Download Complex numbers

Manipulate real and complex
numbers and solve equations
AS 91577
Worksheet 1
Quadratics
General formula:
y = ax + bx + c
2
General solution:
-b ± b - 4ac
x=
2a
2
Example 1
y = x + 3x + 3
2
Equation cannot be factorised.
x + 3x + 3 = 0
2
Using quadratic formula
We use the substitution
-
i =1
2
A complex number
x + 3x + 3 = 0
2
The equation has 2 complex solutions
3
3i
x= ±
2
2
Real
Imaginary
Equation has 2 complex
solutions.
b - 4ac < 0
2
Example 2
x - 2x + 5 = 0
2
2± 2 -4´5
Þx=
=
2
2
Example 2
x - 2x + 5 = 0
2
2 ± 2 - 4 ´ 5 2 ± -16
Þx=
=
2
2
2
Example 2
x - 2x + 5 = 0
2
2 ± 2 - 4 ´ 5 2 ± -16
Þx=
=
2
2
2 ± 4i
Þx=
= 1± 2i
2
2
Adding complex numbers
Subtracting complex numbers
Example
z = 3+ i
w = -1+ 2i
Example
z = 3+ i
w = -1+ 2i
Þ z + w = ( 3 -1) + i(1+ 2)
= 2 + 3i
Multiplying Complex Numbers
(x + yi)(u + vi) = (xu –
yv) + (xv + yu)i.
Example
(3 + 4i)(5 - 2i)
=
Example
( 3 + 4i)(5 - 2i)
= ( 3 ´ 5 + 4 ´ 2) + ( 3´ 2 + 4 ´ 5)i
= 23 + 14i
Example 2
(3x + 2y)( 3x - 2y) = 9x
2
(3x + 2iy)(3x - 2iy) = 9x
- 4y
2
2
+ 4y
2
Conjugate
If
z = 3+ 4i
The conjugate of z is
If
z = 3- 4i
z = 2 - 5i
The conjugate of z is
z = 2 + 5i
Dividing Complex Numbers
a + ib a + ib c - id
=
´
=
c + id c + id c - id
(ac + bd) + i(bc - ad)
2
2
c +d
Example
z = 2 - 3i
z
2 - 3i
=
w -1+ 4i
w = -1+ 4i
Example
z = 2 - 3i
w = -1+ 4i
z
2 - 3i -1- 4i
=
´
w -1+ 4i -1- 4i
Example
z = 2 - 3i
w = -1+ 4i
z
2 - 3i -1- 4i
=
´
w -1+ 4i -1- 4i
-2 -12) + i(-8 + 3) -14 5
(
=
=
- i
2
2
1 +4
17 17
Solving by matching terms
x + 3+ 2iy = 7i
Match real and imaginary
-
x + 3 = 0 Þ x= 3
Imaginary 2y = 7 Þ y = 3.5
Real
Solving polynomials
Quadratics: 2 solutions
2 real roots
2 complex roots
If coefficients are all real, imaginary roots are
in conjugate pairs
y = x - 2x + 5 = 0
2
If coefficients are all real, imaginary roots are
in conjugate pairs
y = x - 2x + 5 = 0
2
x =1- 2i
and
x = 1+ 2i
Cubic
Cubics: 3 solutions
3 real roots
1 real and 2 complex roots
Quartic
Quartic: 4 solutions
2 real and 2
imaginary roots
4 real roots
4 imaginary
roots
Solving a cubic
x + x -2 =0
3
2
This cubic must have at least 1 real solutions
f (1) = 0 Þ x -1 is a factor and x =1 is a solution
Form the quadratic.
( x -1)(x
2
+ 2x + 2) = 0
x = 1, -1 - i, 1 + i
Solve the quadratic for
the other solutions
Finding other solutions when you are given
one solution.
If z = 1+ i is a root of the equation
z 3 - 5z 2 + 8z - 6 = 0, find the other roots
Because coefficients are real, roots come in conjugate
pairs so
z =1- i is also a root of the equation
Form the quadratic i.e.
z - (1+ i + 1- i)z + (1+ i)(1- i)
2
z - 2z + 2
2
Form the cubic:
(z - 2z + 2)(z - 3) = 0
2
Argand Diagram
z = 3- 2i
Just mark the spot
with a cross
Plot z = 3 + i
z
Adding z = 3+ i and w = -1+ 2i
z=i
z = -1
z =1
z = -i
Multiplying a complex number
by a real number.
