1
Expanded Course Notes on Complex Numbers from the Cambridge Text by Arnold and Arnold
By Mr Scovell
2
Complex Numbers
Explanation & Arithmetic of complex numbers.
Why do we need complex numbers?
The need for complex numbers arises when we solve equations that previously have had no solution.
Consider the solution to x2 – 4x + 13 = 0
x = 2
364
To solve this equation we would need to extend the number system to include numbers with negative
squares. To do this we let
i = 1 i2 = -1
Then every real number would have two square roots.
For example: – 4 could be written as 4 i2 then – 4 has two square roots, 2 i and - 2 i
We can now solve more complicated equations
Example: Solve x2 – 4x + 13 = 0
Solution
x =2
364
= 2
1364
=2
164
=2
64 i
= 2 + 3i or 2 – 3i
3
The notation of complex numbers.
Complex numbers are symbolized by the pronumeral z. They comprise of two parts:
z = x + iy where x,y are elements of ℝ
The Real part of z, written Re (z) = x
The Imaginary part of z, written Im (z) = y
Note the description „complex‟ does not imply non-real.
For example: the number 3 is a complex number and can be written as z = 3 + 0i
The number 3i is also a complex number z = 0 + 3i but is considered non real.
The following Venn diagram shows the relationship between:
ℂ ~ The set of complex numbers
ℝ ~ The set of real numbers
ℚ ~ The set of rational numbers
Z ~ The set of integers
M ~ The set of imaginary numbers
4
Arithmetic of complex numbers.
Equality
a + bi = c + di a = c b = d
Conjugate
If z = x + iy then the complex conjugate
z = x – iy
Examples: Find the complex conjugate of
I) 2 – 3i II) i III) 2
Solution
z = 2 + 3i z = 0 - i z = 2 – 0i
Addition z1 + z2
(a + bi) + (c + di) = (a + c) + i(b + d)
Example: If z1 = 3 + 2i and z2 = 4 – 3i evaluate z1 + z2
Solution
z1 + z2 = 3 + 2i + 4 – 3i
= 7 – i
Multiplication z1.z2
(a + bi) (c + di) = (ac - bd) + i(ad + bc)
Example: If z1 = 3 + 2i and z2 = 4 – 3i evaluate z1 z2
Solution
z1 z2 = (3 + 2i)(4 – 3i)
= 18 – i
5
Reciprocals & realising the denominator
The reciprocal of z is z
1 , usually the reciprocal is written in the form x + iy.
This involves a procedure similar to rationalising the denominator.
Example: If z = 4 + 3i find z
1 in the form a + ib
Solution
z
1 =
i34
1
= i34
1
i
i
34
34
= 2916
34
i
i
i
2 = -1 so
= 916
34
i
= 25
34 i
= 25
4 -
25
3i
Square roots of complex numbers
Example: If z2 = 3 + 4i find z
Solution
Let z = x + iy where x,y are ℝ
Then z2 = (x + iy)
2
(x + iy)2
= 3 + 4i
x2 + 2xyi + y
2i2 = 3 + 4i
x2 + 2xyi + y
2(-1) = 3 + 4i
6
x2 – y
2 + 2xyi = 3 +4i
Equating real and imaginary parts
x2 – y
2 = 3 -----1 and
2xy = 4 -----2
Solving simultaneously
in order to get a y2 to substitute into
4x2y
2 = 16
x2y
2 = 4
y2 =
2
4
x
Substitute y2 into
x2 -
2
4
x = 3
x4 – 4 = 3x
2
x4 – 3x
2 − 4 = 0
Let m = x2 then
m2 – 3m – 4 = 0
(m – 4)(m + 1) = 0
m = 4 or m = −1
x2 = 4 x
2 = −1
x = 2 x = i (but x is an ℝ so x i)
If x = 2 y = 1
x = -2 y = -1
z = 2 + i or z = -2 – i
7
Examples:
I If z1 = 2 + 3i and z2 = 1 + i find
a) Re (z1 +z2) b) Im (z1 2z )
Solution
z1 + z2 = (2 +3i) + (1+ i) z1 2z = (2 + 3i)(1 – i)
= 3 + 4i = 2 – 2i + 3i – 3i2
= 2 + i + 3
= 5 + i
Re (z1 +z2) = 3 Im (z1 2z ) = 1
II Show that zz = 2 2x y
Solution
Let z = x + iy
z = x – iy
z z = (x + iy)(x – iy)
= x2 – y
2i2
= x2 + y
2
III [HSC 2000 3 mks]
Find all pairs of integers x and y that satisfy (x + iy)2 = 24 + 10i
Solution
Let z = x + iy
z2 = 24 + 10i You can see that we are just finding the square root of 24 + 10i
Solving
(x + iy)2 = 24 + 10i
8
x2 + 2xyi + y
2i2 = 24 + 10i
x2 – y
2 + 2xyi = 24 + 10i
Equating real and imaginary parts
x2 – y
2 = 24 -------① and
2xy = 10 ------ ②
Rearranging ② to make y2 the subject
xy = 5
x2y
2 = 25
y2 =
2
25
x
Substituting y2 into ①
x2 -
2
25
x = 24
x4 – 25 = 24x
2
x4 – 24x
2 – 25 = 0
Let m = x2
m2 – 24m – 25 = 0
(m – 25)(m + 1) = 0
m = 25 or m = -1
x2 = 25 x
2 = -1
x = 5 x I since x ℝ
When x = 5 y = 1
x = -5 y = -1
i.e z = 5 + i or z = -5 – i
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IV [HSC 1990 2 marks]
Express i
z
53 in the from x + iy , where x and y are real.
