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Chapter 6: Complex Numbers and Vectors Chapter 6 Complex Numbers and Vectors Part I: Complex Numbers If you were asked to solve the quadratic equation x 2 6 x 34 , you would start be applying the b b 2 4ac x quadratic formula 2a with a 1 , b 6 and c 34 Hence, 6 100 6 36 136 x x 2 2 . But, how do we find 100 ? 100 cannot be represented by a number, as there is no real number whose square is negative so therefore we define 100 j10 . Example: Find Solution: 49 j 7 . Example: Find Solution: TRY: Find 49 . 90.25 . 90.25 j 9.5 . 169 . 1/17 Chapter 6: Complex Numbers and Vectors Powers of j j 1 j 2 1 j 3 j 2 j 1 j j j j 4 1 2 2 Example: Find Solution: j 16 Solution: TRY: j . 2 3 j 13 j j 16 . j Example: Find 25 j 1 7 j j 7 Example: Find Solution: j 2 j 4 4 14 1 25 . j Find j 34 . Find j 81 . 4 6 j 16 j j Find j 48 . Find j 123 . 2/17 Chapter 6: Complex Numbers and Vectors Complex Numbers Lets refer back to our original problem of solving x 2 6 x 34 using the quadratic formula. We found that x 6 100 and we said that 100 j10 . 2 6 j10 Thus x 2 and therefore: 6 j10 6 j10 3 j5 x 3 j 5 and x 2 2 The result x 3 j 5 consists of 2 separate terms that cannot be combined any further. In the expression x 3 j5 3 is called the real part of x 5 is called the imaginary part of x the two combined form a complex number. Hence, a complex number = real part + j (imaginary part) Addition and Subtraction of Complex Numbers In order to add and subtract complex number, we match like with like and then add and subtract as normal. Example: Find 4 j 2 5 j 6 Solution: Rewrite as 4 j 2 5 j 6 4 5 j 2 j6 4 5 j2 6 9 j8 3/17 Chapter 6: Complex Numbers and Vectors Example: Solution: Find 6 j 2 7 j3 Find 3 j 2 2 j6 6 j 2 7 j3 6 7 j 2 j3 6 7 j2 3 13 j Example: Solution: 3 j 2 2 j 6 3 2 j 2 j6 3 2 j2 6 1 8 j TRY: 4 j7 8 j 2 . Find 4 j5 3 j 2 . Find 3 j5 5 j 4 2 j3 . Find 4/17 Chapter 6: Complex Numbers and Vectors Multiplication of complex numbers Example: Find 3 j 42 j5 Solution: 3 j 42 j5 (3 2) (3 j5) ( j 4 2) ( j 4 j5) 6 j15 j8 ( j 4 j5) 6 j 23 j 2 20 2 We know that j 1 , therefore 6 j 23 j 2 20 6 j 23 120 6 j 23 20 14 j 23 Example: Find 4 j53 j 2 Solution: 4 j53 j 2 (4 3) (4 j 2) ( j5 3) ( j5 j 2) 12 j8 j15 ( j5 j 2) 12 j 7 j 2 10 We know that j 1 , therefore 12 j 7 j 2 10 12 j 7 110 12 j 7 10 22 j 7 . 2 Example: Find 5 j 27 j36 j 4 Solution: 5 j 27 j36 j 4 5 7 5 j3 j 2 7 j 2 j36 j 4 35 j15 j14 j 2 66 j 4 35 j 29 66 j 4 29 j 296 j 4 Now 29 j 296 j 4 29 6 29 j 4 j 29 6 j 29 j 4 174 j116 j174 j 2 116 174 j58 116 290 j58. 5/17 Chapter 6: Complex Numbers and Vectors TRY: Find 6 j 710 j3 Find 3 j96 j 2 Find 12 j314 j5 Find 312 j 26 j Find 2 j34 j 6 Find 7 j 2 j32 j Example: Find Solution: 4 j 24 j 2 4 j 24 j 2 (4 4) (4 j 2) ( j 2 4) ( j 2 j 2) 16 j8 j8 ( j 2 j 2) 16 j 2 4 16 (1)2 16 4 20 In this example, our result contains no imaginary part, only a real number. This is an exceptional case. Look at the 2 complex numbers that we have multiplied together; they are the same except for the middle sign. A pair of complex numbers like this are known are conjugate complex numbers and when two conjugate complex numbers are multiplied together, you always get only a real number. TRY: Find 2 j 72 j 7 6/17 Chapter 6: Complex Numbers and Vectors Graphical Representation of a Complex Number Although we cannot evaluate a complex number as a real number, we can represent it diagrammatically and a diagram representing a complex number is known as an Argand Diagram. On an Argand diagram, the x-axis represents the real part of the complex number the y-axis represents the imaginary part of the complex number. Therefore we can graphically represent the complex number 6 j7 . Argand diagram for Y (Imaginary) 8 6+j7 a 6 4 b 2 1 2 3 4 5 6 7 X (Real) TRY: Draw an Argand diagram for 6 j 7 . Draw an Argand diagram for 2 j 4 . Find and draw an Argand diagram for 2 j3 6 j 2 . Find and draw an Argand diagram for 3 j9 12 j3 . 