Complex NumbersBasicsDefinition Of πββ1=ππ!=β1Calculating Powers Of π: Use π!=β1 and indices rule (π₯")#=π₯"#π$%=(π!)!$=(β1)!$=1π!&=(π!)'(π=(β1)π=βπCalculating Square Rootsββ10ββ40=ββ1β10ββ1β40=πβ10πβ40=π!β400=β20.Do not make the mistake of saying ββ10ββ40=β400=20Cartesian FormCartesian FormWe usually use the letter z to denote a complex numberπ§=real part Β±π(imaginary part)Note: This can also be written as π§=real part Β±(imaginarypart)πWe usually use the letters πand πor π₯and π¦π§=a+ππor π§=π₯+ππ¦π
π(π§)means the real part of π§πΌπ(π§)means the imaginary part of π§Complex ConjugateWe swap the sign of the imaginary part for the complex conjugateπ§=π+ππβΉπ§β=πβπππ§=πβππβΉπ§β=π+ππWe use the notation π§βπππ§Μ
to denote the complex conjugate of z.Representation On An Argand DiagramAdding and Subtracting When we add/subtract we combinethe real parts together and the imaginary parts(π+ππ)+(π+ππ)=(π+π)+(π+π)πMultiplying When we multiply, we expand the brackets as normal and then collect like terms. Remember that π!can be replaced with β1=12+9πβ8πβ6π!=12β6(β1)+π=12+6+π=18+πDividing Multiply numerator and denominator by the complex conjugate. !"#$%"&$=!"#$%"&$Γ%'&$%'&$=!%'!&$"#%$'#&$!%!'%&$"%&$'&!$!=!%'#&$!'!&$"#%$%!'&!$!Finding The Modulus And ArgumentThe modulusof a complex number denoted, |π§|,is the distance from the origin to that numberon an argand diagram.The argument of a complex number, arg π§, is the angle between the positive real axis and the line joining the number to the origin on an Argand diagram. Method to calculate modulus and argument:Given π§=π₯+π¦πβMmodulus=|z|=Tπ₯!+π¦!argument=argπ§=π‘ππ+'X,#,"YwhereβΟ<π<πNote: The red parts are always a plusNext step for argument: Draw π₯+π¦πout to know which quadrant youβre in, start from positive x axis and find the anti-clockwise angle to find the value of thetaPropertiesOf Modulus And Complex Conjugate:β’|π§'π§!|=|π§'||π§!|β’`-"-!`=|-"||-!|β’(π§Β±π€)β=π§βΒ±π€ββ’(π§π€)β=π§βπ€ββ’'$%(β=$β%βif π€β 0β’π§Γπ§β=|π§|&Use of factor theoremand polynomial divisionFactorisingQuadratics: Use quadratic formula and work backwardsUnderstanding Roots of Quadratics and Cubics:β’A cubic with real coefficients either has:oall three roots realoone root real and the other two form a complex conjugate pairthe middle terms cancel out
β’A quarticwith real coefficients either has:oall fourroots realotwo roots real and the other two form a complex conjugate pairotwo roots form a conjugate pair and the other two roots also form a conjugate pairFactorisingCubics, Quarticsand Above:Use the factor theorem to find one of the factors and then use algebraic division or comparing coefficients until we have all the factors/roots.Recall that for 2 roots of the polynomial πandπ, then we have factors β(π§βπ)β and β(π§βπ)β, and can multiply/expand them to get another factor: π§!β(π+π)π§+(ππ)In other words when we have 2 roots we can build the equation, π§!β(sumofroots)π§+(productofroots)Remember that complex number roots occur in conjugate pairs, so if we know one root, then the conjugate is necessarily another root.Solving/Finding RootsOf Quadratics:We can use the quadratic formula. For an equation ππ§!+ππ§+π=0, we getπ§(='#"β#!'*!%+!and π§+='#'β#!'*!%+!Solving/Finding RootsOf Cubics, Quartics And Above:To solve polynomial equations that may have complex roots, we can use the same approach as above to factoriseand then we just go one step further bysettingthefactors equal to 0 after. Remember that complex number roots occur in conjugate pairs, so if we know one root, then the complex conjugate is necessarily another root.Given Some Of The Roots, Find The EquationWe must use the fact that complex number roots occur in conjugate pairs, so if we know one root, then the conjugate is necessarily another root and then build the equation with the roots πand πas:π§!