Advanced Math 45S
Introduction
Complex Numbers
Complex Numbers
Up until now you have been unable to find the square root of a negative number. In
more advance mathematics you can take the square root of a negative number and
you get a new type of number, that is a Complex Number.
Complex numbers were first invented / discovered by an Italian mathematician
Gerolamo Cardano in the sixteenth century when he attempted to solve equations
that required intermediate steps containing square roots of negative number. At the
time nobody thought there could be any “real world” use for this new type of
number, other than helping in the solution of some equations.
In fact complex numbers are used in electrical circuits involving alternating current,
telecommunications (AM, FM and CB), control systems used in auto-pilots in
airplanes, car power steering and missile guidance systems as well as many areas in
mechanical, civil and aeronautical engineering.
Anyway this new number is called i where i stands for imaginary. You have only
dealt with real numbers before, i.e. numbers that do not involve i.
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Advanced Math 45S
Complex Numbers
1. Definition
A complex number is one that can be represented in the form
where:
a + bi
a and b are real numbers
i = √−1 or i2 = −1
a is referred to as the real part and b as the imaginary part of the complex number.
Since
0∙𝑖 =0
every real number is a complex number as a + 0i = a
Example 1: Simplify √−9
Write the number under the radicand as a product of −1 and a non-negative
number.
√−9 = √9 × −1
Then write as product of radicals.
√9 × −1 = √9√−1
Find the roots.
so
√9√−1 = 3i
√−9 = 3𝑖
Example 2: Simplify i4
Rewrite as product of i2’s and i’s.
i4 = i2  i2
Substitute.
= (-1)  (-1)
Simplify.
=1
Example 3: Simplify i7
Rewrite as product of i2’s and i’s.
i7 = i2  i2  i2  i
Substitute.
= (-1)  (-1)  (-1)  i
Simplify.
= -i
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Advanced Math 45S
Complex Numbers
Example 4: Simplify √81 + √−36
Write the number under the radicand of the imaginary part of the complex number
as a product of −1 and a non-negative number.
√81 + √−36 = √81 + √36 × −1
Write as product of radicals.
√81 + √36 × −1 = √81 + √36√−1
Find the roots of both parts of complex number.
√81 + √36√−1 = 9 + 6𝑖
Assignment #1
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Advanced Math 45S
Complex Numbers
2. Representing Complex Numbers Geometrically
Any real number can be represented as a point on a line. Since a complex number is
represented by two real numbers a and b ( i.e. an ordered pair (a, b)) , the ordered
pair (a, b) can determine a point on a plane. This plane is called an Argand or
Complex Plane.
Assignment #2
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Advanced Math 45S
Complex Numbers
3. Absolute Value of a Complex Number
The absolute value of a real number is the distance from the origin on a real number
line. Similarly the absolute value of a complex number
a + bi is the distance from the origin on the Argand Plane.
