Further Pure 1

1
Further Pure 1

• Lesson 6 Complex Numbers

2
Modulus/Argument of a complex number

• The position of a complex number (z) can be
  represented by the distance that z is from the
  origin (r) and the angle made with the positive
  real axis (?).
• Distance is given by r z
• r z v(x2y2)
• r is known as the modulus of a complex number.

Im
r
y
?
x
Re
3
Argument of a complex number










Im

• Now lets look at the angle (?) that the line
  makes with the positive real axis.
• NOTE ? is measured in radians in an
  anticlockwise direction from the positive real
  axis.
• ? is measured from p to p and is known as the
  principal argument of z.
• Argument z arg z ?
• y r sin?, x r cos?
• tan ? y/x
• ? inv tan (y/x)

r
y
?
x
Re
4
Modulus/Argument of a complex number

• Calculate the modulus and argument of the
  following complex numbers. (Hint, it helps to
  draw a diagram)
• 1) 3 4j z v(3242) 5
• arg z inv tan (4/3)
• 0.927
• 2) 5 - 5j z v(5252) 5v2
• arg z inv tan (5/-5)
• -p/4
• 3) -2v3 2j z v((2v3)222) 4
• arg z inv tan (2/-2v3)
• 5p/6

5
Modulus/Argument of a complex number

• Note, inv tan (y/x) will return answer in
  first/fourth quadrant.
• Last example on previous slide inv tan (2/-2v3)
  on your calculator will return, -p/6, however the
  answer we want is 5p/6.
• In some circumstances you way need to add or
  subtract p.
• This is why a diagram is useful.

6
Modulus-Argument form of a complex number

• So far we have plotted the position of a complex
  number on the Argand diagram by going
  horizontally on the real axis and vertically on
  the imaginary.
• (This is just like plotting co-ordinates on an
  x,y axis)
• However it is also possible to locate the
  position of a complex number by the distance
  travelled from the origin (pole), and the angle
  turned through from the positive x-axis.
• (This is sometimes known as Polar co-ordinates
  and can be studied in another course)

7
Modulus-Argument form of a complex number

• (x,y)

• (r, ?)
• cos? x/r, sin? y/r
• x r cos?, y r sin?,

y






Im






Im
?
y
x
r
y
?
x
Re
Re
8
Converting

• Converting from Cartesian to Polar
• (x,y) v(x2y2),(inv tan (y/x)) (r, ?)
• Convert the following from Cartesian to Polar
• i) (1,1) (v2,p/4)
• ii) (-v3,1) (2,5p/6)
• iii) (-4,-4v3) (8,-2p/3)

Im
r
y
?
x
Re
9
Converting

• Converting from Polar to Cartesian
• (r, ?) (r cos?, r sin ?)
• Convert the following from Polar to Cartesian
• i) (4,p/3) (2,2v3)
• ii) (3v2,-p/4) (3,-3)
• iii) (6v2,3p/4) (-6,6)

Im
r
y
?
x
Re
10
Using the Calculator

• It is possible to convert from Cartesian to Polar
  and back using your calculator.
• (Casio fx-83MS other calculators may be
  different)
• Make sure that your calculator is in Radians.
• The polar coordinates (v2,p/4) equal the
  Cartesian co-ordinates (1,1)
• On the calculator press
• The answer shown is the first Cartesian
  co-ordinate x. To view the y press RCL F. To view
  the x again press RCL E.
• Now try and convert back from the Cartesian to
  the Polar.

Shift
Rec(
v
2
,
p

4
)
11
Modulus-Argument form of a complex number

• Now we can define the Modulus-Argument form of a
  complex number to be
• z r (cos? j sin?)
• Here r z and ? arg z
• When writing a complex number in Modulus-Argument
  form it can be helpful to draw a diagram.
• Remember that ? will take values between -p and
  p.

