Further pure 1
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Further Pure 1. 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) Irrational – Any number that can`t be expressed as a fraction. ( π , √2, √3 + 1)

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Further Pure 1

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Further pure 1

Further Pure 1

Complex Numbers


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)

    • 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

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.


Complex numbers1

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).


Solving quadratics

Solving Quadratics

  • Use the knowledge you have gained in the last few slides to solve the quadratic equation

    z2 + 6z + 25 = 0

  • Remember

  • Solution


Solving quadratics1

Solving Quadratics


Addition and subtraction

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


Multiplying

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?


Multiplying1

Multiplying

  • Alternatively you could use the box method.

  • (x1+y1j)(x2+y2j) = (x1x2 - y1y2) + (x1y2 + x2y1)j


Questions

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


Complex conjugates

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*


Activity

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


Division

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


Division1

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


Division2

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)


Argand diagrams

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


Modulus of a complex number

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


Modulus of a complex number1

Modulus of a complex number

  • Find

    a)|3 + 4j| = 5

    b)|5 - 12j| = 13

    c)|6 - 8j| = 10

    d)|-24 - 10j| = 26


Sum of complex numbers

z2

z1

z1 +z2

Sum of complex numbers

  • z1 + z 2 =

Im

Re


Difference of complex numbers

z1

z2

z2 –z1

Difference of complex numbers

  • z2 -z1 =

Im

Re


Sets of points in argand diagram

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


Examples

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


Examples1

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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