Solution
Let z = a + ib
Then i
z
53 =
i
iba
53
=i
iba
53
i
i
53
53
= 2
2
259
5353
i
bibiaia
= 259
)53(53
abiba
= 34
)53(53 abiba
= 34
53 ba +
34
)53( ab i
V [HSC 2003 2 marks]
Let z = 2+ i and w = 1 – i. Find in the form x + iy,
Solution
(i) z w ii) z
4
z w = (2 + i)(1 + i) z
4 =
i2
4
= 2 + 2i + i + i2 =
i2
4
i
i
2
2
= 1 + 3i = 24
48
i
i
= 5
48 i
= 5
8-
5
4i
Students attempt CAMBRIDGE EXERCISE 21 on p. 31
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Geometrical representation of complex numbers.
The Argand diagram
Complex numbers can be plotted on an Argand diagram.
z = x + iy (x, y)
Example: If z = 3 + 2i , plot z and z on an Argand diagram
Solution
z = 3 + 2i A (3, 2)
z = 3 - 2i B (3, -2)
Radians (background knowledge).
Geometry uses degrees, minutes & seconds to measure an angle.
Another way to measure angles is in radians. A radian is the angle
an arc of 1 unit subtends at the centre of the circle of radius 1 unit.
π radians = 180
1 = radians180
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To convert degrees to radians:
Divide the angle (in degrees) by 180 (usually resulting in a fraction) and multiply by π.
Example: Convert to radians a) 30 b) -120
Solution
= 180
30 =
180
120
= 6
=
3
2
Polar co-ordinates.
The point A with the Cartesian co-ordinates (a, b) on an Argand diagram can also be specified by polar
co-ordinates (r, θ).
(r, θ) : r is the length the point is from the origin, r > 0
θ is the angle from the positive direction of the x-axis to the ray r.
For the point A to have unique polar co-ordinates (r, θ) it is necessary to restrict θ to 2π.
Otherwise points will be indistinguishable. For example (2, 6
) ≡ (2,
6
7). We will make the
restriction -π < θ ≤ π , as illustrated below .
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Examples: Plot on an Argand diagram the following polar co-ordinates:
I) (1, 4
)
II) (-2, 3
2)
III) (1, 3
)
IV) (2, 2
)
V) (1, 6
5)
Modulus – Argument form.
With regard to the polar co-ordinate (r, θ):
r is called the modulus of a complex number and is written │z│
│z│ = │x + iy│ = 22 yx = r
θ is called the argument (or principal argument) and is written as arg z
arg z = tan-1
x
y = θ
A
B
C
D
E
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Examples: Find │z│ and arg z of
Solution
I) z = 2 + 2i │z│ = 22 22
= 22
arg z = tan-1
2
2
= 45
= 4
II) z = √3 + i │z│ = 22
13
= 2
arg z = tan-1
3
1
= 30
= 6
III) z = -√3 + i │z│ = 22
13
= 2
arg z = π - tan-1
3
1
= 180 - 30
= π - 6
= 6
5
(2, 2)
2
2
1
√3
(√3, 1)
1
√3
(-√3, 1)
θ = arg z
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IV) z = 2 + 4i │z│ = 22 42
= 52
arg z = tan-1
2
4
= 6326‟ (in radians ??)
= tan-1
2
V) z = 1 - i │z│ = 22 )1(1
= 1
arg z = 0 - tan-1
1
1
= 0 - 45
= 4
VI) z = √3 i │z│ = 22
)1(3
= 2
arg z = -π + tan-1
3
1
= -180 + 30
= -π + 6
= 6
5
(4, 2)
4
2
(1, -1)
1
1
1
√3
(-√3, 1)
θ = arg z
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Modulus-Argument form.
From trigonometry
x = r cosθ
y = r sinθ
z = x + iy
= rcosθ + i.rsinθ
= r(cosθ + isinθ) [this is called modulus-argument form]
[can be abbreviated to rcisθ ]
where r = │z│ θ = arg z in radians
Example: [HSC 1997 2 marks] Express √3 – i in modulus-argument form
Solution
r = 22
)1(3
= 2
tanθ = 3
1
θ = 6
So, √3 – i = 2(cos
6
+ isin
6
)
arg z = 6
= 2(cos
6
- isin
6
) *
r
θ
y
x
(r, θ)
x
y
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Vectors.
A vector is a graphical representation of a magnitude and a direction.
Thus z = x + iy
= OP
Magnitude of OP = length OP
= │z│
Direction of OP = θ
= arg z
z as a vector can be written in Cartesian form (x + iy)
Modulus-argument form (rcisθ)
Examples: Sketch the following vectors on an Argand diagram
Solution
I) z = 3 – i II) z = 2 cis3
2
r
θ
y
x
(x, y)
x
y
0
P
y
P (3, -1)
x 0
x
0
y
Q
2
3
2
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Since a vector is just a magnitude and a direction these vectors can be reproduced by translation.