Hence
|𝑎 + 𝑏𝑖| = √𝑎2 + 𝑏 2
Example:
Find the absolute value of 3 + 4i
Solution:
|3 + 4𝑖| = √32 + 42
= √25
= 5
Example: Find the absolute value of -6i
Solution:
|0 − 6𝑖| = √02 + (−6)2
= √36
= 6
Assignment #3
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Advanced Math 45S
Complex Numbers
4. Adding and Subtracting Complex Numbers
Addition: (a + bi) + (c + di) = (a + c) + (b + d)i
Examples
(2 + 3i) + (1 + 4i) = (2 + 1) + (3 + 4)i
= 3 + 7i
(7 + 4i) + (-3 + 1i) = (7 + -3) + (4 + 1)i
= 4 + 5i
(3 + i) + (4 - 5i) = (3 + 4) + (1 + -5)i
= 7 – 4i
Subtraction: (a + bi) - (c + di) = (a + bi) + (-c - di)
= (a - c) + (b - d)i
Examples
(2 + 3i) - (1 + 4i)
= (2 – 1) + (3 – 4)i
= 1–i
(2 + 6i) - (-1 + 4i) = (2 – (-1)) + (6 – 4)i
= 3 + 2i
(5 + 3i) - (1 - 4i)
= (5 – 1) + (3 – (-4))i
= 4 +7i
Assignment #4
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Advanced Math 45S
Complex Numbers
5. Multiplication of Complex Numbers
(a + bi ) ∙ (c + di ) = (ac - bd) + (ad + bc)i
Note the multiplication of complex numbers is similar to that of polynomials, just
remember that i2 can be replaced with -1
Example:
(2 – 3i)(4 – i) = 8 – 2i - 12i +3i2
= 8 – 14i + 3(-1)
= 5 – 14i
Example:
(3 + i)(4 - 3i) = 12 – 9i + 4i – 3i2
= 12 – 5i – 3(-1)
= 15 – 5i
Assignment 5
Page 7 of 25
Advanced Math 45S
Complex Numbers
6. The Complex Conjugate
For denominators that involve complex numbers with real and imaginary parts we
have to use the complex conjugate.
a – bi is the complex conjugate of a + bi
The product of a complex number and its conjugate is a real number. Symbolically if
z is a complex number, then 𝑧̅ is its conjugate.
Example:
(5 – 2i)(5 + 2i)
= 25 – 4i2
= 25 – 4(-1)
= 29
No assignment
7. Dividing Complex Numbers.
Example:
5
𝑖
5 5 𝑖
= ∙
𝑖
𝑖 𝑖
5𝑖
= 2
𝑖
5𝑖
=
−1
= −5𝑖
Multiple the numerator and denominator by the conjugate of the denominator
Example:
3 − 2𝑖 3 − 2𝑖 5 − 3𝑖
=
×
5 + 3𝑖 5 + 3𝑖 5 − 3𝑖
=
15 − 19𝑖 + 6𝑖 2
25 − 9𝑖 2
=
9 − 19𝑖
34
Assignment #6
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Advanced Math 45S
Complex Numbers
8. Proving Properties of Complex Numbers
1. If z1 and z2 are complex numbers then
̅̅̅̅̅̅̅̅̅
𝑧1 + 𝑧2 = 𝑧̅1 + 𝑧̅2
Proof: Let z1 = a + bi and z2 = c + di
so
𝑧1 + 𝑧2 = 𝑎 + 𝑏𝑖 + 𝑐 + 𝑑𝑖 = (𝑎 + 𝑐) + (𝑏 + 𝑑)𝑖
̅̅̅̅̅̅̅̅̅
𝑧1 + 𝑧2 = (𝑎 + 𝑐) − (𝑏 + 𝑑)𝑖
and since
𝑧̅1 = 𝑎 − 𝑏𝑖 𝑎𝑛𝑑 𝑧̅2 = 𝑐 − 𝑑𝑖
𝑧̅1 + 𝑧̅2 = 𝑎 − 𝑏𝑖 + 𝑐 − 𝑑𝑖
= (𝑎 + 𝑐) − (𝑏 + 𝑑)𝑖
hence
̅̅̅̅̅̅̅̅̅
𝑧1 + 𝑧2 = 𝑧̅1 + 𝑧̅2
2. If z1 and z2 are complex numbers then
𝑧1 ⋅ 𝑧2 = 𝑧̅1 ⋅ 𝑧̅2
̅̅̅̅̅̅̅̅
Proof: Let z1 = a + bi and z2 = c + di
So
𝑧1 ⋅ 𝑧2 = (𝑎 + 𝑏𝑖)(𝑐 + 𝑑𝑖)
= 𝑎𝑐 + 𝑎𝑑𝑖 + 𝑏𝑐𝑖 + 𝑏𝑑𝑖 2
= (𝑎𝑐 − 𝑏𝑑) + (𝑎𝑑 + 𝑏𝑐)𝑖
so
𝑧1 ⋅ 𝑧2 = (𝑎𝑐 − 𝑏𝑑) − (𝑎𝑑 + 𝑏𝑐)𝑖
̅̅̅̅̅̅̅̅
and since
𝑧̅1 = 𝑎 − 𝑏𝑖 𝑎𝑛𝑑 𝑧̅2 = 𝑐 − 𝑑𝑖
Hence
𝑧̅1 ⋅ 𝑧̅2 = (𝑎 − 𝑏𝑖)(𝑐 − 𝑑𝑖)
= 𝑎𝑐 − 𝑎𝑑𝑖 − 𝑏𝑐𝑖 + 𝑏𝑑𝑖 2
= (𝑎𝑐 − 𝑏𝑑) − (𝑎𝑑 + 𝑏𝑐)𝑖
𝑧1 ⋅ 𝑧2 = 𝑧̅1 ⋅ 𝑧̅2
̅̅̅̅̅̅̅̅
No assignment
Page 9 of 25
Advanced Math 45S
Complex Numbers
9. Solving Quadratic Equations with Real Coefficients and having complex
roots.