7/17 Chapter 6: Complex Numbers and Vectors Polar Form of a Complex Number It is sometimes necessary, or more convenient, to express a complex number in Polar form. Argand diagram for Y (Imaginary) 8 6+j7 a 6 4 r b 2 1 θ2 3 4 5 6 7 X (Real) (With respect to the Argand diagram). We can see that, r 2 a2 b2 , b a a r cos and b r sin z a jb , this can be rewritten as z r cos jr sin z r cos j sin tan As r a2 b2 b tan 1 a 8/17 Chapter 6: Complex Numbers and Vectors This is known as the polar form of the complex number b 1 tan a jb where r a b and a 2 2 r is known as the modulus of the complex number z and can be written as mod z is known as the argument of the complex number and can be written as arg z Example: Express z 4 j 3 in polar form Solution: We know that for a jb we have b r a. Therefore for 4 j 3 we have a 4, b 3 so we 1 a b and tan 2 2 2 2 calculate: r 4 3 16 9 25 5 1 tan and So 3 θ 37degrees 4 So z 5cos37 jsin37 Example: Given arg z . z 2 j 5 , find mod z and Solution: For 2 j 5 we have a 2, b 5 so we 2 2 calculate: mod z r 2 5 4 25 29 1 5 arg z tan and 2 . (So θ 68degrees) 9/17 Chapter 6: Complex Numbers and Vectors z 4 j 7 , find mod z and arg z . Given z 2 j 4 , find mod z and arg z . TRY: Given Given z 9 j 4 , find mod z and arg z . 10/17 Chapter 6: Complex Numbers and Vectors Part II: Vectors Physical quantities can be divided into 2 main groups: A scalar quantity is one that is defined completely by a single number with appropriate units e.g. length, area, volume, mass, time etc. When the units are stated the quantity is denoted entirely by its size or magnitude e.g. speed of 10 km/hour is a scalar quantity. A vector quantity is defined completely when we know not only its magnitude (units) but also the direction in which it operates e.g. force, velocity, acceleration. A vector quantity involves direction as well as magnitude e.g. a velocity of 10 km/hour due north is a vector quantity. Representing a vector. A vector quantity can be represented graphically by a line drawn so that its length represents the magnitude of the vector and the direction of the line is the direction of the vector. A vector quantity AB is referred to as AB or bold type a. The magnitude of the vector quantity is defined as AB or a , or simply AB or a. Note that BA 11/17 Chapter 6: Complex Numbers and Vectors would represent a vector quantity of the same magnitude but the opposite direction to AB . Introduction to 3D vectors. The axes of reference Ox, Oy and Oz are chosen so that they form a right handed set. Vector OP is defined by its components: A along Ox, B along Oy, C along Oz. The symbols i, j, k denote unit vectors. Let i = unit vector in Ox direction, j = unit vector in Oy direction, k = unit vector in Oz direction. Then OP ai bj ck . Magnitude of a vector For the vector quantity OP r ai bj ck , then the magnitude of the vector in terms of the unit vectors 2 2 2 OP r a b c is . Example: Given PQ 4 i 3 j 2k find PQ . Solution: If PQ 4i 3 j 2k , then PQ TRY: 4 2 3 2 2 2 16 9 4 29 . Given OA 2i 4 j-3k find OA . Given OB 1i-2 j 3k find OB . 12/17 Chapter 6: Complex Numbers and Vectors Addition and Subtraction. Example: Let a 2i 3 j 5k and b 4i 1j 6k , Find a b . Solution: a b 2i 3 j 5k 4i 1j 6k 6 i 4 j 11k . Example: Let a 2i 3 j 5k and b 4 i 1j 6k , Find a b . a b 2i 3 j 5k 4i 1j 6k Solution: 2i 3 j 5k - 4i 1j 6k . 