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 + 2 i mean? • Even worse, Leibniz (one of the founders of calculus), found that √ √ √ � � 1 + 3 i + 1 − 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.]
• 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!
Claim: Complex numbers arise naturally when solving quadratic equations. Here are three examples of quadratic equations that we may solve. x 2 − 4 x + 3 = ( x − 3)( x − 1) 0 = ⇒ x = 1 , 3 x 2 − 4 x + 4 = ( x − 2)( x − 2) 0 = ⇒ x = 2 , 2 5 x 2 − 4 x + 5 = ( x − 2) 2 + 1 0 = 4 ( x − 2) 2 = − 1 ⇒ 3 x − 2 = ±√− 1 ⇒ 2 x = 2 ± √− 1 ⇒ 1 0 − 1 x 0 0 . 5 1 1 . 5 2 2 . 5 3 3 . 5 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 j 2 = − 1 has to be used. Addition and Subtraction: We collect like terms. For example (1 + 5 j ) + (2 − 3 j ) = (1 + 2) + (5 − 3) j = 3 + 2 j, and (1 + 5 j ) − (2 − 3 j ) = (1 − 2) + (5 + 3) j = − 1 + 8 j. So these work in the same way as ( x + 5 y ) + (2 x − 3 y ) = (1 + 2) x + (5 − 3) y = 3 x + 2 y and ( x + 5 y ) − (2 x − 3 y ) = (1 − 2) x + (5 + 3) y = − x + 8 y.
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 j 2 = − 1 . So ( a + bj )( c + dj ) = ac + bd j 2 + adj + bcj = ( ac − bd ) + ( ad + bc ) j. A numerical example: (1 + 2 j )(3 + 4 j ) = (1 × 3) + (2 j × 4 j ) + (1 × 4 j ) + (2 j × 3) = 3 + 8 j 2 + 4 j + 6 j = 3 − 8 + 10 j = − 5 + 10 j.
Note: Integer powers of complex numbers (like squares and cubes and 10th powers etc.) may be evaluated this way. j 2000 = ( j 4 ) 500 = (1) 500 = 1 . (because j 2 = − 1 ⇒ j 4 = 1 ) Powers of j : (1 + j ) 2 = 1 + j 2 + 2 j = 2 j (1 + j ) 3 = − 2 + 2 j (1 + j ) 4 = − 4 (1 + j ) 5 = − 4 − 4 j (1 + j ) 6 = − 8 j (1 + j ) 7 = 8 − 8 j (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.
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 , 5 j = − 5 j, 3 + 4 j = 3 − 4 j. A very useful result. It is also brilliant: zz = ( x + yj )( x − yj ) by definition = x 2 − y 2 j 2 + ✟✟ xyj − ✟✟ ✟ xyj ✟ multiplying out = x 2 + y 2 using j 2 = − 1 . Given that zz is real, we now have all the tools we need to be able to divide by a complex number.
Example 1: 1 3 + 4 j × 3 − 4 j 1 3 − 4 j = = 3 + 4 j 3 − 4 j (3 + 4 j )(3 − 4 j ) 3 − 4 j 3 4 = = 25 − 25 j. 25 In general: 1 x − yj x − yj = = x 2 + y 2 . x + yj ( x + yj )( x − yj ) Example 2: 2 + j (2 + j )(1 − j ) 3 − j = (1 + j )(1 − j ) = . 1 + j 2 Note: This idea of using a complex conjugate helps us elsewhere: √ √ √ 8 8 × (1 − 5) 8(1 − 5) = = = 2( 5 − 1) . √ √ √ 1 − 5 1 + 5 (1 + 5) (1 − 5)
2.4 Geometrical Interpretation The complex number z = a + bj may be represented as coordinates in the complex plane : Imag b • z = a + bj r θ Real a 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 = √ a 2 + b 2 = √ 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 .
Imag b • z = a + bj r θ Real a − θ r • z = a − bj − b 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 ) .
Imag − 1 + j • • 1 + j Real • 1 − j √ θ = arg( z ) = tan − 1 1 = 1 4 π (not 5 If z = 1 + j then | z | = 2 and 4 π ). √ θ = arg( z ) = tan − 1 ( − 1) = 3 4 π (not − 1 If z = − 1 + j then | z | = 2 , and 4 π ). √ θ = arg( z ) = tan − 1 ( − 1) = − 1 4 π (not 3 If z = 1 − j then | z | = 2 , and 4 π ).
Addition: Looks as though if follows the same rule as the addition of two vectors in that the sum is obtained by completing the rhombus. Imag b + d • z 1 + z 2 • z 2 = c + dj • z 1 = a + bj Real a + c Figure 2.3. Showing how two complex numbers, z 1 = a + bj and z 2 = 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.
Multiplication: For example, we have (2 + 4 j ) × (1 + j ) = − 2 + 6 j . Imag − 2 + 6 j • • 2 + 4 j • 1 + j Real Figure 2.4. Showing the multiplication of the two given complex numbers. Can we make sense of this? Let’s check the moduli: √ √ √ √ √ | 2 + 4 j | = 20 , | 1 + j | = 2 , | − 2 + 6 j | = 40 = 2 × 20 . So the modulus of the product is equal to the product of the moduli.
Imag b • z = a + bj r θ Real a 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 z 1 = r 1 (cos θ 1 + j sin θ 1 ) and z 2 = r 2 (cos θ 2 + j sin θ 2 ) . ⇒ z 1 z 2 = r 1 r 2 (cos θ 1 + j sin θ 1 )(cos θ 2 + j sin θ 2 ) , � � = r 1 r 2 (cos θ 1 cos θ 2 − sin θ 1 sin θ 2 ) + j (sin θ 1 cos θ 2 + cos θ 1 sin θ 2 ) , � � = r 1 r 2 cos( θ 1 + θ 2 ) + j sin( θ 1 + θ 2 ) using multiple-angle formulae.
So z 1 = r 1 (cos θ 1 + j sin θ 1 ) and z 2 = r 2 (cos θ 2 + j sin θ 2 ) ⇒ � � z 1 z 2 = r 1 r 2 cos( θ 1 + θ 2 ) + j sin( θ 1 + θ 2 ) . 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 0 + 180 ◦ = 180 ◦ 2 × ( − 2) = − 4 2 × 2 = 4 90 ◦ + 90 ◦ = 180 ◦ j × j = − 1 1 × 1 = 1 √ √ 45 ◦ + 45 ◦ = 90 ◦ (1 + j ) × (1 + j ) = 2 j 2 × 2 = 2 √ √ √ 63 . 435 ◦ + 45 ◦ = 108 . 435 ◦ (2 + 4 j ) × (1 + j ) = − 2 + 6 j 20 × 2 = 40 This reminds me of what happens with exponentials: ap b × cp d = ac p b + d .
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 ) = re jθ . 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: Complex exponential form: z = re jθ . Cartesian form: z = a + bj or 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 re jθ , the angle, θ , must always be measured in radians. The same is true for sine and cosine.
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 ( a b ) c = a bc to e θj : ( e θj ) n = e nθj � n � ⇒ cos θ + j sin θ = cos( nθ ) + j sin( nθ ) . This may be used recover the various multiple angle formulae from trigonometry. When n = 2 : � 2 � cos 2 θ + j sin 2 θ = cos θ + j sin θ = [cos 2 θ − sin 2 θ ] + j [2 sin θ cos θ ] . On equating real and imaginary parts we get cos 2 θ = cos 2 θ − sin 2 θ sin 2 θ = 2 sin θ cos θ
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