Example: Solve 3x2 – 4x + 10 = 0
Using the quadratic equation
𝑥=
so
𝑥=
4 ± √(−4)2 − 4(3)(10)
2(3)
4 ± √16 − 120
=
6
4 ± √−104
=
6
4 ± 2𝑖√26
=
6
2 ± 𝑖√26
=
3
2 + 𝑖√26
2 − 𝑖√26
,𝑥 =
3
3
Note that complex roots always come in pairs.
Assignment #7 Questions 1-3
Page 10 of 25
Advanced Math 45S
Complex Numbers
10. Solving Quadratic Equations with Complex Coefficients
We will restrict our examples to those that meet the following conditions:
a) a, b, and c are imaginary
b) a and c are imaginary and b is real
c) b is imaginary and a and c are real.
We will deal with the other possibilities later when we can find the square root of a
complex number!
Example 1: Solve x2 + 3ix – 4 = 0
𝑥=
−3𝑖 ± √(3𝑖)2 − 4(1)(−4)
2(1)
=
−3𝑖 ± √−9 + 16
2
=
−3𝑖 ± √7
2
so
x=
(b is imaginary, a and c are real)
−3𝑖 + √7
−3𝑖 − √7
, 𝑥=
2
2
Example 2: Solve 5ix2 – 2ix + 3i = 0
(a,b,c are imaginary)
2𝑖 ± √(−2𝑖)2 − 4(5𝑖)(3𝑖)
𝑥=
2(5𝑖)
=
2𝑖 ± √−4 + 60
10𝑖
=
2𝑖 ± √56
10𝑖
=
2𝑖 ± 2√14
10𝑖
=
𝑖 ± √14
5𝑖
so
𝑥=
𝑖 + √14
𝑖 − √14
,𝑥 =
5𝑖
5𝑖
or
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Advanced Math 45S
Complex Numbers
1 𝑖√14
1 𝑖√14
+
, 𝑥= −
5
5
5
5
𝑥=
An alternative solution to this would be to factor i out of the original equation.
i.e. 5ix2 – 2ix + 3i = 0
i(5x2 – 2x + 3) = 0
since i ≠ 0 , 5x2 – 2x + 3 = 0
so
x=
=
2 ± √(−2)2 − 4(5)(3)
2(5)
2 ± √4 − 60
10
=
2 ± √−56
10
=
2 ± 2√(−1)(14)
10
=
1 ± 𝑖√(14)
5
so
𝑥=
1 𝑖√14
1 𝑖√14
+
, 𝑥= −
5
5
5
5
Example 3: Solve 2ix2 - 6x + 3i = 0
(a and c are imaginary, b is real)
6 ± √(−6)2 − 4(2𝑖)(3𝑖)
2(2𝑖)