Consider z = 1 + i marked by OP below.
the vector OP ≡ CD ≡ AB
√2
1
1
x
y
0
P
√2
1
1
x
y
0
P
D (-1, -1)
C (-2, -2)
√2
√2
A (2, 1)
B (3, 2)
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Operations with vectors. “tip to tail”
Addition p + q = z
Subtraction p - q = z
Examples: If z1 = 3 – 2i and z2 = -1 + 4i sketch z1, z2, z1 + z2
Solution
z
Z
p
q
q p
p
q
- q
p
x
y
z1
z2
z1 + z2
z2 moved “tip to tail”
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In the above example, could you easily sketch a vector equivalent to z1 – z2 ?
Example: In the following diagram find vectors if OP = z
a) BP b) CP c) PA d) DP
Solution
Some other vectors we can easily find are
BO = 4 OC = 3i OA = 1
a) BP = BO + OP
= 4 + z
b) CP = OP – OC
= z – 3i
c) PA = OA – OP
= 1 – z
d) DP = DO + OP
= 3i + z
x
y
z1
z2
z1 - z2
x
y
A (1,0)
P
B (-4,0) O
C (0,3)
D (0,-3)
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Multiplication & division
Consider the following vectors & z1.z2
∆OAB Ⅲ ∆OCD arg z3 = arg z1 + arg z2
OA
OD=
OB
OC = arg (z1.z2)
1
3
z
z=
1
2z
│z3│ = │z2││z1│
= │z1.z2│
z3 = z1.z2
Rule: When multiplying ~ multiply moduli and add arguments
When dividing ~ divide moduli and subtract arguments
x
z3
1 B (1, 0)
z2
z1
A
C
D
y
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Example: If z1 = 1 + i and z2 = √3 – i
a) find 2
1
z
z
b) find
2
1argz
z
c) hence find the smallest positive integer n such that if z = i
i
3
1, zn is real and evaluate zn
Solution
a) │z1│ = 22 11 │z2│ = 2
2
)1(3
= √2 = 2
2
1
z
z =
2
1
z
z
= 2
2
b) arg z1 = tan-1
1
1 arg z2 = - tan
-1
3
1
= 4
= -
6
arg z =
2
1argz
z(from pt b) =
4
-
6
= 12
5
x
y
z1
z2
1
1
√3
1 Note the – sign as the arg is
below the x – axis.
Subtracting the arg‟s as two
vectors are being divided.
22
c) If zn is real then arg z
n = 0, π, 2π, 3π, … i.e graphically z
n is
= k π , where k = 0, ± 1, ± 2 …
Now arg zn = n arg z (prove later with de Moivre)
So k π = n 12
5 (from above)
n = k π 12
5
= 5
12k , where k = 0, ± 1, ± 2 …
Since n is an integer > 0 the first value of k that makes n > 0 is k = 5
n = 12
Because zn is real it lies on the x axis.
│z12
│ = │z│12
arg z12
= 5π
=
12
2
2
= 64
1
z12
= 64
1cis5π
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Triangular inequality
Draw any triangle ABC
What can you say about the relationship
between a,b & c?
a + b ≥ c a – b ≤ c [triangular inequality]
Example: If z1 = 3 + 4i and │z2│ = 13
a) find the greatest value of │z1 + z2│.
b) If │z1 + z2│ takes its greatest value, express z2 in the form a + i b
Solution
a) │z1│ + │z2│ ≥ │z1 + z2│ (triangle inequality)
│z1│ = 22 43
= 5
So, 5 + 13 ≥ │z1 + z2│
18 ≥ │z1 + z2│
thus a maximum value of 18 is obtained.
b) │z2│= k│z1│, where k ∈ ℝ z2 = k.z1
13 = k.5 = 5
13(3 + 4i)
k = 5
13 =
5
39 +
5
42i
(when │z1 + z2│ is a max)
A
B C
c
a
b
Students attempt CAMBRIDGE EXERCISE 22 & 23 on p. 43 & 54
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Powers & roots of complex numbers.
