# Further Pure 1 - PowerPoint PPT Presentation

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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

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)

• Natural numbers – The counting numbers (1, 2, 3, ….)

• Integers – All whole numbers ( -5, -1, 6)

• Complex Numbers (Imaginary 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 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).

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

z2 + 6z + 25 = 0

• Remember

• Solution

### 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 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?

### Multiplying

• Alternatively you could use the box method.

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

### 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

• 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

• 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

• 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

### 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

### 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

• 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

• 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 number

• Find

a)|3 + 4j| = 5

b)|5 - 12j| = 13

c)|6 - 8j| = 10

d)|-24 - 10j| = 26

z2

z1

z1 +z2

• z1 + z 2 =

Im

Re

z1

z2

z2 –z1

• z2 -z1 =

Im

Re

### 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

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

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