#### Transcript Decision Maths - Haringeymath's Blog

Further Pure 1 Lesson 5 – Complex Numbers Numbers Wiltshire What types of numbers do we already know? Real numbers – All numbers ( 2, 3.15, π ,√2) Rational – Any number that can be expressed as a fraction – (4, 2.5, 1/3) Irrational – Any number that can`t be expressed as a fraction. ( π ,√2, √3 + 1) Natural numbers – The counting numbers (1, 2, 3, ….) Integers – All whole numbers ( -5, -1, 6) Complex Numbers (Imaginary numbers) Complex Numbers Wiltshire What is the √(-1)? We define the √(-1) to be the imaginary number j. (Hence j2 = -1) Note that lots of other courses use the letter i, but we are going to use j. We can now use this to calculate a whole new range of square roots. What is √(-144)? Answer is +12j and -12j, or ±12j. Complex Numbers Wiltshire Now we can define a complex number (z) to be a number that is made up of real and imaginary parts. z=x+yj Here x and y are real numbers. x is said to be the real part of z, or Re(z). y is said to be the imaginary part of z, or Im(z). Solving Quadratics Wiltshire Use the knowledge you have gained in the last few slides to solve the quadratic equation z2 + 6z + 25 = 0 Remember Solution b b 2 4ac z 2a 6 62 4 1 25 z 2 1 Solving Quadratics 6 36 100 z 2 6 64 z 2 68 j z 2 z 3 8 j Wiltshire Addition and Subtraction Wiltshire To Add and subtract complex numbers all you have to do is add/subtract the real and imaginary parts of the number. (x1+y1j) + (x2+y2j) = x 1 + x 2 + y 1j + y 2j = (x1 + x2)+ (y1 + y2)j (x1+y1j) – (x2+y2j) = x 1 - x 2 + y 1j - y 2j = (x1 - x2)+ (y1 - y2)j Multiplying Wiltshire Multiplying two complex numbers is just like multiplying out two brackets. You can use the FOIL method. First Outside Inside Last. Remember j2 = -1 (x1+y1j)(x2+y2j) = x1x2 + x1y2j + x2y1j + y1y2j2 = x 1x 2 + x 1y 2j + x 2y 1j - y 1y 2 = x 1x 2 - y 1y 2 + x 1y 2j + x 2y 1j = (x1x2 - y1y2) + (x1y2 + x2y1)j What is j3, j4, j5? Multiplying Wiltshire Alternatively you could use the box method. x1 y1j x2 x1x2 x2y1j y2j x1y2j - y1y2 (x1+y1j)(x2+y2j) = (x1x2 - y1y2) + (x1y2 + x2y1)j Questions If z1 = 5 + 4j z2 = 3 + j Find a) z1 + z3 = 12 + 2j b) z1 - z2 = 3 + 3j c) z1 – z3 = -2 + 6j d) z1 × z2 = 11 + 17j e) z1 × z3 = 43 + 18j Wiltshire z3 = 7 – 2j Complex Conjugates Wiltshire The complex conjugate of z = (x + yj) is z* = (x – yj) If you remember the two solutions to the quadratic from a few slides back then they where complex conjugates. z = -3 + 8j & z = -3 – 8j In fact all complex solutions to quadratics will be complex conjugates. If z = 5 + 4j What is z + z* What is z × z* Activity Wiltshire Prove that for any complex number z = x + yj, that z + z* and z × z* are real numbers. First z + z* = (x + yj) + (x – yj) = x + x + yj – yj = 2x = Real Now z × z* = (x + yj)(x – yj) = x2 – xyj + xyj – y2j2 = x2 – y2(-1) = x2 + y2 = Real Now complete Ex 2A pg 50 Division Wiltshire There are two ways two solve problems involving division with complex numbers. First you need to know that if two complex numbers are equal then the real parts are identical and so are there imaginary parts. If we want to solve a question like 1 ÷ (2 + 4j) we first write it equal to a complex number p + qj. 1 p qj 24 j Now we re-arrange the equation to find p and q. (p + qj)(2 + 4j) = 1 Division Wiltshire Expanding the equation gives 2p – 4q + 2qj + 4pj = 1 The number 1 can be written as 1 + 0j So (2p – 4q) + (2q + 4p)j = 1 + 0j Now we can equate real and imaginary parts. 2p – 4q = 1 4p + 2q = 0 Solve these equations p = 1/10 & q = -1/5 Therefore 1 ÷ (2 + 4j) = 0.1 – 0.2j Division Wiltshire The second method is similar to rationalising the denominator in C1. 1 1 24 j 24 j 1 1 j 24 j 24 j 24 j 20 10 5 The 20 on the bottom comes from the algebra we proved a few slides back. (x + yj)(x – yj) = x2 + y2 Now see if you can find (3 - 5j) ÷ (2+9j) 3 5 j 3 5 j 2 9 j 6 10 j 27 j 45 39 37 j 29 j 29 j 29 j 4 81 85 Now complete Ex 2B pg 53 Argand Diagrams Complex numbers can be shown Geometrically on an Argand diagram The real part of the number is represented on the x-axis and the imaginary part on the y. -3 -4j 3 + 2j 2 – 2j Wiltshire Im Re Modulus of a complex number Wiltshire A complex number can be represented by the position vector. Im x y The Modulus of a complex number is the distance from the origin to the point. |z| = √(x2+y2) Note |x| = x y x Re Modulus of a complex number Wiltshire Find a) |3 + 4j| = 5 b) |5 - 12j| = 13 c) |6 - 8j| = 10 d) |-24 - 10j| = 26 Sum of complex numbers z1 + z 2 = Im x1 x2 x1 x2 y1 y2 y1 y2 2 5 2 5 7 5 1 5 1 6 Wiltshire z2 z1 z1 + z2 Re Difference of complex numbers Wiltshire z2 - z1 = Im x2 x1 x2 x1 y2 y1 y2 y1 z1 7 2 7 2 5 6 5 6 5 1 z2 z2 – z1 Now complete Ex 2C pg 57 Re Sets of points in Argand diagram Wiltshire What does |z2 – z1| represent? If z1 = x1 +y1j & z2 = x2 +y2j Then z2 – z1 = (x2 – x1) + (y2 – y1)j So |z2 – z1| = √((x2 – x1)2 + (y2 – y1)2) This represents the distance between to complex numbers z1 & z2. Im (x2,y2) y2- y1 (x1,y1) x2- x1 Re Examples Draw an argand diagram showing the set of points for which |z – 3 – 4j| = 5 Solution First re-arrange the question |z – (3 + 4j)| = 5 From the previous slide this represents a constant distance of 5 between the point (3,4) and z. This will give a circle centre (3,4) Now do Ex 2D pg 60 Wiltshire Im Re Examples Wiltshire How would you show the sets of points for which: i) |z – 3 – 4j)| ≤ 5 ii) |z – 3 – 4j)| < 5 iii) |z – 3 – 4j)| ≥ 5 Im Im Im Re Re Re