Example: Prove De Moivre’s theorem by mathematical induction
Solution
S(n): (cosθ + i sinθ)n = cosnθ + i sinnθ), n = 1, 2, 3, …
S(1): (cosθ + i sinθ)1 = cosθ + i sinθ TRUE
If S(k) is true, (cosθ + i sinθ)k = coskθ + i sinkθ
S(k + 1):
(cosθ + i sinθ)k + 1
= (cosθ + i sinθ)k (cosθ + i sinθ)
= (coskθ + isinkθ) (cosθ + i sinθ)
= 2cos cos cos sin cos sin sin sink i k i k i k
= cos cos sin sin cos sin cos sink k i k k
= cos sink i k
= cos(k + 1)θ + i sin(k + 1)θ TRUE
If S(1) is True, S(k) is ture & S(k + 1) is true then
(cosθ + i sinθ)n = cosnθ + i sinnθ) by induction
De Moivre’s theorem (cosθ + i sinθ)n = cosnθ + i sinnθ
i.e. show (cosθ + i sinθ)k + 1
= cos(k + 1)θ + i sin(k + 1)θ
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Example: Show that nn zz , where n is an integral
Solution
Let z = r(cosθ + i sinθ)
then z = r(cosθ – i sinθ)
zn = nr (cosθ + i sinθ)
n
nz = nr (cosθ − i sinθ)n
LHS nz = nr (cosθ − i sinθ)
n
= nr (cosnθ − i sinnθ) [De Moivre‟s theorem]
RHS nz = r
n(cosθ - i sinθ)
n
= rn
(cosnθ - i sinnθ) [De Moivre‟s theorem]
LHS = RHS
26
Example: Express (√3 + i)8 + (√3 - i)8 in the form a + i b
Solution
Let z = √3 + i z = √3 – i
Now │z│ = 22
13 arg z = tan-1
3
1
= 2 = 30
= 6
So z = r (cosθ + isinθ) z = r (cosθ – i sinθ)
= 2 (cos6
+ isin
6
) = 2 (cos
6
– i sin
6
)
Now z8 = 2
8 (cos
6
+ i sin
6
)8
= 2
8 (cos
3
4 + i sin
3
4) [De Moivre‟s theorem]
And from previous example nn zz
So z8 = 8z = 2
8(cos
6
- i sin
6
)8
= 28 (cos
3
4 - i sin
3
4) [De Moivre‟s theorem]
Now (√3 + i)8 + (√3 - i)
8 = z
8 + z
8 = 2
8 (cos
3
4 + i sin
3
4) + 2
8 (cos
3
4 - i sin
3
4)
= 28 cos
3
4 + 2
8 cos
3
4
= 256 - 05 + 256 - 05
= - 256 + 0i
1
√3
(√3, 1)
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Example: a) By expressing cos4θ, sin4θ in terms and powers of cosθ and sinθ show that
tan4θ =
42
3
tantan61
tan4tan4
Solution
Let z = cosθ + isinθ
z4 = (cosθ + isinθ)
4
= cos4θ + isin4θ [De moivre‟s theorem]
and z4 = (cosθ + isinθ)
4 =
n
k 0
nCk an-k bk
=
4
0k
nCk (cosθ)
n-k (i sinθ)
k [polynomial expansion]
= 4C0(cosθ)
4 (i sinθ)
0 +
4C1(cosθ)
3 (i sinθ)
1 +
4C2(cosθ)
2 (i sinθ)
2 +
4C1(cosθ)
1 (i sinθ)
3 +
4C4(cosθ)
0 (i sinθ)
4
= cos4θ + 4cos
3θ i sinθ + 6cos
2θ i
2sin
2θ + 4cosθ i
3 sin
3θ + i
4 sin
4θ
= cos4θ + 4cos
3θ i sinθ + 6cos
2θ (-1)sin
2θ + 4cosθ (-1i) sin
3θ + (+1)sin
4θ
= cos4θ + 4icos
3θsinθ - 6cos
2θ sin
2θ - 4icosθsin
3θ + sin
4θ
= cos4θ - 6cos
2θ sin
2θ + sin
4θ + i(4cos
3θsinθ - 4cosθsin
3θ) [in the form x + i y ]
Equating real and imaginary parts
cos4θ = cos4θ - 6cos
2θ sin
2θ + sin
4θ and
sin4θ = 4cos3θsinθ - 4cosθsin
3θ
tan4θ =
4224
33
sinsincos6cos
sincos4sin4cos
=
42
3
tantan61
tan4tan4 when divided thru by cos
4θ
28
b) Hence solve the equation t4 + 4t3 – 6t2 – 4t + 1 = 0
Solution
Rearranging the equation
t4 + 4t
3 – 6t
2 – 4t + 1 = 0
t4 – 6t
2 + 1 = 4t − 4t
3 dividing thru by LHS
1 = 16
4424
3
tt
tt
Let t = tanθ then
16
4424
3
tt
tt =
42
3
tantan61
tan4tan4 = tan4θ = 1
solving tan4θ = 1 let v = 4θ
tan v = 1
= 4
,
4
5,
4
9,
4
13, …
But v = 4θ = 4
,
4
5,
4
9,
4
13, …
θ = 16
,
16
5,
16
9,
16
13, … finding a general rule
θ = 16
)14( k where k = 0, 1, 2, 3 …
29
Examining solutions remembering that the solution t = tanθ
k 0 1 2 3 4 5 6
θ
16
16
5
16
9
16
13
16
17
16
21
16
25
t 019… 149… -502… -066… 019… 149… -502…
We can see that for values above k = 3 the solution repeats.
