 = 1

9 To work out i n for any n ; divide n by 4 and check the remainders Remainder of 1 : the answer is i Remainder of 2: the answer is -1 Remainder of 3: the answer is -i Example 2 Simplify i 2011? Solution 2011 mod 4 3 i 2011 = -i

10 Representing the Complex Number Complex Numbers are represented in the Complex Plane Real numbers are represented in a Cartesian (XY) Plane The Complex Plane is an adapted Cartesian Plane where the horizontal axis is the real axis and the vertical axis as the imaginary axis A complex number can be represented as: An ordered pair (a, b) A vector a + ib Where a is the real part, b is the imaginary part

11 Operations With Complex Numbers Addition a + bi + c + di = a + c + i b + d Subtraction a + bi c + di = a c + i b d Multiplication a + bi c + di = ac bd + i ad + bc Division a+bi c+di = ac+bd bc ad i c 2 +d2 + c 2 +d 2

12 Conjugate of a Complex Number The conjugate of a complex number z = x + iy is represented as z = x iy Equality a + bi = c + di if an only if a = c and b = d

13 Example 3 Given z 1 = 3 + i and z 2 = 1 + 2i Solution Find i z 1 + z 2 ii z 1 z 2 iii z 1 z 2 z iv 1 z 2 v Re z 1 vi Im(z 2 )

14 Example 3 Given z 1 = 3 + i and z 2 = 1 + 2i Find i z 1 + z 2 ii z 1 z 2 iii z 1 z 2 z iv 1 z 2 v Re z 1 vi Im(z 2 ) Solution i. 4+3i

15 Example 3 Given z 1 = 3 + i and z 2 = 1 + 2i Find i z 1 + z 2 ii z 1 z 2 iii z 1 z 2 z iv 1 z 2 v Re z 1 vi Im(z 2 ) Solution ii. 2-i

16 Example 3 Given z 1 = 3 + i and z 2 = 1 + 2i Find i z 1 + z 2 ii z 1 z 2 iii z 1 z 2 z iv 1 z 2 v Re z 1 vi Im(z 2 ) Solution iii. 3 + i 1 + 2i = (3 + 2i 2 ) + (6i + i) = 1 + 7i

17 Example 3 Given z 1 = 3 + i and z 2 = 1 + 2i Find i z 1 + z 2 ii z 1 z 2 iii z 1 z 2 z iv 1 z 2 v Re z 1 vi Im(z 2 ) Solution iv. 3 + i 1 + 2i = 3+i 1+2i 1 2i 1 2i = 3 2i2 6i+i = 5 5i 5 = 1 i

18 Example 3 Given z 1 = 3 + i and z 2 = 1 + 2i Find i z 1 + z 2 ii z 1 z 2 iii z 1 z 2 z iv 1 z 2 v Re z 1 vi Im(z 2 ) Solution v. Re z 1 =3

19 Example 3 Given z 1 = 3 + i and z 2 = 1 + 2i Find i z 1 + z 2 ii z 1 z 2 iii z 1 z 2 z iv 1 z 2 v Re z 1 vi Im(z 2 ) Solution v. Im z 1 =2

20 Example 4 Given z 1 = 3 + i and z 2 = 2 i Solution Find i. z 1 + z 2 ii. z 1 z 2 iii. z 1 z 2 z iv. 1 z 2 2 v. z 1

21 Example 4 Given z 1 = 3 + i and z 2 = 2 i Find i. z 1 + z 2 ii. z 1 z 2 iii. z 1 z 2 z iv. 1 z 2 2 v. z 1 Solution i. 5 ii i

22 Example 4 Given z 1 = 3 + i and z 2 = 2 i Find i. z 1 + z 2 ii. z 1 z 2 iii. z 1 z 2 z iv. 1 z 2 2 v. z 1 Solution iii. 3 + i 2 i = 6 i 2 3i + 2i = 7 i

