Complex Numbers

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Complex Numbers
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Book References forComplex Numbers

• The following books (in the UCL Watson library)
  have sections on complex numbers
  ------------------------------------------------
  ------
• 1) Advanced engineering mathematics - K. A.
  Stroud and Dexter J. Booth.
• 2) Modern engineering mathematics - Glyn James
• 3) Mathematics for engineers a modern
  interactive approach - Anthony Croft, Robert
  Davison

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Book References forComplex Numbers

• 4) Mathematical methods for physicists - George
  B. Arfken, Hans J. Weber
• 5) Mathematical methods for physicists a
  concise introduction - Tai L. Chow.
• 6) Schaum's outline of theory and problems of
  complex variables with an introduction to
  conformal mapping and its application - Murray R.
  Spiegel
• Note some cover more detail than others. Make
  sure you read what is necessary

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For week 9 and 10

• Imaginary numbers and complex numbers
• Operations on complex numbers
• Equal complex numbers
• Complex roots of quadratic equations
• Cubic equations with complex roots
• Argand diagram

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For week 9 and 10

• Modulus/argument form of complex number
• Product and quotient of complex numbers
• DeMoivres theorem
• Cube roots of complex numbers of unity
• nth root of a complex number
• Locii problems (if time allows)

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Imaginary and Complex Nos

• We have always assumed that to take a square root
  we need x2 0
• but what of x2 0 ? Specifically how would we
  find
• v-4 v-8 v-17 etc ?
• Well consider v-4 v(4 x 1) 2v-1

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Imaginary and Complex Nos

• And in general v(-n2) v(n2 x 1) nv-1
• and we symbolise v-1 as i
• i is called imaginary number
• Hence v-4 2i, v-8 2iv2, v-17 17i
• and in general a complex number is
• a bi
• a the real part and b the imaginary part

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Operations on Complex Nos

• Addition and subtraction
• simply add respective real and imaginary parts
• (23i) (-1 - i) (2-1) (3-1)i
• 1 2i
• In general
• (abi) (c di) (ac) (bd)i

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Operations on Complex Nos

• Multiplication
• multiply in the usual way and use property of
• i 2 -1
• (23i).(-1 - i) 2(-1) 2i 3i - 3i2
• (-23) (23)i
• 1 5i

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Operations on Complex Nos

• Multiplication
• more generally
• (abi).(c di) a.c (a.db.c)i b.di2
• (ac - bd) (a.db.c)i
• consider also
• (abi).(a - bi) a2 (a.b-a.b)i a.bi2
• a2 b2

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Operations on Complex Nos

• Complex conjugates
• Complex numbers of the form
• (abi) and (a - bi)
• are called complex conjugate, and (a - bi) is
  the conjugate of (a bi).
• In general complex number are denoted z
• Complex conjugate is then denoted z or z

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Operations on Complex Nos

• Division of complex numbers
• We use the aspect of complex conjugates in order
  to divide two complex numbers
• Consider 2 9i
• 5 2i
• We can rewrite this as (2 9i)(52i)
• (5 2i)(52i)

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Operations on Complex Nos

• Division of complex numbers
• hence
• (2 9i)(52i) 10-18 49i
• (5 2i)(52i) 25 4
• -8/29 49/29 i

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Zero and equal Complex Nos

• The zero complex number
• A complex number can only be zero if both Re and
  Im parts are zero
• a bi 0 gt a 0 and b 0
• Equal complex numbers
• Two complex numbers are equal if the respective
  Re and Im parts are equal

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Zero and equal Complex Nos

• I.e.
• abi cdi
• (a-c) (b-d)i 0
• a c 0 and b d 0
• hence ac , bd
• Then Re(abi) Re(cdi)
• Im(abi) Im(cdi)

? abicdi
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Zero and equal Complex Nos

• Examples
• See lectures i) alternative form for dividing
  complex number
• ii) finding the square root of a
  complex number

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Complex roots of a quadratic equation

• We may now solve all types of quadratic
  equations.
• As such consider solving x22x20 the formula
  gives x (-2 v-4)/2
• which now gives x -1 i
• Hence
• x22x2 (x (1i)).(x (1 - i))

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Complex roots of a quadratic equation

• In general for ax2bxc0 with complex roots
• b2 4ac lt 0
• and be end up with complex conjugates as roots
  p qi and p qi
• where p-b/(2a) and qv(b2 4ac) / (2a)

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Cubic equations with complex roots

• In solving the previous quadratic we saw that the
  roots occurred in complex conjugate pairs
• We also know that when solving a cubic equation
  we always end up with three roots
• This means the the roots of a cubic will have the
  form 3 real roots or 1 real root and a complex
  conjugate pair

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Cubic equations with complex roots

• Hence consider cubic f(x)ax3bx2cxd0
• By the factor theorem we have that f(k)0, hence
  k is a root of f(x).
• Hence (x-k)(px2qxr) 0 and we may solve the
  quadratic to obtain either real roots or complex
  conjugates