Hence the solutions are
t = tan16
, tan
16
5, tan
16
9, tan
16
13
30
Example: Show that zn + z -n = 2cosnθ
Let z = cosθ + i sinθ
zn = (cosθ + i sinθ)
n
= cosnθ + i sinnθ [De Moivre‟s theorem]
z -n
= (cosθ + i sinθ)-n
= ni )sin (cos
1
= nin sincos
1
= nin sincos
1
nin
nin
sincos
sincos realizing the denominator
=
nin
nin22 sincos
sincos
= cosnθ - i sinnθ
So zn + z
–n = cosnθ + i sinnθ + cosnθ - i sinnθ
= 2cosnθ
{students can prove z n
– z –n
= 2i sinnθ }
31
Example: Show that sin3θ
= ¼ (3sinθ – sin3θ) *reasonably hard question
Method 1. Using the above results
Let z = cosθ + i sinθ
z1 – z
–1 = 2i sinθ [from identity z
n – z
–n = 2i sinnθ ]
(z1 – z
–1)3 = (2i sinθ)
3
= 8i3 sin
3θ
= - 8i sin3θ
z3 – z
–3 = 2i sin3θ [from identity z
n – z
–n = 2i sinnθ ]
Now (z1 – z
–1)3 =
n
k 0
nCk an-k bk
=
3
0k
nCk (z)
n-k (-z
-1)k [polynomial expansion]
= 3C0 (z)
3 (-z
-1)0 +
3C1 (z)
2 (-z
-1)1 +
3C2 (z)
1 (-z
-1)2 +
3C3 (z)
0 (-z
-1)3
= z3 - 3 z
2 z
-1 + 3 z
1 z
-2 - z
–3
= z3 - 3z + 3 z
-1 - z
–3
= z3 - z
–3 - 3z + 3 z
-1
= z3 - z
–3 - 3(z - z
–1)
Substituting the identities above
(z1 – z
–1)3 = z
3 - z
–3 - 3(z - z
–1)
- 8i sin3θ = 2i sin3θ - 3(2i sinθ)
- 8i sin3θ = 2i sin3θ - 6i sinθ dividing through by –8i
sin3θ =
i
i
i
i
8
sin6
8
3sin2
32
sin3θ =
4
sin3
4
3sin
sin3θ =
4
1(-sin3θ + 3sinθ)
= 4
1(3sinθ – sin3θ) = RHS
Method 2. Mr. Scovell’s way
Note: Use this (as it is the most consistent method in attacking problems) way unless
Asked to prove z n
– z –n
= 2i sinnθ or zn + z
-n = 2cosnθ in a previous question
You encounter a problem (see question re: 16cos4θ = 2cos4θ + 8cos2θ + 6)
The method illustrated below is consistent with HSC questions. See Q.2 part d) 2003 HSC.
Let z = cosθ + i sinθ
z3 = (cosθ + i sinθ)
3
= cos3θ + i sin3θ [De Moivre‟s theorem]
also
(cosθ + i sinθ)3 =
n
k 0
nCk an-k bk
=
3
0k
nCk (cosθ)
n-k (i sinθ)
k [polynomial expansion]
= 3C0(cosθ)
3 (i sinθ)
0 +
3C1(cosθ)
2 (i sinθ)
1 +
3C2(cosθ)
1 (i sinθ)
2 +
3C1(cosθ)
0 (i sinθ)
3
= cos3θ + 3cos
2θ i sinθ + 3cosθ i
2sin
2θ + i
3 sin
3θ
= cos3θ + 3cos
2θ i sinθ - 3cosθ sin
2θ - i sin
3θ
= cos3θ - 3cosθ sin
2θ + i 3cos
2θsinθ - i sin
3θ
= cos3θ - 3cosθ sin
2θ + i (3cos
2θsinθ - sin
3θ)
33
So cos3θ + i sin3θ = cos3θ - 3cosθ sin
2θ + i (3cos
2θsinθ - sin
3θ)
Equating imaginary parts
sin3θ = 3cos2θsinθ - sin
3θ
sin3θ = 3cos
2θsinθ – sin3θ
= 3(1 – sin2θ)sinθ – sin3θ
= 3sinθ – 3sin3θ - sin3θ
4sin3θ = 3sinθ – sin3θ
sin3θ =
4
1(3sinθ – sin3θ) = RHS
Students can attempt showing 16cos4θ = 2cos4θ + 8cos2θ + 6
Note: Mr. Scovell‟s method will fail here due to 8cos2θ which won‟t occur in the expansion.
34
Roots of unity & complex roots
Discussion
If z = rcisθ
zn = r
ncisnθ
What about the reverse situation; given zn find z ?
