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Further Pure 1

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Further Pure 1

Complex 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)

- 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.

- 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

- 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 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?

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

- 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

- 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*

- 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

- 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

- 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

- 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)

- 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

- 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

- 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

- 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

- 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

- 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