1 / 23

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
Download Presentation

Further Pure 1

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related