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Complex Numbers Summary Academic Skills Advice What does a complex number mean? A complex number has a βrealβ part and an βimaginaryβ part (the imaginary part involves the square root of a negative number). e.g. π = π + ππ We use Z to denote a complex number: Real Example: Imaginary You might see the π before or after itβs number - it doesnβt matter which. Z = 4 + 3i Re(Z) = 4 Im(Z) = 3 Sometimes (especially in engineering) a j is used instead of π β they mean the same thing. Powers of i π stands for ββ1 so: 2 π 2 = (ββ1) = -1 π 4 = (π 2)2 = (-1)2 = 1 For any power of π take out as many π 4βs and π 2βs as possible and they will all end up as ±π or ±1. Example: Summary: π = ββπ ππ = βπ ππ = π πβπ = βπ πβπ = π π 11 = (π 4 )2 π 2 π = 12 × β1 × π = βπ OR: just take out π 2βs if you find it easier to remember. Example: π 33 = (π 2 )16 π = (β1)16 π = π Adding & Subtracting This is easy β just add or subtract the real part and add or subtract the imaginary parts: Examples: (4 + 3π ) + (2 + 6π ) = (6 + 9π ) (3 + 7π ) β (1 β 3π ) = (2 + 10π ) Multiplying Multiply out the 2 brackets. Example: (3 + 5π )(4 β 2π ) = 12 β 6π +20π β 10π 2 = 12 + 14π β 10 (-1) = 22 + 14π © H Jackson 2010 / 2015 / Academic Skills 1 Complex Conjugate The conjugate is exactly the same as the complex number but with the opposite sign in the middle. When multiplied together they always produce a real number because the middle terms disappear (like the difference of 2 squares with quadratics). Example: (4 + 6π )(4 β 6π ) = 16 β 24π + 24π β 36π 2 = 16 β 36(-1) = 16 + 36 = 52 Dividing Dividing by a real number: divide the real part and divide the imaginary part. Dividing by a complex number: Multiply top and bottom of the fraction by the complex conjugate of the denominator so that it becomes real, then do as above. Examples: 3+4π 2 4β5π 3+2π = 3 = 4β5π 2 4 + π = 1.5 + 2π 2 3+2π × 3β2π 3β2π = 12β8πβ15π+10π 2 9β6π+6πβ4π 2 = 12β23π+10(β1) 9β4(β1) = 2β23π 13 = π ππ β ππ ππ π Graphical Representation A complex number can be represented on an Argand diagram by plotting the real part on the π₯-axis and the imaginary part on the y-axis. Example: Imaginary P (π§ = π₯ + π¦π) y 0 Modulus: Argument: Nb Tan(ΞΈ) = π¦ π₯ so π π½ = πππβπ ( ) π π π₯ Real is written as |π| and is the length of OP, therefore |π| = βππ + ππ is the angle ΞΈ that is made with the horizontal axis (denoted by β ). Polar & Exponential Form As well as the basic form (π§ = π₯ + ππ¦) there are 2 more ways of writing a complex number: Polar: π = π(ππππ½ + πππππ½) Exponential: π = ππππ½ Where π is the length of the line and π is the angle it makes with the π₯-axis (this should be in radians for the exponential form). Remember: To find the modulus (length), π: use Pythagoras π¦ To find the argument (angle), π: use π‘ππβ1 (π₯ ) © H Jackson 2010 / 2015 / Academic Skills 2 Converting between the different forms: Basic Need to find π and π π = βπ₯ 2 + π¦ 2 Polar or Exponential π¦ π = π‘ππβ1 ( ) π₯ Polar or Exponential Example: Need to find π₯ and π¦ π₯ = ππππ π π¦ = ππ πππ Basic Express π§ = 3 + 4π in polar and exponential form imaginary π 0 4 π 3 Modulus: π = β32 + 42 = β16 + 9 = β25 = 5 Argument: π = π‘ππβ1 (3) = 53.1o Polar form: π§ = 5(cos(53.1) + ππ ππ(53.1)) Exp form: π§ = 5π 0.927π real 4 (in radians π = 0.927π ) Nb always do a quick sketch of the complex number and if itβs in a different quadrant adjust the angle as necessary. Example: Express π§ = 7π π π3 in basic form π π₯ = ππππ π β΄ π₯ = 7 cos (3 ) = 3.5 π¦ = ππ πππ β΄ π¦ = 7 sin ( 3 ) = 6.1 π Basic form: π§ = 3.5 + 6.1π A reminder of the 3 forms: Basic π = π + ππ Conversions: π₯ = ππππ π π¦ = ππ πππ © H Jackson 2010 / 2015 / Academic Skills Polar π = π(ππππ½ + πππππ½) Exponential π = ππππ½ π = βπ₯ 2 + π¦ 2 π¦ π = π‘ππβ1 ( ) π₯ 3 Multiplying with Polar or Exponential form Let π§1 = π§2 π§3 This means, when multiplying 2 complex numbers: Multiply the πβ²π Add the angles (πβ²π ) Then |π§1 | = |π§2 | × |π§3 | And β π§1 = β π§2 + β π§3 π π Example: If π§1 = 5π 2 π and π§2 = 3π 3 π find π§1 π§2 New modulus: New angle: 5 × 3 = 15 π π 5π + = 2 3 6 5π β΄ π§1 π§2 = 15π 6 π Dividing with Polar or Exponential form Let π§1 = Then |π§1 | = And π§2 This means, when dividing 2 complex numbers: π§3 Divide the πβ²π Subtract the angles (πβ²π ) |π§2 | |π§3 | β π§1 = β π§2 β β π§3 π π π π π§ Example: if π§1 = 5 (πππ ( 2 ) + ππ ππ ( 2 )) and π§2 = 3 (πππ ( 3 ) + ππ ππ ( 3 )) find π§1 New modulus: 5 ÷ 3 = 2 π New angle: 2 β΄ π β3 = π§1 = π§2 5 3 5 3 π 6 π π (πππ (6 ) + ππ ππ (6 )) De Moivres Theorem: Think: Raise π to the power of π and multiply the angle by π. Is used for raising a complex number to a power. ππ = ππ (πππ(ππ½) + ππππ(ππ½)) e.g π e.g.2: (1 + π)100 π If π§ = 3 (πππ (3 ) + ππ ππ (3 )) 5 5 then π§ = 3 (πππ 5π 3 + ππ ππ 5π 3 We could use De Moivres or: (1 + π)100 = ((1 + π)2 )50 ) 1 The same method can be used for a root (e.g. π§ π ). However, there will be π answers, all with the same modulus but with different arguments. To find the arguments you need to keep 2π adding π to your previous answer. © H Jackson 2010 / 2015 / Academic Skills 4