β(π+π)π§+(ππ)Now we can use the comparing coefficients methodto find the unknownsEquating Real and Imaginary Coefficients In Order To:β’Find unknownsin equationsβ’Find square rootsβ’Solve equationsProving purely real or purely imaginaryModulus Argument FormConverting cartesian to modulus argument form:π+ππβπ(ππππ½+πππππ½)=πππππ½We just need to find πππππ. To find πand πwe use the formulaπ+ππβlπ=βπ!+π!π=π‘ππ+'X`#"`Yand then draw the angle πin the quadrant where the complex number π+ππliesRead off πby starting on the positive π₯axis (like when you solve for trig using the CAST diagram, but remember: βπβ€π<πwhich means we can only go 180Β°in either a clockwise or anti clockwise direction. Representation On An Argand DiagramLength and angleComplex Conjugateπ§β=π§Μ
=π(cosπβπsinπ)De Moivreβs Theorem(πππ π₯+ππ πππ₯)/=cosππ₯+ππ ππππ₯Useful the following follow-on results:β’π§+/=π+/(cosππβπsinππ)β’π§+'-=2cosπβ’π§β'-=2πsinπβ’π§/+'-'=π§/+π§+/=2cosππ.Rearranging βcosππ=-',-('!β’π§/β'-'=π§/βπ§+/=2πsinππ. Rearranging βsinππ=-'+-('!0Multiplying and DividingMultiplying(multiply the moduli and add the arguments)[π'(cosπ'+πsinπ')][π!(cosπ!+πsinπ!)]= π'π![cos(π'+π!)+πsin(π'+π!)]=π'π!π0(2",2!)Dividing(divide the moduli and subtract the arguments)4"(5672",07892")4!(5672!,07892!)can be divided quickly and is 4"4![cos(π'βπ!)+πsin(π'βπ!)]Representing π§=π(cosπ+πsinπ)as π§=π(cos(π+2ππ)+πsin(π+2ππ))Finding cube roots and above(solutions to π§/=π )and representation on an argand diagram(we use De Moivreβs Theorem)π§/=π βΉπ§=π'/tcosπ+2πππ+πsinπ+2πππuSolutions to π§/=1π§=cos!:;/+πsin!:;/for π=1,2,3,4,...πnth roots of unity 1,π,π!,π(,...,π/+'(solutions to π§/=1where n is positiveinteger)and properties:β’1,π,π!,π(,...,π/+'form the vertices of a regular n-gonwith centre at the originβ’1+π+π!+π(+β―+π/+'=0If we know one solution to π§/=π (call it π§')and the solutions to π§/=1(roots of unity) then the roots of π§/=π are π§',π§'π,π§'π!,...π§'π/+'
Eulerβs FormConvertingModulus Argument Form To Eulerβs Form π(πππ π+ππ πππ)=ππ02Note: If given Cartesian form we must turn it into modulus argument form firstConvertingCartesian Form To Eulerβs Form:π+ππβππππ½Turn it into modulus argument form and then into Eulers formMultiplying and Dividing:Multiplying (multiply the moduli and add the arguments)π'π2"Γπ!π2!=π'π!π0(2",2!)Dividing (divide the moduli and subtract the arguments)π'π2"π!π2!=π'π!π0(2"+2!)Inequality Propertiesβ’|π
π(π§)|β€|π§|and |πΌπ(π§)|β€|π§|β’|π§+π€|β€|π§|+|π€|β’|π§+π€|β₯|π§|β|π€|Loci|π§|=πβcircle centre origin and radius π|π§|<πβInside of circle centre origin and radius π|π§|β₯πβoutside of circle including circumference centre origin and radius π|π§βπ|=πβcircle centre πand radius π|π§βπ|<πβInside of circle centre πand radius π|π§βπ|β₯πβoutside of circle including circumference centre πand radius π|π§βπ|=|π§βπ|βlet π§=π₯+ππ¦and use |π+ππ|=βπ!+π!and see which equation you getarg(π§βπ)=πis a line from π₯=πon the π₯axis with angle πfrom the positive π₯axisTrig Powers and Linear FunctionsWriting Trig Powers In Terms of Linear Functions Of Trig:To write powers of πππ in terms of cos and sinUseXπ§+'-Y/=(2πππ π₯)/to write πππ /π₯in terms of cosππ₯and/orsinππ₯Do binomial on LHS and then group using π§/Β±'-'results using π§/+'-'=2cosππUse indices rule (π₯/)>to simplify RHSRearrange for power termTo write powers of π ππin terms of cos and sinXπ§β'-Y/=(2πsinπ₯)/to write π ππ/π₯in terms of cosππ₯and/orsinππ₯Do binomial on LHS and then group using π§/Β±'-'results using π§/β'-'=2πsinππUse indices rules (π₯/)>on RHSRearrange for power termWriting Linear Functions Of Trig In Terms Of Trig Powers:use(πππ π₯+ππ πππ₯)/=cosππ₯+ππ ππππ₯to write cosππ₯or sinππ₯in terms of π ππ>π₯and/orπππ >π₯Note: The equality is true because of Deβ Moivres theoremUse binomial expansion on LHSEquate LHS with the real part of RHS want πππ ππ₯Equate LHS the imaginary part of RHS want π ππππ₯Sum of SeriesUse results about sum of geometric series with complex numbers(sum and sum to infinity)