2i 2 j 1k Scalar products (and dot products). If a and b are 2 vectors, the scalar product of a and b is defined as the scalar (number) abcosθ where a and b are the magnitudes of the vectors a and b and θ is the angle between them. The scalar product is denoted by a.b (often called the dot product): a.b abcosθ = a =b projection of b on a projection of a on b In both cases the result is a scalar quantity. 13/17 Chapter 6: Complex Numbers and Vectors For a a1i a2 j a3 k , b b1i b2 j b3 k then a.b a1i a 2 j a3 k b1i b2 j b3 k a1ib1i a1ib2 j a1ib3 k a 2 jb1i a 2 jb2 j a 2 jb3 k a3 kb1i a3 kb2 j a3 kb3 k a1b1ii a1b2 ij a1b3 ik a 2 b1 ji a 2 b2 jj a 2 b3 jk a3 b1ki a3 b2 kj a3 b3 kk. Note: i.i 11cos 0 1 , similarly j.j 1 , k.k 1 . Also: i.j 11cos 90 0 , similarly j.k 0 , k.i 0 . Putting these altogether in the formulation above gives: a.b a1b1 1 a1b2 0 a1b3 0 a2 b1 0 a2 b2 1 a2 b3 0 a3 b1 0 a3 b2 k 0 a3 b3 1 Therefore to find the scalar product we just sum the products of the coefficients of the unit vectors along the corresponding axes: a.b a1b1 a2 b2 a3 b3 . Example: Given a 2i 3 j 5k and b 4i 1j 6k find the dot product (or scalar product) a.b . Solution: Comparing a 2i 3 j 5k and b 4i 1j 6k to a a1i a2 j a3 k and b b1i b2 j b3 k respectively we can see that a1 2, a 2 3, a3 5, b1 4, b2 1, b3 6. Therefore we can calculate a.b a1b1 a2 b2 a3 b3 . 2 4 3 1 5 6 8 3 30 41 14/17 Chapter 6: Complex Numbers and Vectors Example: Given a 3i-2 j 1k and b 2i 3 j-4k find the dot product (or scalar product) a.b . Solution: Comparing a 3i-2 j 1k and b 2i 3 j-4k to a a 1i a 2 j a 3k and b b1i b 2 j b 3k respectively we can see that a1 3, a 2 2, a3 1, b1 2, b2 3, b3 4. Therefore we can calculate a.q a 1b1 a 2b 2 a 3b 3 . 3 2 2 3 1 4 6 6 4 4 TRY: Given a 4 i 1j 3k and b 1i 3 j-7 k , find a b , a b , a b . Given OA a i j 4k OB b 8i 2k OC c 5i 2 j 11k Find: a) b - a b) c - b c) the magnitude of the vector OB , i.e. OB d) e) f) the magnitude of the vector OC , i.e. OC the scalar product a.b the scalar product b.c 15/17 Chapter 6: Complex Numbers and Vectors Vector products (and cross products). If a and b are 2 vectors, the vector product of a and b (a×b) is defined as a vector having magnitude absin θ where a and b are the magnitudes of the vectors a and b and θ is the angle between them. The product vector acts in a direction perpendicular to both a and b in the sence that a, b and a×b form a right-handed set in that order. |a×b| = absin θ Note that |b×a| reverses the direction. The vector product is denoted by a×b (often called the cross product): For a a1i a2 j a3 k , b b1i b2 j b3 k then a b a1i a2 j a3 k b1i b2 j b3 k a1b1i i a1b2i j a1b3i k a2b1 j i a2b2 j j a2b3 j k a3b1k i a3b2 k j a3b3 k k Note i×i = j×j= k×k = 0 and i×j=k, j×k=i, k×i=j Also note i×j = -(j×i), j×k= -(k×j), k×i= -(i×k) Putting these altogether in the formulation above gives: a b a1b1 0 a1b 2 k a1b3 j a 2 b1 k a 2 b 2 0 a 2 b3 i a 3 b1 j a 3 b 2 i a 3 b3 Therefore the vector product is given by: 16/17 Chapter 6: Complex Numbers and Vectors a b (a 2 b 3 a 3 b 2 )i (a 1 b 3 a 3 b1 ) j (a 1 b 2 a 2 b1 )k The cross product can be visualised as follows: a1 a2 a3 a1 a2 a3 a1 a2 a3 b1 b2 b3 + i b1 b 2 b3 b1 b 2 b3 - j + k Example: Given a 2i 3 j 5k and b 4i 1j 6k find the cross product (or vector product) a b . Solution: Comparing a 2i 3 j 5k and b 4i 1j 6k to a a1i a2 j a3 k and b b1i b2 j b3 k respectively we can see that: a1 2, a 2 3, a3 5, b1 4, b2 1, b3 6. Therefore we can calculate a b (a 2 b 3 a 3 b 2 )i (a 1b 3 a 3 b1 ) j (a 1b 2 a 2 b1 )k a b (3 6 5 1)i (2 6 5 4) j (2 1 3 4)k a b (13)i (8) j (10)k a b 13i 8 j 10k TRY: Given a 4i 1j 3k and b 1i 3 j-7 k , find the vector product a b . 17/17