23 Example 4 Given z 1 = 3 + i and z 2 = 2 i Find i. z 1 + z 2 ii. z 1 z 2 iii. z 1 z 2 z iv. 1 z 2 2 v. z 1 Solution iv. 3+i 2 i = 3 + i 2 i 2 + i 2 + i = 6 + i2 + 3i + 2i = 5 + 5i 5 = 1 + i

24 Example 4 Given z 1 = 3 + i and z 2 = 2 i Find i. z 1 + z 2 ii. z 1 z 2 iii. z 1 z 2 z iv. 1 z 2 2 v. z 1 Solution v. 3 + i 2 = 9 + 6i + i 2 = 8 + 6i

25 Example 5 Give that z = 5, find i. Re(z) ii. Im(z) Solution i. 5 ii. 0

26 Example 6 If z such that Im z = 2 and that z 2 is imaginary. Find z? Solution Let z = x + iy z 2 = x + iy 2 = x 2 y 2 + 2xyi Since Z 2 imaginary x 2 y 2 = 0 y = Im z = 2 x 2 4 = 0 x = ±2 z = ±2 + 2i

27 Example 7 Solve the quadratic equation x 2 + x + 1 = 0? x 2 + x x x = 1 2 = 3 4 = 3 4 i2 2 1 x = ± 3 2 i x = 1 2 ± 3 2 i

28 Complex Plane Complex numbers are represented as vectors in the complex plane. This representation is referred to as an Argand Diagram The vertical axis is the imaginary axis The horizontal axis is the real axis

29 Example 8 z = 3 + 4i Represented as a vector from the Origin (0,0) to the point (3,4)

30 Complex Numbers like vectors with magnitude and direction The magnitude is the length of the vector The direction is the angle it makes anticlockwise

31 Two forms for writing Complex Numbers Cartesian Form: z = x + iy Mod-Arg Form: z = z (cosθ + isinθ) where θ = tan 1 y x and z = x 2 + y 2 Acceptable Shorthand: z = z cisθ where cisθ = (cosθ + isinθ)

32 Mod Arg form Mod stands for Modulus being, which is used in mathematics to convey length Arg stands for Argument, which refers to the angle anticlockwise from the horizontal Mod z: z = x + iy = x 2 + y 2 Arg: θ = tan 1 y x Mod-Arg Form: z = z (cosθ + isinθ) where θ = tan 1 y x and z = x 2 + y 2

33 Example 9 z = 3 + 4i, Write in mod-arg form? mod z = 3 + 4i = = 5 arg z = tan 1 y x = tan z = 5(cosθ + isinθ) where θ = tan This is 5 cos53 + isin53 to the nearest degree

34 Example 10 Convert the following to mod-arg form 1. z = 3 + 4i 2. z = 3 4i 3. z = 3 4i

35 Use of Radian Measurement 1 Radian = degree 180 π π 2 π 4 π 3 = 90 degrees = 45 degrees = 60 degrees π 6 = 30 degrees

36 Example 11 Convert the following to mod-arg form, leave answer in radians? i. 1 i ii. 1 3i iii. 3 iv. 3i

37 Example 12 Express the following in Cartesian form? i. 2cis30 ii. 4cis 5π 6

38 Example 13 Consider the complex numbers below (note it is in Cartesian form) Let z 1 = x 1 + iy 1, z 2 = x 2 + iy 2, z 3 = x 3 + iy 3 Show that (z 1 z 2 ) z 3 = z 1 (z 2 z 3 )? Solution Consider the L.H.S. Firstly z 1 z 2 = (x 1 x 2 y 1 y 2 ) + i(x 1 y 2 + x 2 y 1 ) And (z 1 z 2 ) z 3 = x 1 x 2 y 1 y 2 + i(x 1 y 2 + x 2 y 1 ) ( x 3 + iy 3 ) Hence the L.H.S. = (x 1 x 2 x 3 x 1 y 2 y 3 x 2 y 1 y 3 x 3 y 1 y 2 ) + i(x 1 x 2 y 3 + x 1 x 3 y 2 + x 2 x 3 y 1 y 1 y 2 y 3 )