Example: Solve z3 = 1
Let z = r(cosθ + i sinθ)
z3 = r
3(cosθ + i sinθ)
3
= r3 cos3θ + i sin3θ
r3 cos3θ + i sin3θ = 1
r3 cos3θ + i sin3θ = 1 + 0 i
Now │z3│ = 22 01
= 1
│z3│ = 1 then
│z│ = 1
Arg z3 = 0
This means z3 = cos3θ + i sin3θ = cos0 + i sin0
Equating real & imaginary parts
cos3θ = 1 and sin3θ = 0
Solving for θ
3θ = 0, 2π, 4π, 6π, 8π, …
35
θ = 0, 2
3
,
4
3
, …
Finding a general rule
θ = 3
2 k, where k = 0, 1, 2, 3, …
[Note that there a 3 distinct k values to correspond with 3 distinct solutions. We know there are 3
solutions because we are solving z3]
Now substituting what we know to evaluate the solutions of z
│z│ = r = 1 θ = 3
2 k
z = r(cosθ + i sinθ)
z = cos3
2 k + i sin
3
2 k
When k = 0
z1 = cos 0 + i sin0
= 1
When k = +1
z2 = cos3
2 + i sin
3
2
= 2
1 +
2
3i [In the form x + iy]
z1
1
z2
2
3
2
1
3
2 = 120
36
When k = - 1
z3 = cos3
2 + i sin
3
2
= 2
1 -
2
3i
[Note that if you had let k = 0, 1, 2 then]
For k = 2
z = cos3
4 + i sin
3
4
= 2
1 -
2
3i
= z3
Also note that for z3 = 1 there are the equally spaced solutions around the unit circle, namely
1, cos3
2 + i sin
3
2, cos
3
2 + i sin
3
2
z3
2
3
2
1
3
2 = 120
z3
z2
z1
3
2
37
Example: If w is a non-real cube root of unity show that 1 + w + w2 = 0
w3 = 1
w3 – 1 = 0
(w – 1)(w2 + w + 1) = 0
So w – 1 = 0 or w2 + w + 1 = 0
Since w is a NON-REAL root
w ≠ 1 w2 + w + 1 = 0
Example: a) Find the values for which z6 = -64
Let z = r(cosθ + i sinθ)
z6
= r6(cos6θ + i sin6θ) [De Moivre‟s theorem]
z6
= r6(cos6θ + i sin6θ) = - 64 + 0i
│z6│ = 22 064
= 64 = r6
│z│ = 6 64
= 2 = r
Arg z6 = π
Rearranging - 64 + 0i into mod-arg form
z6
= r6(cos6θ + i sin6θ) = 64 (cosπ + i sinπ)
6θ = π, 3π, 5π, …
θ = 6
,
6
3,
6
5, …
arg z6 = π
z6
38
Finding a general rule
θ = 6
+
6
2 k
= 6
)12( k where k = 0, ±1, ±2, - 3
Substituting into z = r(cosθ + i sinθ)
r = 2 θ = 6
)12( k where k = 0, ±1, ±2, - 3
z = 2(cos6
)12( k + i sin
6
)12( k)
When k = 0,
z1 = 2(cos6
+ i sin
6
)
= 2(2
3 +
2
1 i )
= i3
When k = -1
z2 = 2(cos6
+ i sin
6
)
= i3
When k = 1
z3 = 2(cos6
3 + i sin
6
3)
= 2i
39
When k = -2
z4 = 2(cos6
3 + i sin
6
3)
= - 2i
When k = 2
z5 = 2(cos6
5 + i sin
6
5)
= - i3
When k = -3
z6 = 2(cos6
5 + i sin
6
5)
= - i3
Plotting the points
Note: z2 = 1z z4 = 3z z6 = 5z
z1
z2
z3
z4
z5
z6
40
b) Express z6 + 64 = 0 as a product of quadratic factors
z6 + 64 = 0
(z – z1)(z – z2)(z – z3)(z – z4)(z – z5)(z – z6) = 0
From above: z2 = 1z z4 = 3z z6 = 5z
[(z – z1)(z – 1z )] [(z – z3)(z – 3z )] [(z – z5)(z – 5z )] = 0
[z2 – z.z1 - z 1z + z1 1z ] [z
2 – z.z3 - z 3z + z3 3z ] [z
2 – z.z5 - z 5z + z5 5z ] = 0
[z2 – z(z1 + 1z ) + z1 1z ] [z
2 – z(z3 + 3z ) + z3 3z ] [z
2 – z(z5 - 5z ) + z5 5z ] = 0
From above z1 = i3 z3 = 2i z5 = - i3
1z = i3 3z = -2i 5z = - i3
So
[z2 – z( i3 + i3 ) + ( i3 )( i3 )][z
2 – z(2i + -2i) + (2i.-2i)][z
2 – z(- i3 + - i3 ) +
(- i3 ).( - i3 )] = 0
[z2 – 2 3 z + 3 - i
2] [z
2 - 4i
2] [z
2 +2 3 z + 3 - i
2] = 0
[z2 – 2 3 z + 4] [z
2 + 4] [z
2 +2 3 z + 4] = 0
Note: Some questions are more easily answered or require an answer using modulus-argument form
for z1, z2, z3
41
Example: Factorise z5 – 1 into real linear and quadratic factors.
First step would be to solve z5 – 1 = 0
The working for the solutions is not shown here {students can solve for themselves as practice later ~
also in Ex 24 as a q}.
The 5 solutions would be
z1 = 1 when k = 0
z2 = cos5
2 + i sin
5
2 when k = +1
z3 = cos5
2 + i sin
5
2 when k = -1
z4 = cos5
4 + i sin
5
4 when k = 2
z5 = cos5
4 + i sin
5
4 when k = -2
Note: z3 = 2z z5 = 4z
Now z5 – 1 = (z – z1)(z – z2)(z – z3)(z – z4)(z – z5)
= (z – z1)(z – z2)(z – 2z )(z – z4)(z – 4z )
= (z – z1)[z2 – z(z2 + 2z ) + z2 2z ] [z
2 – z(z4 + 4z ) + z4 4z ]
Note the identities: In x + i y [we could have used these identities in the previous Q]
z + z = 2x z. z = x2 + y
2 = 1 in the case where r = 1
r2 otherwise
= (z – 1)[z2 – z(2 cos
5
2 ) +1] [z
2 – z(2 cos
5
4) + 1]
Students attempt CAMBRIDGE EXERCISE 24 on p. 60
42
Curves & regions in the Argand diagram.
