Download Senior Math Circles Whi so Comple x Part IA complex

Faculty of Mathematics
Waterloo, Ontario N2L 3G1
Centre for Education in
Mathematics and Computing
Senior Math Circles
March 23, 2016
Whi so Complex Part I
Alain Gamache
A complex situation
−1 = i2
= i×i
√ √
−1 −1
=
√
=
−1 × −1
√
=
1
= 1
Sets as we know them
These sets can be thought of as helping us to solve polynomial equations. However, x2 +1 = 0
has no solution in any of these sets.
1
Defining a complex number
A complex numbers (in standard form) is an expression of the form x + yi where x, y ∈ R
and i is the imaginary unit. We often use the letter z to express a complex number.
Examples :
• 5 − 6i
•
√
3−i
• −8
•
2i
5
A new set!
How to represent a complex number?
We know that every point on the number line represents a real number. But how can we
represent 3 + 4i?
Solution: we plot z = x + yi as a point (x, y) in a cartesian plane, called complex plane,
where the x − axis is the real axis and the y − axis is the imaginary axis. Such a diagram
is called an Argand Diagram.
2
The Argand Diagram
The Modulus
The modulus (or absolute value) of a complex number z = x + yi is the nonnegative real
number :
p
|z| = |x + yi| = x2 + y 2
Compute the modulus of the following complex numbers:
1. z1 = 3 + 4i
√
√
Solution: |z1 | = 32 + 42 = 25 = 5
2. z2 = −4 − 2i
p
√
√
Solution: |z2 | = (−4)2 + (−2)2 = 20 = 2 5
3
What is the geometrical interpretation of the modulus?
The Complex Conjugate
The complex conjugate of a complex number z = x + yi is the complex number :
z = x − yi
Detemine the conjugate of each of those complex numbers:
1. z1 = 3 + 4i
Solution: z1 = 3 − 4i
2. z2 = −4 − 2i
Solution: z2 = −4 + 2i
4
What is the geometrical interpretation of the complex conjugate?
Complex Arithmetic
Let z = a + bi and w = c + di. We define :
• z + w = (a + bi) + (c + di) = (a + c) + (b + d)i
• z − w = (a + bi) − (c + di) = (a − c) + (b − d)i
• z · w = (a + bi) · (c + di) = (ac − bd) + (ad + bc)i
Using the definition of the multiplication, show that i2 = −1
i2 = (0 + i)(0 + i) = (0 · 0 − 1 · 1) + (0 · 1 + 1 · 0) = −1 + 0i = −1
Hence i is one complex square root of −1. The other square root of −1 is −i.
5
Examples : Let z1 = 3 + 4i and z2 = −4 − 2i, express each of the following in standard
form:
1. z1 + z2
(3 + 4i) + (−4 − 2i) = (3 − 4) + (4 − 2)i = −1 + 2i
2. z1 − z2
(3 + 4i) − (−4 − 2i) = (3 + 4) + (4 + 2)i = 7 + 6i
3. z1 z2
(3+4i)(−4−2i) = (3(−4)−4(−2))+(3(−2)+4(−4))i = (−12+8)(−6−16)i = −4−22i
4. z1 z22
(z1 z2 )z2 = (−4 − 22i)(−4 − 2i) = (16 − 44) + (8 + 88)i = −28 + 96i
Consider :
3 + 4i
−4 − 2i
What does it mean ?
3 + 4i −4 + 2i
−20 − 10i
−2 − i
·
=
=
−4 − 2i −4 + 2i
20
2
Complex KenKen
4×
3+
i–
1+ i –
i–
4×
3+
1– i –
i–
2×
8×
2×
i–
(d)
(e)
Solve using 1,
2, 1 + i, 1 − i
Figure 4. Solve using the four numbers 1, 1 + i, 1 − i, and 2.
Source: The College Mathematics Journal, Volume 43, Number 1, January 2012,
pp. 37-42(6)
1 + i and 1 − i since (1 + i) − (1 − i) = 2i. So, the only options for this cage are for
it to contain 1 and 1 + i (since (1 + i) − 1 = i), or for it to contain 1 and 1 − i (since
1 − (1 − i) = i). Similarly, the ‘1 + i−’ cage must contain a 2 (since, otherwise, the
1s completely cancel during subtraction). Thus, the only possibility is for this cage
to contain 2 and 1 − i (since 2 − (1 − i) = 1 + i). The Gaussian integers, as lattice
points in the complex plane, are also vectors. For example, 1 + i can be thought of as
the vector from the origin (0, 0) to (1, 1). Geometrically, the difference α − β between
any two complex numbers
6 α and β is just the vector from point β to point α. Thus, in
puzzle (d), it is easy to spot which pairs of numbers can go in the ‘i−’ cage because the
i vector can only be drawn from (1, 0) to (1, 1), or from (1, −1) to (1, 0); similarly,
we can determine the pair of numbers that must go in the ‘1 + i−’ cage merely by
observing that we can only draw a 1 + i vector from (1, −1) to (2, 0).
