SLIDE 1 2 Complex Numbers
2.1 Historical Note [For interest only]
- There is much that may be said about the genesis of complex numbers.
[See: “An Imaginary Tale: The Story of √−1”, Paul J. Nahin.]
- They form an extension of the number line to a 2D plane.
- Although complex numbers use i = √−1, which makes no physical sense, they form a consistent
system to use mathematically. What does 3 + 2i mean?
- Even worse, Leibniz (one of the founders of calculus), found that
- 1 +
√ 3 i +
√ 3 i = √ 6.
- In 1572 Rafael Bombelli observed that, if you ignore what such expressions might mean and just do
the sums, you get the correct real solutions. Eventually this line of thinking led to the creation of the system of complex numbers, in which −1 has a square root. [From: Significant Figures by Ian Stewart.]
SLIDE 2
- The tale of how to solve the cubic is full of intrigue, betrayal, plagiarism and has a cast of colourful
- characters. Worth reading Nahin’s book.
- Cardan published how to solve a general cubic — see the main printed notes.
- The supreme irony is that, when a cubic polynomial has three real roots, then Cardan’s formula
requires one to find square roots of complex numbers!
- When a cubic polynomial has a single real root, the same formula doesn’t require square roots of
complex numbers..... ..... but the cubic will have two complex roots as well as the real one. If complex numbers are 2D, then Hamilton’s quaternions are 4D, octonions are 8D, sedenions are 16D and trigintaduonions are 32D. These are of decreasing utility!
SLIDE 3
Claim: Complex numbers arise naturally when solving quadratic equations. Here are three examples of quadratic equations that we may solve. 0 = x2 − 4x + 3 = (x − 3)(x − 1) ⇒ x = 1, 3 0 = x2 − 4x + 4 = (x − 2)(x − 2) ⇒ x = 2, 2 0 = x2 − 4x + 5 = (x − 2)2 + 1 ⇒ (x − 2)2 = −1 ⇒ x − 2 = ±√−1 ⇒ x = 2 ± √−1
1 2 3 4 0.5 1.5 2.5 3.5 −1 1 2 3 4 5 x
SLIDE 4
Notation: In Mathematics: √−1 = i. In Engineering: √−1 = j. Sorry — it’s not my fault....
2.3 Manipulation of Complex Numbers
The basic rules for adding, subtracting and multiplying complex numbers are the same as for real numbers but sometimes j2 = −1 has to be used. Addition and Subtraction: We collect like terms. For example (1 + 5j) + (2 − 3j) = (1 + 2) + (5 − 3)j = 3 + 2j, and (1 + 5j) − (2 − 3j) = (1 − 2) + (5 + 3)j = −1 + 8j. So these work in the same way as (x + 5y) + (2x − 3y) = (1 + 2)x + (5 − 3)y = 3x + 2y and (x + 5y) − (2x − 3y) = (1 − 2)x + (5 + 3)y = −x + 8y.
SLIDE 5
Multiplication: Here we expand the product in the same way as we do for the product, (a + b)(c + d) = ac + bd + ad + bc, except that a little tidying up takes place afterwards using j2 = −1. So (a + bj)(c + dj) = ac + bd j2 + adj + bcj = (ac − bd) + (ad + bc)j. A numerical example: (1 + 2j)(3 + 4j) = (1 × 3) + (2j × 4j) + (1 × 4j) + (2j × 3) = 3 + 8 j2 + 4j + 6j = 3 − 8 + 10j = −5 + 10j.
SLIDE 6
Note: Integer powers of complex numbers (like squares and cubes and 10th powers etc.) may be evaluated this way. Powers of j: j2000 = (j4)500 = (1)500 = 1. (because j2 = −1 ⇒ j4 = 1) (1 + j)2 = 1 + j2 + 2j = 2j (1 + j)3 = −2 + 2j (1 + j)4 = −4 (1 + j)5 = −4 − 4j (1 + j)6 = −8j (1 + j)7 = 8 − 8j (1 + j)8 = 16 .....hmm, so this appears to imply that the 8th root of 16 is 1 + j? But surely it is √ 2? More on this later.
SLIDE 7
Division: This is the most complicated operation of the four. We need the complex conjugate. If z = x + yj, then the complex conjugate of z is z = x − yj. “z bar” Examples: 6 = 6, 5j = −5j, 3 + 4j = 3 − 4j. A very useful result. It is also brilliant: zz = (x + yj)(x − yj) by definition = x2 − y2 j2 + ✟✟
✟
xyj − ✟✟
✟
xyj multiplying out = x2 + y2 using j2 = −1. Given that zz is real, we now have all the tools we need to be able to divide by a complex number.
SLIDE 8 Example 1: 1 3 + 4j = 1 3 + 4j × 3 − 4j 3 − 4j = 3 − 4j (3 + 4j)(3 − 4j) = 3 − 4j 25 =
3 25 − 4 25j.
In general: 1 x + yj = x − yj (x + yj)(x − yj) = x − yj x2 + y2. Example 2: 2 + j 1 + j = (2 + j)(1 − j) (1 + j)(1 − j) = 3 − j 2 . Note: This idea of using a complex conjugate helps us elsewhere: 8 1 + √ 5 = 8 (1 + √ 5) × (1 − √ 5) (1 − √ 5) = 8(1 − √ 5) 1 − 5 = 2( √ 5 − 1).
SLIDE 9 2.4 Geometrical Interpretation
The complex number z = a + bj may be represented as coordinates in the complex plane: Imag Real a b
θ r Figure 2.1. Showing the location of the complex number, z = a + bj. Also known as the Argand diagram. The modulus of z: |z| = r = √a2 + b2 = √ zz. The argument of z: arg(z) = θ = tan−1(b/a). Measure anti-clockwise. Be careful.... Real(z) = a is the real part of z. Or Re(z) = a. Imag(z) = b is the imaginary part of z. Or Im(z) = b.