(x + yi) u = xu + yu i.
Multiplying a complex number by i.
z i = (x + yi) i = –y + xi.
Reciprocal of z
1
1
x - iy
x - iy
z = =
´
= 2
z x + iy x - iy x + y 2
-1
Conjugate
If z = x + iy, then z = x - iy
Rectangular to polar form
Using Pythagoras
Modulus is the length
x 2 + y 2 = r2
z = x2 + y2 = r
Argument is the angle
y
tan q =
x
Check the quadrant of the complex number
Modulus is the length
Example 1
Rectangular form
z = 3+ 4i
Modulus = z = 32 + 4 2 = 5
4
tan q = Þ q = 0.927 (53.1°)
3
Polar form
z = 5cos(0.927) + 5isin(0.927)
= 5cis(0.927)
Example 2
z = 3- 4i
Modulus = z = 32 + 4 2 = 5
-4
tan q =
Þ q = 0.927 ( 53.1°)
3
z = 5cis( 0.927)
-
-
-
= 5cos( 0.927) + 5isin( 0.927)
Example 3
z = -3- 4i
Modulus = z = 3 + 4 = 5
2
2
-4
tan q =
Þ q = p + 0.927 (= 4.069)
-3
z = 5cis( 4.069)
= 5cos(4.069) + 5isin(4.069)
Converting from polar to rectangular
æpö
æpö
æpö
z = 6cis ç ÷ = 6 ´ cos ç ÷ + 6 ´ sin ç ÷ i
è 6ø
è 6ø
è 6ø
z = 3 + 3 3i
Multiplying numbers in polar form
Example 1
æp ö
æp ö
z = 3cisç ÷ and w = 2cisç ÷
è6ø
è4ø
æp p ö
æ 5p ö
zw = 3 ´ 2cisç + ÷ = 6cisç ÷
è6 4ø
è 12 ø
Multiplying numbers in polar form
Example 2
Take out
multiples of
2p
æ 5p ö
æ 7p ö
z = 5cisç ÷ and w = 6cisç ÷
è 4 ø
è 3ø
æ 5p 7p ö
æ 43p ö
zw = 5 ´ 6cisç
+
÷ = 30cisç
÷
è 4
è 12 ø
3ø
æ19p ö
zw = 30cisç
÷
è 12 ø
Remove all multiples of
2p (360°)
De Moivre’s Theorem
(rcisq)
Example 1
n
= r cis(nq )
n
æ
pö
p
3p
3
ç2cis ÷ = 2 cis(3 ´ ) = 8cis
è
4ø
4
4
3
De Moivre’s Theorem
(rcisq)
Example 2
n
= r cis(nq )
n
Take out
multiples of
2p
æ
5p ö
5p
4
ç 3cis ÷ = 3 cis(4 ´ ) = 81cis5p = 81cis(p )
è
4ø
4
4
Solving equations using De Moivre’s Theorem
z + 81i = 0 Þ z = -81i
4
4
æ -p ö
z = 81cisç ÷
1. Put into polar form
è 2 ø
æ
ö
p
4
2. Add in multiples of 2p z = 81cisç2np - ÷
è
2ø
4th root 81
3. Fourth root
æ 2np p ö
z = 3cisç
- ÷
è 4
8ø
4
4. Generate solutions
-p
Divide angle by 4
z1 = 3cis
Letting n = 0, 1, 2, 3
8
z3 = 3cis
7p
8
3p
z2 = 3cis
8
11p
z4 = 3cis
8
Take note:
24i = 24cis
-24 = 24cisp
p
2
24 = 24cis0
-p
-24i = 24cis
2
Useful websites
Good general level
http://www.clarku.edu/~djoyce/complex/
Advanced level
http://mathworld.wolfram.com/ComplexNumber.html
Good general level
http://www.purplemath.com/modules/complex.htm
Good general level- Also gives proofs
http://www.sosmath.com/complex/complex.html
Problems at 3 levels
http://www.ping.be/~ping1339/Pcomplex.htm#READTHIS-FIRST