Polar Coordinates
A point in the plane corresponds to a length and an angle:
π
Example : Let P (3, ). What are the cartesian, (x, y) coordinates of P ?
4
√ !
√
π
π
3 2 3 2
P (x, y) = P 3 cos , 3 sin
=P
,
4
4
2
2
Polar Coordinates in the Complex Plane
What would be the polar coordinate P (r, θ) for the complex number z = −1 + i?
r=
p
√
(−1)2 + 12 = 2
θ = arctan
1
−1
= arctan(−1) =
7
3π
4
Complex Numbers in Polar Form
This polar representation gives another way of expressing a complex number.
Let z = x + yi, then:
z = x + yi
= r cos θ + r sin θi
= r(cos θ + i sin θ)
z = r(cos θ + i sin θ), r > 0 is called the polar form of z, where r is the modulus of z and
the angle θ is called the argument of z. Express each complex number in polar form:
1. z = i
√
r = 12 + 02 , θ =
z = cos π2 + i sin π2
π
2
2. w = −4
r = 16, θ = π
w = 16(cos π + i sin π)
√
3. v = 3 − 3i
√
√
√
r = 3 + 9 = 12 = 2 3, θ =
√
v = 2 3 cos 5π
+ i sin 5π
3
3
5π
3
Comparing Complex Numbers
For two complex numbers z = x + yi = r1 (cos θ + i sin θ) and w = a + bi = r2 (cos φ + i sin φ),
z = w only if :
• x = a and y = b or
• r1 = r2 and θ = φ + 2πk, k ∈ Z
Example : For which value of a and b is the following true:
(a + b)(5 − i) = 10a + b + 1 − (2a + 1)i
5a + 5b − ai − bi = 10a + b + 1 − (2a + 1)i
5a + 5b − (a + b)i = 10a + b + 1 − (2a + 1)i
8
Multiplication of Complex Numbers in Polar Form
π
2π
π
Let z = 2 cos 2π
and
w
=
3
cos
.
+
i
sin
+
i
sin
3
3
6
6
What is the polar form of zw?
Consider v = r1 (cos θ + i sin θ) and t = r2 (cos γ + i sin γ), then :
vt = r1 (cos θ + i sin θ) · r2 (cos γ + i sin γ)
= r1 r2 (cos θ + i sin θ)(cos γ + i sin γ)
= r1 r2 (cos θ cos γ + i sin θ cos γ + i cos θ sin γ + i2 sin θ sin γ)
= r1 r2 (cos θ cos γ − sin θ sin γ + i sin θ cos γ + i cos θ sin γ)
= r1 r2 (cos θ cos γ − sin θ sin γ + i(sin θ cos γ + cos θ sin γ))
Using the trigonometric identities for cos(A + B) and sin(A + B), we get :
vt = r1 r2 (cos(θ + γ) + i sin(θ + γ))
In short, to multiply two complex numbers in polar forms, we multiply the modulus and we
2π
π
π
+
i
sin
and
w
=
3
cos
+
i
sin
. What is the
add the arguments! Let z = 2 cos 2π
3
3
6
6
polar form of zw?
2π π
2π π
+
+
zw = 6 cos
+ i sin
3
6
3
6
5π
5π
+ i sin
zw = 6 cos
6
6
If z = r(cos θ + i sin θ) is a complex number, geometrically, where would iz be in relations
to z ? i = 0 + 1i in polar form :
• r=1
• θ=
π
2
i = cos
π
π
+ i sin
2
2
So :
π
π
π
π iz = cos + i sin
· r(cos θ + i sin θ) = r cos θ +
+ i sin θ +
2
2
2
2
9
It’s a 90 degrees rotation!
De Moivre’s Theorem
If z = r(cos θ + i sin θ), then z n = rn (cos nθ + i sin nθ) .
1000
3π
3π
Example : What is the value of cos
+ i sin
4
4
1000
3π
3π
3000π
3000π
cos
+ i sin
=
cos
+ i sin
4
4
4
4
= (cos 750π + i sin 750π)
= cos 0 + i sin 0
= 1
10