6 ± √36 + 24
=
4𝑖
6 ± √60
=
4𝑖
6 ± 2√15
=
4𝑖
3 ± √15
=
2𝑖
𝑥=
so
x=
3 + √15
,
2𝑖
x=
3 − √15
2𝑖
Page 12 of 25
Advanced Math 45S
Complex Numbers
or
(−3 + √15 )𝑖
,
2
Assignment #7 Questions 4-10
x=
x=
(−3 − √15)𝑖
2
11. Forming Equations given Complex Roots.
Example 1: roots 2 ± i
Equation:
[x – (2 + i)] [x – (2 - i)] = 0
[ (x – 2) - i ] [ (x – 2) + i ] = 0
(x – 2)2 - i2 = 0
(x – 2)2 + 1 = 0
x2 – 4x + 4 + 1 = 0
x2 – 4x + 5 = 0
NB: When the roots are a conjugate pair the resulting equation with have real
coefficients
Example 2: Roots 3 + i , -2 – i
Equation:
[ x - (3 + i) ] [x – (-2 – i ) = 0
[ (x – 3) – i ] [ (x + 2) + i ) = 0
(x – 3)(x + 2) + (x – 3)i - (x + 2)i – i2 = 0
x2 – x – 6 +(x – 3 - x – 2)i + 1 = 0
x2 – x – 6 - 5i + 1 = 0
x2 – x – (5 +5i) = 0
NB: When the roots are not a conjugate pair, the resulting equation must have one
imaginary coefficient.
Example 3: Roots (2 ± i), -1
Equation:
[ x – (2 + i)] [x – (2 – i)] [ x + 1] = 0
[ (x – 2) – i ] [(x – 2) + i ] [x + 1] = 0
[ (x – 2)2 – i2] [x + 1] = 0
(x2 - 4x + 4 + 1)(x + 1) = 0
x3 – 4x2 + 5x + x2 − 4x + 5 = 0
x3 – 3x2 + x + 5 = 0
Assignment #8
(3 roots so cubic equation)
Page 13 of 25
Advanced Math 45S
Complex Numbers
12. Plotting points on a Polar Co-ordinate system.
Many systems and styles of measure are in common use today. When graphing on a
flat surface, the rectangular coordinate system and the polar coordinate system are
the two most popular methods for drawing the graphs of relations. Polar
coordinates are best used when periodic functions are considered. Although either
system can usually be used, polar coordinates are especially useful under certain
conditions.
The rectangular coordinate system is the most widely used coordinate system.
Second in importance is the polar coordinate system. It consists of a fixed point 0
called the pole, or origin. Extending from this point is a ray called the polar axis.
This ray usually is situated horizontally and to the right of the pole. Any point, P, in
the plane can be located by specifying an angle and a distance. The angle, θ, is
measured from the polar axis to a line that passes through the point and the pole. If
the angle is measured in a counterclockwise direction, the angle is positive. If the
angle is measured in a clockwise direction, the angle is negative. The directed
distance, r, is measured from the pole to point P. If point P is on the terminal side of
angle θ, then the value of r is positive. If point P is on the opposite side of the pole,
then the value of r is negative. The polar coordinates of a point can be written as an
ordered pair ( r, θ). The location of a point can be named using many different pairs
of polar coordinates. Figure 1 illustrates three different sets of polar coordinates for
the point P (4,50°).
Assignment #9 Qu 1
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Advanced Math 45S
Complex Numbers
13. Convert Rectangular Coordinates to Polar Coordinates and Vice Versa.
Rectangular Coordinates are in terms of x and y i.e. (x, y)
Polar Coordinates are in terms of r and θ i.e. (r, θ)
The following relationships between rectangular and polar coordinates can be used.
Converting from Rectangular to Polar
𝑟 = √𝑥 2 + 𝑦 2
𝑦
tan 𝜃 =
𝑥
Example: Convert the point rectangular coordinate (-2, 3) to polar form.