Discussion.
Consider z to be a complex number that satisfies Re (z) = 3
On the Argand diagram Re z would be a locus as illustrated below.
All the dotted lines represent the vectors z where Re z = 3. There are infinite amount of vectors
terminating along the bold line.
Sketching a curve
Example: a) Sketch the curve defined by the equation Im (z – 1 + 3i) = 4
Let z = x + iy
Then z – 1 + 3i = x + iy - 1 + 3i
= x - 1 + i( y + 3)
Since Im (z – 1 + 3i) = 4
( y + 3) = 4
y = 1
All the dotted vectors z have Im z = 1 and terminate along the bold line. You don‟t have to draw the
dotted vectors. They are there simply to illustrate the concept.
│z│
1 P
3
43
b) Find the minimum value of │z│ for which Im (z – 1 + 3i) = 4 occurs and state the vector z
for this condition.
z will have a minimum value when z is perpendicular to y = 1
z = 0 + i
= i
Example: Sketch the region in the Argand diagram defined by
6 ≤ Re [(2 – 3i) z ] < 12 and Re z . Im z ≥ 0
Let z = x + iy
Then [(2 – 3i) z ] = (2 – 3i)( x + iy)
= 2x + 2iy – 3ix – 3i2y
= 2x + 3y + i (2y – 3x)
So Re [(2 – 3i) z ] = 2x + 3y
We need to graph 6 ≤ 2x + 3y < 12
And the intersection of
Re z . Im z = x . y
Which is the graph of x . y ≥ 0
[Note xy ≥ 0 are all points in the first quadrant]
│z│
1
2x +3y = 6
2
3
2x +3y = 12
The shaded area represents
6 ≤ 2x + 3y < 12
and the intersection with
x.y ≥ 0
44
Example: z satisfies │z – 1 + 2i│ = │z + 3│
a) Sketch the locus of the point P representing z
ALGEBRAIC SOLUTION
Let z = x + iy
Then │z – 1 + 2i│ = │z + 3│ becomes
│ x + iy – 1 + 2i│ = │ x + iy + 3│
│x – 1 + i(y + 2)│ = │x + 3 + iy│
Finding the modulus of each side
22 )2()1( yx = 22)3( yx
(x – 1)2 + (y + 2)
2 = (x + 3)
2 + y
2
x2 – 2x + 1 + y
2 + 4y + 4 = x
2 + 6x + 9 + y
2
4y = 8x + 4
y = 2x + 1
GEOMETRICAL SOLUTION
Let the point P be the locus of z
Rearranging │z – 1 + 2i│ = │z + 3│
│z – (1 - 2i)│ = │z – (- 3)│
│ z – z1 │ = │ z – z2│
Geometrically this means that the distance of z
from the point (1, -2) is equal the distance from
the point (-3, 0).
Since z is equidistant from the two points the locus of z
must be the perpendicular bisector of the two points.
Note: the vectors z – z1 & z – z2
Start from the respective points and end along the bold line.
If they have trouble with the above geometry- can use below to reinforce the geometrical concepts.
1
x
y
x
y
(1, -2)
-3
P
z – z2
z – z1
45
z starts from the origin and terminates in infinite places along the dotted line
z2 starts from the orgin and ends at the point (-3, 0)
z – z2 is found with vector subtraction
b) Find the minimum value of │z│
The minimum value of │z│ is the perpendicular distance from the origin to the locus.
d = 22
11
ba
cbyax
= 22 )1(2
1)0(1)0(2
= 5
1 units
x
y
-3
P
z – z2
z2
z
x
y
-3
P
z2
z z – z2
x
1
y
46
Discussion
Sketch the region defined by │z│ ≤ 2
The shaded region defines all vectors with a modulus ≤ 2
Example: If z satisfies │z – 2 – 2i│ = 2
a) Sketch the locus of point P representing z on an Argand diagram.
b) Find the maximum value for │z│
c) Find the range of values for arg z
ALGEBRAIC SOLUTION
a) Let z = x + iy
│z – 2 – 2i│ = 2
│x + iy – 2 – 2i│ = 2
│x – 2 + i(y – 2 )│ = 2
z z
z
2 -2
-2
2
z z
47
│ x – 2 + i(y – 2 )│= 22 )2()2( yx = 2
(x – 2)2 + (y – 2)
2 = 2
This is a circle with centre (2, 2) and a radius of 2 units
GEOMETRICAL SOLUTION
Let z = 2 +2i
│z – 2 – 2i│ = 2
│z – (2 + 2i)│ = 2
│z – z1│ = 2
Constructing the answer in parts
z1 is a vector starting at the origin and
terminating at (2, 2)
Since we are dealing with the modulus │z – z1│
(2, 2)
x
y
2 zmax
z
zmin
(2, 2)
z1
48
y
(2, 2)
x
2
zmax
1
1
(3, 3)
The vector z – z1 has a distance of 2 units
z – z1 will be a vector originating at (2, 2) and meeting vector z 2 units away (in all directions).
b) The maximun value for │z│
z has a maximum modulus when z
is furthest from the origin. Remember z
starts at the origin and terminates at any
point on the circle. The maximum
value of │z│ occurs at the point (3, 3).