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Cubic equations with complex roots

• Examples
• See lecture i) cube roots of unity
• ii) other examples

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The Argand diagram

• Complex number can actually be represented on a
  graph with the x-axis as the Re axis and the
  y-axis as the Im axis

z a bi
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The modulus/argument of a complex numbers

• From the Argand diagram we may now consider the
  complex number of length r and angle ? with the
  Im axis

z a bi
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The modulus/argument of a complex numbers

• Length OZ r is called the modulus of the
  complex number and is denoted z

z r v(a2 b2)
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The modulus/argument of a complex numbers

• Angle ? is called the argument of the complex
  number, denoted Arg(z)

Arg(z) ? tan-1 b/a
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The modulus/argument of a complex numbers

• When solving for ? we must choose the principle
  argument only
• Examples
• See lecture

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The modulus/argument (or polar form) of a complex
numbers

• Consider the general complex number xyi on an
  Argand diagram

x r.cos? y r.sin?
xyi r.cos? ir.sin?
xyi r.(cos? i.sin?)
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The modulus/argument (or polar form) of a complex
numbers

• This is the modulus-argument form of a complex
  number
• and expresses a complex number in polar
  co-ordinates (r, ?)
• Examples
• See lecture

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Products and quotients in polar form

• Consider now two complex numbers in polar
  co-ordinates (r, ?)
• z1 r1.(cos?1 i.sin?1)
• z2 r2.(cos?2 i.sin?2)
• multiplying
• z1z2 r1r2.(cos?1 i.sin?1)(cos?2 i.sin?2)
• r1r2 .cos(?1?2) i.sin (?1?2)

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Products and quotients in polar form

• So z1z2 r1r2 and Arg(z1z2) ?1?2
• Similarly for quotients of complex numbers in
  polar form
• z1 r1.(cos?1 i.sin?1)
• z2 r2.(cos?2 i.sin?2)
• r1/r2 .cos(?1- ?2) i.sin (?1- ?2)

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Products and quotients in polar form

• So z1/z2 r1/r2 and Arg(z1/z2) ?1- ?2
• So
• z1.z2 z1.z2
• Arg(z1.z2) arg (z1) arg(z2)
• z1/z2 z1 / z2
• Arg(z1/z2) arg(z1) arg(z2)

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Products and quotients in polar form

• Examples
• See lecture

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DeMoivres Theorem

• We know from the product of two complex numbers
  in polar form that
• z1z2 r1r2 .cos(?1?2) i.sin (?1?2) ()
• Consider a complex number z of unit modulus
• hence z2 r2.(cos2? i.sin2?) from ()
• And z2 z.z 1, Arg(z2)arg(z)arg(z)

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DeMoivres Theorem

• hence z2 r2.(cos2? i.sin2?) from ()
• Similarly it can be shown that
• z3 r3.(cos3? i.sin3?)
• This could also be shown to be the case by
  explicitly multiplying out
• z3 r.(cos? i.sin?)3

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DeMoivres Theorem

• hence
• z2 z.z 1
• Arg(z2)arg(z)arg(z) 2?
• Continuing this way, for z3 we have
• z3 z2.z 1
• Arg(z3)arg(z2)arg(z) 3?

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DeMoivres Theorem

• Similarly it can be shown that
• z4 r4.(cos4? i.sin4?)
• This seems to suggest that
• zn rn.(cos n? i.sin n?)
• DeMoivres theorem states that
• zn rn.(cos n? i.sin n?)

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DeMoivres Theorem

• This can be proved by induction
• therefore let Pn be the statement
• Pn (cos? i.sin?)n (cos n? i.sin n?)
• Show that his is true for n1 ?

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DeMoivres Theorem

• Then for 1 lt k lt n
• Pk (cos? i.sin?)k (cos k? i.sin k?)
• We want to show that Pk1 is like Pk but replaced
  with (k1)
• hence
• Pk1 (cos? i.sin?)k1
• (cos? i.sin?)k.(cos? i.sin?)

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DeMoivres Theorem

• i.e.
• Pk1 (cos? i.sin?)k1
• (cosk? i.sink?).(cos? i.sin?)
• After some multiplication and simplification we
  obtain
• Pk1 (cos? i.sin?)k1
• (cosk?.cos? sink?.sin?).
• i.(sink?.cos? cosk?.sin?)

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DeMoivres Theorem

• Hence
• Pk1 (cos? i.sin?)k1
• (cos(k1)? i.sin(k1)?)
• Hence Pk gt Pk1 for any value of n

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DeMoivres Theorem

• For proof of Demoivres theorem when n is ve and
  when n is rational
• Application of DeMoivres Theorem
• See lecture

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DeMoivres Theorem

• Properties of z and 1/z
• We have seen that for a complex number of unit
  length
• z cos? i.sin?
• 1/z cos? - i.sin?
• then
• z 1/z 2.cos?
• z 1/z 2i.sin?