𝑟 = √(−2)2 + (3)2 = √13
3
3
tan 𝜃 =
, 𝜃 = tan−1
= −56.3°
−2
−2
Rectangular Coordinates (-2, 3) are the same as Polar Coordinates (√13, 303.7°)
Assignment 9 Qu 2
Converting from Polar to Rectangular:
x = r cos θ
y= r sin θ
Example: Convert the polar coordinates (5, π) to rectangular coordinate.
Note that in this example the angle has been given in radians. The value of the angle
in Polar coordinates can be given in degrees or radians so just be careful to make
sure your calculator is in the correct mode .
x = 5 cos π = -5
y = 5 sin π = 0
so the equivalent rectangular coordinates are (-5, 0)
Assignment 9 Qu 3
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Advanced Math 45S
Complex Numbers
14. Converting Equations from Rectangular Form to Polar Form and Vice
Versa
Example 1: Convert the equation x2 + y2 = 2y to polar form.
We know that x2 + y2 = r2 and y = r sin θ
So
x2 + y2 = 2y
becomes
r2 = 2r sin θ
then, divide both side by r
r = 2 sin θ
Example 2: Convert the equation r2 cos θ sin θ = 1 to rectangular form
𝑥
Since that x2 + y2 = r2 , cos 𝜃 = 𝑟 𝑎𝑛𝑑 sin 𝜃 =
𝑦
𝑟
r2 cos θ sin θ = 1
becomes
𝑥
𝑦
(𝑥 2 + 𝑦 2 ) ( ) ( ) = 1
𝑟
𝑟
𝑥𝑦
(𝑥 2 + 𝑦 2 ) ( 2 ) = 1
𝑟
(𝑥 2 + 𝑦 2 ) (
𝑥𝑦
𝑥 2 +𝑦 2
)=1
xy = 1
An alternative method would be
r2 cos θ sin θ = 1
(r cos θ)(r sin θ) = 1 but x = r cos θ and y = r sin θ
so
xy = 1
Assignment 10
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Advanced Math 45S
Complex Numbers
15. Graphing Equations in Polar form on a Polar Coordinate System
Lines:
Circles:
Cardioids:
Page 17 of 25
Advanced Math 45S
Complex Numbers
Archimedes Spiral:
Three Leaved Roses
No assignment
Page 18 of 25
Advanced Math 45S
Complex Numbers
16. Expressing a Complex Number in Polar Form
Our aim in this section is to write complex numbers in terms of a distance from the
origin and a direction (or angle) from the positive horizontal axis.
We find the real (horizontal) and imaginary
(vertical) components in terms of r (the length
of the vector) and θ (the angle made with the
real axis):
From Pythagoras, we have: r2 = a2 + b2
and basic trigonometry gives us:
tan θ = b/a
a = r cos θ
b = r sin θ
Multiplying the last expression throughout by i gives us:
bi=ir sinθ
So we can write the polar form of a complex number as:
a + bi=r cos r + ir sinθ
or
a + ib = r (cos θ+i sin θ)
r is the absolute value (or modulus) of the complex number
θ is the argument of the complex number.
Another ways of writing the polar form of a complex number:
r cis θ
This is really just a shorthand for r(cos θ+i sin θ)
Example: Convert 3 – 4i to polar form.
