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Further Pure 1

Further Pure 1. Lesson 5 – Complex Numbers. Numbers. 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)

jeremy-brown
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Further Pure 1

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  1. Further Pure 1 Lesson 5 – Complex Numbers

  2. Numbers • 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)

  3. Complex Numbers • 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.

  4. Complex Numbers • Now we can define a complex number (z) to be a number that is made up of real and imaginary parts. z = x + y j • 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).

  5. Solving Quadratics • Use the knowledge you have gained in the last few slides to solve the quadratic equation z2 + 6z + 25 = 0 • Remember • Solution

  6. Solving Quadratics

  7. Addition and Subtraction • 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) = x1 + x2 + y1j + y2j = (x1 + x2)+ (y1 + y2)j • (x1+y1j) – (x2+y2j) = x1 - x2 + y1j - y2j = (x1 - x2)+ (y1 - y2)j

  8. Multiplying • 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 = x1x2 + x1y2j + x2y1j - y1y2 = x1x2 - y1y2 + x1y2j + x2y1j = (x1x2 - y1y2) + (x1y2 + x2y1)j • What is j3, j4, j5?

  9. Multiplying • Alternatively you could use the box method. • (x1+y1j)(x2+y2j) = (x1x2 - y1y2) + (x1y2 + x2y1)j

  10. Questions • If z1 = 5 + 4j z2 = 3 + j z3 = 7 – 2j • 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

  11. Complex Conjugates • 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*

  12. Activity • 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

  13. Division • 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. • Now we re-arrange the equation to find p and q. (p + qj)(2 + 4j) = 1

  14. Division • 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

  15. Division • The second method is similar to rationalising the denominator in C1. • 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) • Now complete Ex 2B pg 53

  16. 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 Im Re

  17. Modulus of a complex number • A complex number can be represented by the position vector. • The Modulus of a complex number is the distance from the origin to the point. • |z| = √(x2+y2) • Note |x| = x Im y x Re

  18. Modulus of a complex number • Find a) |3 + 4j| = 5 b) |5 - 12j| = 13 c) |6 - 8j| = 10 d) |-24 - 10j| = 26

  19. z2 z1 z1 +z2 Sum of complex numbers • z1 + z 2 = Im Re

  20. z1 z2 z2 –z1 Difference of complex numbers • z2 -z1 = • Now complete Ex 2C pg 57 Im Re

  21. Sets of points in Argand diagram • 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

  22. Im 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

  23. Im Im Im Re Re Re Re Examples • 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

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