So z = 3 + 3i
(2, 2)
x
y
│z – z1│= 2
z
z1
z – z1
49
c) The range of values for arg z is found by COD ≤ arg z ≤ AOD
Finding COD & AOD
OB = 2 + 2 = 22
BOD = 45
= 4
AOB = BOC = sinα = 22
2
= 2
1
= 6
Now COD = BOD – BOC (α )
= 6
-
4
= 12
And AOD = BOD + AOB
= 6
+
4
= 12
5
the range of values for arg z is 12
≤ arg z ≤
12
5
x
(2, 2)
y
A
z
z1
2
B
C
0
z
2
D
22
α
50
Discussion
Consider the Argand diagram below and the equation of the locus z
arg z = 3
The gradient of OP is = tan3
= - 3
the equation of the locus is y = - 3 x where x >0 [in the form y = mx + b]
Note since the locus has an argument z ≠ 0 as the vector z = 0 has no argument.
Example: Shade in the region in the Argand diagram defined by
4
≤ arg z ≤
2
OR* │z│< 2
* Note OR and not AND. This means you shade in the union of both regions and not the intersection.
z
-3
P
O
4
2
2
-2
-2
51
Example: Sketch the locus
a) arg (z – 1 – 2i) = 3
When dealing with arguments a GEOMETRICAL SOLUTION is sometimes easier.
GEOMETRICAL SOLUTION
arg (z – 1 – 2i) = 3
arg (z – [1 + 2i]) = 3
Let z1 = 1 + 2i [see z1 below]
arg (z – z1) = 3
Note the point from which the locus
originates is (1, 2). The locus then
heads in the direction of 3
.
Note that it does not include the
point A
z1 z - z1
3
z
z
A
52
b) i) sketch arg (z – 1 – 2i) = 6
arg (z – 1 – 2i) = 6
arg (z – [1 + 2i]) = 6
Let z1 = 1 + 2i [see z1 below]
arg (z – z1) = 6
ii) What is the Cartesian equation of the locus?
We know a point (1, 2) and the gradient can be found using
m = tanα y – y1 = m(x – x1)
= tan6
y – 2 =
3
1(x – 1)
= 3
1 y =
3
1x -
3
1+ 2
y = 3
1x +
3
132 where x > 1
z1
z - z1
6
z
A
53
Example: z is a complex number that satisfies
2 ≤ │z + 3│≤ 3 and 0 ≤ arg (z + 3) ≤ 3
Find the area & perimeter of the region so formed by these conditions.
GEOMETRICAL SOLUTION
2 ≤ │z + 3│≤ 3
2 ≤ │z - (- 3)│≤ 3
Let z1 = - 3
2 ≤ │z – z1│≤ 3 is a circle with centre (-3, 0) and radii 2 & 3 respectively
0 ≤ arg (z + 3) ≤ 3
0 ≤ arg (z - z1) ≤ 3
is the direction of the vector z – z1 from the point (-3, 0)
(-3, 0)
3
-1 0
54
Area = 6
1π (R
2 – r
2) Perimeter =
6
12πR +
6
12πr + 1 + 1
= 6
(3
2 – 2
2) =
6
12π(3 + 2) + 2
= 6
5 units
2 = 2 +
3
5 units
Example: If z satisfies arg (z + i) = arg (z – 1), sketch the locus of P representing z in the
Argand diagram. [reasonably hard question]
GEOMETRICAL SOLUTION
arg (z + i) = arg (z – 1)
The vectors (z + i) starts from the point (0, -1)
(z – 1) starts from the point (1, 0)
The vector z could terminate at the points P1, P2, & P3 however…
A (0, -1)
B (1, 0) O
P1
z
P2
P2
-i
+1
55
Case ① P1
If P1 represents z then
AP = z + i
BP = z – 1
And arg AP = arg BP [both vectors head is a positve direction]
[the direction itself does not matter so long as it is the same direction for both vectors]
The interval containing P1 is a possible locus of arg (z + i) = arg (z – 1)
Case ② P2
If P2 represents z then
AP = z + i
BP = z – 1
But arg AP ≠ arg BP [the vectors arguments are opposite]
[AP is in a positive direction while Bpis in a negative]
The interval with P2 can not be a possible locus of arg (z + i) = arg (z – 1)
Case 3 P3
If P3 represents z then
AP = z + i
BP = z – 1
And arg AP = arg BP [both vectors head is a negative direction]
The interval containing P3 is a possible locus of arg (z + i) = arg (z – 1)
Students attempt CAMBRIDGE EXERCISE 25 on p. 71
56
Formula test.
57
58
59
60