𝑟 = √32 + 42 = 5
3
cos 𝜃 = ,
5
3
𝑠𝑜 𝜃 = cos−1 ( ) ≈ 53°
5
Page 19 of 25
Advanced Math 45S
Complex Numbers
but this point is in the 4th quadrant so the angle θ= 360° - 53° = 307°
so
3 – 4i = 5(cos 307° + i sin 307°)
or
3 – 4i = 5 cis 307°
Assignment 12 Qu 1
17. Converting A Complex Number in Polar Form to Rectangular Form
Example: Convert
𝜋
𝜋
𝜋
1
3 (𝑐𝑜𝑠 3 + 𝑖 𝑠𝑖𝑛 3 ) into rectangular form
𝜋
3 (𝑐𝑜𝑠 3 + 𝑖 𝑠𝑖𝑛 3 ) = 3 (2 + 𝑖
=
√3
)
2
3 3√3
+
𝑖
2
2
Assignment 12 Qu 2
Page 20 of 25
Advanced Math 45S
Complex Numbers
18. Multiplying and Dividing Complex Numbers in Polar Form
Multiplication:
r(cos α + i sin α) × s(cos β + i sin β) = rs (cos (α+β) + i sin (α + β))
Proof:
r(cosα + isinα) × s(cosβ + isinβ)
= rs(cosαcosβ+isinαcosβ+isinβcosα + i2sinαsinβ)
= rs((cosαcosβ – sinαsinβ)+i(sinαcosβ+sinβcosα))
= rs (cos (α+β) + i sin (α + β))
Example: 3cis 100° × 5cis 150°
3cis 100° × 5cis 150° = 3(cos 100 + i sin 100) × 5(cos 150 + i sin 150)
= 15(cos 250°+i sin 250°)
= -5.13 -14.10i
Assignment 12 Qu 3
Division:
𝑟
r(cos α + i sin α) ÷ s(cos β + i sin β) = 𝑠 (cos (α - β) + i sin (α - β))
Proof:
𝑟(𝑐𝑜𝑠𝛼 + 𝑖𝑠𝑖𝑛𝛼) 𝑟(𝑐𝑜𝑠𝛼 + 𝑖𝑠𝑖𝑛𝛼) (𝑐𝑜𝑠𝛽 − 𝑖𝑠𝑖𝑛𝛽)
=
×
𝑠(𝑐𝑜𝑠𝛽 + 𝑖𝑠𝑖𝑛𝛽) 𝑠(𝑐𝑜𝑠𝛽 + 𝑖𝑠𝑖𝑛𝛽) (𝑐𝑜𝑠𝛽 − 𝑖𝑠𝑖𝑛𝛽)
𝑟(𝑐𝑜𝑠𝛼𝑐𝑜𝑠𝛽 − 𝑖𝑐𝑜𝑠𝛼𝑠𝑖𝑛𝛽 + 𝑖𝑠𝑖𝑛𝛼𝑐𝑜𝑠𝛽 − 𝑖 2 𝑠𝑖𝑛𝛼𝑠𝑖𝑛𝛽)
=
𝑠(cos 2 𝛽 − 𝑖 2 sin2 𝛽)
=
𝑟((𝑐𝑜𝑠𝛼𝑐𝑜𝑠𝛽 + 𝑠𝑖𝑛𝛼𝑠𝑖𝑛𝛽) + 𝑖(𝑠𝑖𝑛𝛼𝑐𝑜𝑠𝛽 − 𝑠𝑖𝑛𝛼𝑐𝑜𝑠𝛽))
𝑠(cos 2 𝛽 + sin2 𝛽)
𝑟
= (cos(𝛼 − 𝛽) + 𝑖𝑠𝑖𝑛(𝛼 − 𝛽))
𝑠
Page 21 of 25
Advanced Math 45S
Complex Numbers
Example: 3cis 150° ÷ 5cis 100°
3cis 150° ÷ 5cis 100° = 3(cos 150° + isin 150°) ÷ 5(cos 100° + isin 100°)
= 3/5(cos(150-100) + i sin(150-100))
= 3/5 (cos 50° + i sin 50°)
= 0.39 + 0.46i
Assignment 12 Qu 4
Page 22 of 25
Advanced Math 45S
Complex Numbers
19. Computing Powers and Roots of Complex Numbers using DeMoivre’s
Theorem.
DeMoivre’s Theorem:
[𝑟(cos 𝜃 + 𝑖 sin 𝜃)]𝑛 = 𝑟 𝑛 (cos 𝑛𝜃 + 𝑖 sin 𝑛𝜃)
Proof – involves induction – a mathematical method we haven’t done … yet!