12
Modulus-Argument form of a complex number

• Write the following numbers in Modulus-Argument
  form
• i) 5 12j ii) v3 j iii) -4 4j
• Solutions
• i) 13(cos 1.176 j sin 1.176)
• ii) 2(cos (-p/6) j sin (-p/6))
• iii) 4v2(cos(-3p/4) j sin(-3p/4))

13
Modulus-Argument form of a complex number

• When writing a complex number in Modulus-Argument
  form it is important that it is written in the
  exact format as above, not
• z -r (cos? - j sin?)
• (r must be positive)
• You can use the following trig identities to
  ensure that z is written in the correct form.
• cos(p-?) -cos? sin(p-?) sin?
• cos(?-p) -cos? sin(?-p) -sin?
• cos(-?) cos? sin(-?) -sin?

14
Modulus-Argument form of a complex number

• Example Re-write -4(cos? j sin?) in
  Modulus-Argument form
• -4(cos? j sin?) 4(-cos? - j sin?)
• 4(cos(?-p) j sin(?-p))
• This is now in Modulus-Argument form.
• The Modulus is 4 and the Argument is (?-p).
• This may seem a bit confusing, however if you
  draw a diagram and tackle the problems like we
  did a few slides ago it makes more sense.
• Now do Ex 2E pg 66

15
Sets of points using the Modulus-Argument form

• What do you think arg(z1-z2) represents?
• If z1 x1 y1j z2 x2 y2j
• Then z2 z1 (x2 x1) (y2 y1)j
• Now arg(z2 z1) inv tan ((y2 y1)/(x2 x1))
• This represents the angle between the line from
  z1 to z2 and a line parallel to the real axis.
• So arg(z2 z1) a

Im
(x2,y2)
y2- y1
a
(x1,y1)
x2- x1
Re
16
Questions

• Draw Argand diagrams showing the sets of points z
  for which
• i) arg z p/3
• ii) arg (z - 2) p/3
• iii) 0 arg (z 2) p/3
• Now do Ex 2F page 68

17
History of complex numbers

• In the quadratic formula (b2 4ac) is known as
  the discriminant.
• If this is greater than zero the quadratic will
  have 2 distinct solutions.
• If it is equal to zero then the quadratic will
  have 1 repeated root.
• If it is less than zero then there are no
  solutions.
• Complex numbers where invented so that we could
  solve quadratic equations whose discriminant is
  negative.
• This can be extended to solve equations of higher
  order like cubic and quartic.
• In fact complex numbers can be used to find the
  roots of any polyniomial of degree n.

18
History of complex numbers

• In 1629 Albert Girard stated that an nth degree
  polynomial will have n roots, complex or real.
  Taking into account repeated roots.
• For example the fifth order equation
• (z 2)(z-4)2(z29) 0 has five roots.
• 2, 4(twice) 3j and -3j.
• For any polynomial with real coefficients
  solutions will always have complex conjugate
  pairs.
• Many great mathematicians have tried to prove the
  above.
• Gauss achieved it in 1799.

19
Complex numbers and equations

• Given that 1 2j is a root of 4z3 11z2 26z
  15 0
• Find the other roots.
• Since real coefficients, 1 2j must also be a
  root.
• Now z (1 2j) and z (1 2j) will be
  factors.
• So z (1 2j)z (1 2j)
• (z 1) 2j)(z 1) 2j)
• (z 1)2 (2j)2
• z2 2z 1 (-4)
• z2 2z 5

20
Complex numbers and equations

• You can now try to spot the remaining factors by
  comparing coefficients or you can use polynomial
  division.
• 4z 3
• z2 2z 5 4z3 11z2 26z 15 0
• 4z3 8z2 20z
• 3z2 6z 15
• 3z2 6z 15
• 0
• Hence 4z3 11z2 26z 15 (z2 2z 5)(4z
  3)
• The roots are, 1 2j, 1 2j and 3/4

21
Complex numbers and equations

• Given that -2 j is a root of the equation
• z4 az3 bz2 10z 25 0
• Find the values of a and b and solve the
  equation.
• First you need to find the values of z4,z3 z2
  so that you can sub them in to the equation
  above.
• z2 (-2 j)(-2 j) 3 4j
• z3 (3 4j)(-2 j) -2 11j
• Z4 (-2 11j)(-2 j) -7 - 24j
• Now
• (-7 -24j) a(-2 11j) b(3 4j) 10(-2 j)
  25 0