39 Example 13 Consider the complex numbers below (note it is in Cartesian form) Let z 1 = x 1 + iy 1, z 2 = x 2 + iy 2, z 3 = x 3 + iy 3 Show that (z 1 z 2 ) z 3 = z 1 (z 2 z 3 )? Solution continued Now the R.H.S. Firstly z 2 z 3 = (x 2 x 3 y 2 y 3 ) + i(x 2 y 3 + x 3 y 2 ) And z 1 (z 2 z 3 ) = ( x 1 + iy 1 ) x 2 x 3 y 2 y 3 ) + i(x 2 y 3 + x 3 y 2 ) Hence the R.H.S. = (x 1 x 2 x 3 x 1 y 2 y 3 x 2 y 1 y 3 x 3 y 1 y 2 ) + i(x 1 x 2 y 3 + x 1 x 3 y 2 + x 2 x 3 y 1 y 1 y 2 y 3 ) Since L.H.S. = R.H.S. it has is shown

40 Multiplying and dividing in mod-arg form Generally is easier Example 14 z = z (cosθ + isinθ) is mod-arg form Let z 1 = z 1 (cosθ 1 + isinθ 1 ), z 2 = z 2 (cosθ 2 + isinθ 2 ), z 3 = z 3 (cosθ 3 + isinθ 3 ) z n = z n (cosθ n + isinθ n ) Then z 1 z 2 = z 1 (cosθ 1 + isinθ 1 ) z 2 (cosθ 2 + isinθ 2 ) z 1 z 2 = z 1 z 2 cosθ 1 cosθ 2 sinθ 1 sinθ 2 + i(cosθ 1 sinθ 2 + sinθ 1 cosθ 2 )

41 Example 14 Let z 1 = z 1 (cosθ 1 + isinθ 1 ), z 2 = z 2 (cosθ 2 + isinθ 2 ), z 3 = z 3 (cosθ 3 + isinθ 3 ) z n = z n (cosθ n + isinθ n ) Noting that: cosθ 1 cosθ 2 sinθ 1 sinθ 2 = cos(θ 1 ± θ 2 ) cosθ 1 sinθ 2 ± sinθ 1 cosθ 2 = sin(θ 1 ± θ 2 ) Hence z 1 z 2 = z 1 z 2 cos(θ 1 + θ 2 ) + isin(θ 1 + θ 2 ) Further z 1 z 2 z 3 = z 1 z 2 cos(θ 1 + θ 2 ) + isin(θ 1 + θ 2 ) z 3 (cosθ 3 + isinθ 3 ) = z 1 z 2 z 3 cos θ 1 + θ 2 cosθ 3 sin(θ 1 + θ 2 )sinθ 3 + i cos(θ 1 + θ 2 )sinθ 3 + sinθ 1 + θ 2 cosθ 3 Hence z 1 z 2 z 3 = z 1 z 2 z 3 cos(θ 1 + θ 2 + θ 3 ) + isin(θ 1 + θ 2 + θ 3 )

42 This means when you multiply complex numbers : you multiply their lengths together you add their angles together

43 Multiplying and dividing in mod-arg form In General z 1 z 2 z 3 z n = z 1 z 2 z 3 z n cos(θ 1 + θ 2 + θ 3 + θ n ) + isin(θ 1 + θ 2 + θ 3 + θ n ) Or this can be written as z 1 z 2 z 3 z n = r 1 r 2 r 3 r n cos(θ 1 + θ 2 + θ 3 + θ n ) + isin(θ 1 + θ 2 + θ 3 + θ n ) As r is used to represent the length of a vector in this case the length of a complex number That is r = z

44 Example 15 i. Show that z 1 z 2 = z 1 z 2 cos(θ 1 θ 2 ) + isin(θ 1 θ 2 )? ii. Find z and arg z when z = 3 + 2i? iii. iv. Write down the moduli and arguments of 3 + i and 4 + 4i Hence express in modulus/argument form ( 3 + i)(4 + 4i )? State the modulus and argument of 1 + i. Hence write 1 + i 18 in the form a + ib v. If z = 3 2i, mark on an Argand diagram the points representing vi. (i) 2z (ii) 2iz

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