Example 1: Evaluate and express in rectangular form
2𝜋
2𝜋 5
[3 (𝑐𝑜𝑠
+ 𝑖𝑠𝑖𝑛
)]
5
5
2𝜋
2𝜋 5
[3 (𝑐𝑜𝑠
+ 𝑖𝑠𝑖𝑛
)] = 35 (cos 2𝜋 + 𝑖 sin 2𝜋)
5
5
= 35 (1 + 0𝑖)
= 35
Assignment 12 Qu 5
Example 2: Evaluate the fourth roots of −8 + 8√3 𝑖
First convert −8 + 8√3 𝑖 to polar form.
2
𝑟 = √(−8)2 + (8√3) = √64 + 192 = √256 = 16
8√3
𝜃 = tan−1 (
) = 60°
8
but the angle is in quadrant 2 so the angle must be 180° – 60° = 120° and a general
term for all co-terminal angles is ( 120 +360k)° where 𝑘 𝜖 𝐼
so
−8 + 8√3 𝑖 = 16(cos(120 + 360𝑘)° + 𝑖 sin(120 + 360𝑘)°)
where 𝑘 𝜖 𝐼
Hence the question becomes evaluate
1
[16(cos(120 + 360𝑘)° + 𝑖 sin(120 + 360𝑘)°)] ⁄4
1
[16(cos(120 + 360𝑘)° + 𝑖 sin(120 + 360𝑘)°)] ⁄4 =
120 + 360𝑘 °
120 + 360𝑘 °
1⁄
4
16 (cos (
) + 𝑖 𝑠𝑖𝑛 (
))
4
4
Page 23 of 25
Advanced Math 45S
First Root: (k = 0)
Second Root: (k = 1)
Third Root: (k = 2)
Fourth Root: (k = 3)
Complex Numbers
2(cos 30° + 𝑖 𝑠𝑖𝑛 30° ) = √3 + 𝑖
2(cos 120° + 𝑖 𝑠𝑖𝑛 120° ) = −1 + √3𝑖
2(cos 210° + 𝑖 𝑠𝑖𝑛 210° ) = −√3 − 𝑖
2(cos 300° + 𝑖 𝑠𝑖𝑛 300° ) = 1 − √3𝑖
Thus the fourth roots of −8 + 8√3 𝑖 are √3 + 𝑖, −1 + √3𝑖, −√3 − 𝑖, 1 − √3𝑖
Insert diagram????
Assignment 12 Qu 6
Example 3: Solve x2 – 4x +10i = 0
Using the quadratic formula
𝑥=
4 ± √42 − 4(1)(10𝑖)
2(1)
4 ± √16 − 40𝑖
=
2
=
4 ± 2√4 − 10𝑖
2
= 2 ± √4 − 10𝑖
1
use DeMoivre’s theorem to determine (4 − 10𝑖)2
First write 4 – 10i in polar form.
𝑟 = √42 + 102 = √116 = 2√29
10
𝜃 = tan−1 ( ) ≈ 68°
4
angle in 4th quadrant = 360 − 68 = 292
4 − 10𝑖 = 2√29(cos 292° + 𝑖 sin 292)
Using DeMoivres
1
(2√29(cos 292 + 𝑖 𝑠𝑖𝑛292))2
1
2
= (2√29) (𝑐𝑜𝑠
1
292°
292°
+ 𝑖 sin
)
2
2
= (2√29)2 (cos 146 + 𝑖 sin 146)
= -2.72+1.84i
Page 24 of 25
Advanced Math 45S
Complex Numbers
so the solution to the quadratic equation becomes
x = 2 ± (-2.72 + 1.84i)
so
x = -0.72 + 1.84i, x = -4.72 + 1.84i
Assignment 12 Qu 7
Page 25 of 25