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DeMoivres Theorem

• Properties of z and 1/z
• Also
• 1/zn (cos? i.sin?)-n
• (cos(-n?) i.sin(-n?))
• then
• zn 1/zn 2.cos(n?)
• zn 1/zn 2i.sin(n?)

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DeMoivres Theorem

• These properties allow us to express cosn? or
  sinn? in terms of multiple angles
• Examples
• See lecture

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cube roots of complex numbers

• Let us consider the cube root of a complex number
  z3w
• we want to find the cube root of this complex
  number
• DeMoivres theorem can be used to solve this
• Let w r.(cos? i.sin?)

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cube roots of complex numbers

• Let us consider the cube root of a complex number
  z3w
• we want to find the cube root of this complex
  number
• DeMoivres theorem can be used to solve this
• Let w r.(cos? i.sin?)

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cube roots of complex numbers

• If ? is the principle argument then the general
  argument is ?2n?
• Hence w r.cos(?2n?) i.sin(?2n?)
• Hence z w1/3
• r1/3.cos(?/32n?/3) i.sin(?/32n?/3)

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cube roots of complex numbers

• Hence z w1/3
• r1/3.cos(?/32n?/3) i.sin(?/32n?/3)
• The three roots of the cube root are found by
  taking n0, 1, 2
• n0 z1 r1/3.cos(?/3) i.sin(?/3)
• n1 z1 r1/3.cos(?/32?/3) i.sin(?/3
  2?/3)
• n2 z1 r1/3.cos(?/34?/3) i.sin(?/3 4?/3)

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cube roots of complex numbers

• Hence the three roots can be represented as three
  points on the Argand diagram

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Cube root of unity

• We have already seen algebraically how to solve
  z3 1
• But now we know that
• z3 r.cos(?2n?) i.sin(?2n?)
• r1/3.cos(?/32n?/3) i.sin(?/32n?/3)
• ()

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Cube root of unity

• But z 1 and the principle argument of z is 0
  (since 1 is real and lies on the Re axis, then
  ?0)
• From () the other arguments are
• ?0, n0 (?2n?)/3 0
• ?0, n1 (?2n?)/3 2?/3
• ?0, n2 (?2n?)/3 4?/3

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Cube root of unity

• hence
• z1 1
• z2 cos 2? i.sin 2?
• z3 cos 4? i.sin 4?
• Changing these into rectangular coordinates we
  have
• z1 1
• z2 -1/2 i. v3 /2
• z3 -1/2 - i. v3 /2

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Cube root of unity

• Now look at what happens when we square z2
• (z2)2 (-1/2 i. v3 /2)2 -1/2 - i. v3 /2 z3
• Similarly (z2)2 z2
• hence the roots of unity can be denoted as
• 1, ?, ?2
• where ? is a complex root of unity

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Cube root of unity

• It can also be shown that
• 1 ? ?2 0
• Exercise
• If we were to find the 4th roots of unity, what
  relationship would exists between the roots ?
• If we were to find the 5th roots of unity, what
  relationship would exists between the roots ?
• Can u deduce a pattern for the relationship
  between the nth roots of unity ?

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Cube root of unity

• In general the nth roots of unity
• Z 11/n
• will produce n equally spaced roots, separated by
  angles of 2?/n,
• all roots will lie on the unit circle
• The complex numbers form vertices of an n-sided
  polygon

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Cube root of unity

• Examples
• See lecture

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Locii of Complex Numbers

• Q What is a locus ?
• A it is the path traced out by a particle it
  is the direction a particles moves in
• E.g. Consider a football on the floor

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Locii of Complex Numbers

• now let a particle be placed at the centre of
  the football.
• Q What is the path taken by the particle as the
  ball moves ? I.e. what is the locus of the
  particle ?

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Locii of Complex Numbers

• A so the locus of the particle is a straight
  line

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Locii of Complex Numbers

• Q if the particle is now placed on the
  circumference of the circle what is the locus of
  the particle ?

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Locii of Complex Numbers

• A as shown below

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Locii of Complex Numbers

• Other locii include

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Locii of Complex Numbers

• We now want to study the locii of complex number.
• Recall that Z x iy defines the position of a
  point in the Argand diagram.
• If we specify a certain condition we can make
  this point move
• therefore we would want to find the equation of
  the path moved by that point

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Locii of Complex Numbers

• implying calculating the locus of the complex
  number in the complex plane
• the conditions which will control the movement of
  the point will be specified by 1) the modulus
  Z ?x2y2or 2) the argument arg(Z)
  tan-1 (y/x)

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Locii of Complex Numbers

• Examples (see lecture)Given that z x iy
  1) find the locus defined by arg(z) p/42)
  find the equation of the locus z/(z2)
  13) find the equation of the locus arg(z2) p

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Complex Numbers

• The End