SLIDE 10 Imag Real a b −b
z = a − bj θ −θ r r Figure 2.2. Showing the locations of the complex number, z = a + bj and its conjugate, z = a − bj, in the Complex Plane. We see that |z| = |z| and arg(z) = − arg(z).
SLIDE 11 Real Imag
1 − j −1 + j If z = 1 + j then |z| = √ 2 and θ = arg(z) = tan−1 1 = 1
4π (not 5 4π).
If z = −1 + j then |z| = √ 2, and θ = arg(z) = tan−1(−1) = 3
4π (not − 1 4π).
If z = 1 − j then |z| = √ 2, and θ = arg(z) = tan−1(−1) = − 1
4π (not 3 4π).
SLIDE 12 Addition: Looks as though if follows the same rule as the addition of two vectors in that the sum is
- btained by completing the rhombus.
Imag Real
a + c b + d
Figure 2.3. Showing how two complex numbers, z1 = a + bj and z2 = c + dj, may be added by completing the rhombus. Equivalently, we add the real and imaginary parts separately. (a + bj) + (c + dj) = (a + c) + (b + d)j.
SLIDE 13 Multiplication: For example, we have (2 + 4j) × (1 + j) = −2 + 6j. Imag Real
−2 + 6j • Figure 2.4. Showing the multiplication of the two given complex numbers. Can we make sense of this? Let’s check the moduli: |2 + 4j| = √ 20, |1 + j| = √ 2, | − 2 + 6j| = √ 40 = √ 2 × √ 20. So the modulus of the product is equal to the product of the moduli.
SLIDE 14 Imag Real a b
θ r Figure 2.1. Showing the location of the complex number, z = a + bj. Use polar coordinate: a = r cos θ and b = r sin θ ⇒ a + bj = r(cos θ + j sin θ). Let z1 = r1(cos θ1 + j sin θ1) and z2 = r2(cos θ2 + j sin θ2). ⇒ z1z2 = r1r2(cos θ1 + j sin θ1)(cos θ2 + j sin θ2), = r1r2
- (cos θ1 cos θ2 − sin θ1 sin θ2) + j(sin θ1 cos θ2 + cos θ1 sin θ2)
- ,
= r1r2
- cos(θ1 + θ2) + j sin(θ1 + θ2)
- using multiple-angle formulae.
SLIDE 15 So z1 = r1(cos θ1 + j sin θ1) and z2 = r2(cos θ2 + j sin θ2) ⇒ z1z2 = r1r2
- cos(θ1 + θ2) + j sin(θ1 + θ2)
- .
The modulus of the product is equal to the product of the moduli. The argument of the product is equal to the sum of the arguments. Some examples: product moduli argument 2 × (−2) = −4 2 × 2 = 4 0 + 180◦ = 180◦ j × j = −1 1 × 1 = 1 90◦ + 90◦ = 180◦ (1 + j) × (1 + j) = 2j √ 2 × √ 2 = 2 45◦ + 45◦ = 90◦ (2 + 4j) × (1 + j) = −2 + 6j √ 20 × √ 2 = √ 40 63.435◦ + 45◦ = 108.435◦ This reminds me of what happens with exponentials: apb × cpd = ac pb+d.
SLIDE 16 2.5 Complex Exponential form
z = a + bj by definition = (r cos θ) + j(r sin θ) as earlier = r(cos θ + j sin θ) = rejθ. Surely this last step is outrageous? How? Why? We will prove this later using Taylor’s series, but we’ll have to run with it for now.... So complex numbers may be written in two main forms: Cartesian form: z = a + bj
Complex exponential form: z = rejθ. A third notation is z = r∠θ which is often used in Electrical Engineering contexts. A fourth notation is r cis θ, where cis is a shorthand for cos θ + i sin θ — this was invented by Hamilton in 1866, but is rarely used now. Warning: When writing rejθ, the angle, θ, must always be measured in radians. The same is true for sine and cosine.
SLIDE 17 2.6 de Moivre’s Theorem
This is a very useful result and is known as a theorem despite the fact its proof takes only two lines (i.e. just one step). Apply (ab)c = abc to eθj: (eθj)n = enθj ⇒
n = cos(nθ) + j sin(nθ). This may be used recover the various multiple angle formulae from trigonometry. When n = 2: cos 2θ + j sin 2θ =
2 = [cos2 θ − sin2 θ] + j[2 sin θ cos θ]. On equating real and imaginary parts we get cos 2θ = cos2 θ − sin2 θ sin 2θ = 2 sin θ cos θ
SLIDE 18 When n = 3, we have cos 3θ + j sin 3θ =
3 = cos3 θ + 3j cos2 θ sin θ + 3j2 cos θ sin2 θ + j3 sin3 θ = [cos3 θ − 3 cos θ sin2 θ] + j[3 cos2 θ sin θ − sin3 θ]. This may be tidied up to give cos 3θ = 4 cos3 θ − 3 cos θ sin 3θ = 3 sin θ − 4 sin3 θ For n = 6: cos 6θ + j sin 6θ =
6 = cos6 θ + 6j cos5 θ sin θ + 15j2 cos4 θ sin2 θ + 20j3 cos3 sin3 θ + 15j4 cos2 θ sin4 θ + 6j5 cos θ sin5 θ + j6 sin6 θ = [cos6 θ − 15 cos4 θ sin2 θ + 15 cos2 θ sin4 θ − sin6 θ] + [6 cos5 θ sin θ − 20 cos3 θ sin3 θ + 6 cos5 θ sin θ]j This may be tidied to give cos 6θ solely in terms of cos θ or solely in terms of sin θ, and to give sin 6θ.