TypeExplanationsExamplesDefinitionββ1=ππ&=β1We are often asked to do calculate powers of π.Relate to π&using indices rules to deal with. Example 1: π)*=(π&)&)=(β1)&)=1Example 2:π&+=(π&),-π=(β1)π=βπExample 3:ββ10ββ40=ββ1β10ββ1β40=πβ10πβ40=π&β400=β20. Do not make the mistake of saying ββ10ββ40=β400=20Jargon Form: Real + Imaginary so we have π§=π₯+ππ¦with π₯,π¦ββNote: we can also write π§=π₯+π¦πβ’π₯=π
π(π§)means the real part of π§β’π¦=πΌπ(π§)means the imaginary part of π§β’Modulus π+ππ=|π+ππ|=βπ&+π&β’Complex conjugate π§βπππ§Μ
=π₯βππ¦is the complex conjugate of z.Remember if you know one root, then the conjugate is necessarily another root.Properties: o(π§Β±π€)β=π§βΒ±π€βo(π§π€)β=π§βπ€βo'$%(β=$β%βif π€β 0oπ§.π§β=|π§|&β’Argand diagramExample 1:Find the complex conjugate, modulus and state the real and imag parts of 2β3πThe complex conjugate is 2+3π. 2 is the real part, β3is the imaginary partModulus =G2&+(β3)&=β4+9=β13Example 2:π§=2+3π,π€=5β8πFind (π§+π€)β(2β3π)+(5+8π)=7+5πAdding/Subtractingβ’Adding: (π+ππ)+(π+ππ)=(π+π)+π(π+π)β’Subtracting: (π+ππ)β(π+ππ)=(πβπ)+π(πβπ)Example 1:(3β2π)+(4+3π)=(3+4)+(β2+3)π=7+πExample 2:(3β2π)β(4+3π)=(3β4)+(β2β3)π=β1β5πMultiplying/Dividingβ’Multiplying: (π+ππ)(π+ππ)=ππ+πππ+πππ+π&ππ=(ππβππ)+π(ππ+ππ)β’Dividing: /0123014Multiply by complex conjugate πβππ/0123014Γ35143514=/35/41023151"243"51"4"=/35/4102310243"04"and then simplify furtherExample 1:(3β2π)(4+3π)=12+9πβ8πβ6π&=12+9πβ8πβ6(β1)=18+πExample 2:-5&1)0-1=-5&1)0-1Γ)5-1)5-1=,&5615*157,706=75,+1&8=7&8β,+&8πSolving with Complex NumbersWe commonly equate real and imaginary parts in order to solve equations with complex numbers in them.Example 1:Find the values of π₯and π¦if (1βπ)π§=1β3πLet π§=π₯+ππ¦.(1βπ)(π₯+ππ¦)=1β3ππΏπ»π=π₯+ππ¦βππ₯+π¦=(π₯+π¦)+π(βπ₯+π¦)Equating realand imaginaryparts gives π₯+π¦=1and βπ₯+π¦=β3βΉπ₯=2,π¦=β1Example 2:Given that $$5*=β1β2π,find z in the form π+πππ§=(β1β2π)(π§β8)π+ππ=(β1β2π)(π+ππβ8)π+ππ=βπβππ+8β2ππ+2π+16ππ+ππ=(βπ+2π+8)+π(βπβ2π+16)Equating real and imaginary givesβπ+2π+8=β1,βπβ2π+16=β2βΉπ=6,π=2π§=6+2πExample 3:Find the square roots of 8β6πWrite as π§&=8β6πβΊπ§=β8β6π(π₯+ππ¦)&=8β6ππΏπ»π=π₯&+2π₯π¦πβπ¦&=(π₯&βπ¦&)+π(2π₯π¦)Compare coefficients: π₯&βπ¦&=8,2π₯π¦=β6Solving simultaneously gives π₯=Β±3,π¦=β1,so we get π§=3βπ,β3+πFactorising & Solving PolynomialsFactorising: Find a factor and then divide by it like usual.Solving: ΓQuadratics: use quadratic formula as usualΓCubicsand above: Normally given a root. Remember if you know one root, then the conjugate is necessarily another root. It is quicker to usethe equationπ₯&β(π π’πππππ‘π )π₯+πππππ’ππ‘ππππ‘π instead of writing (π₯βππππ‘1)(π₯βππππ‘2)to build an equation based on the 2 conjugate pair roots that we know. We can then divide by this equation to find further roots. These can be hard for students. For more practice, try the following harder examples after seeing the easier examples on the right.β’Factorise the polynomial π(π₯)=π₯)β5π₯-+2π₯&+22π₯β20completely with integer coefficients given 3βπis a root of π(π₯)β’Given that (π§β1β2π)is a factor of 2π§-β3π§&+8π§+5solve the equation 2π§-β3π§&+8π§+5=0over the complex number fieldβ’Let π(π§)=2π§-+ππ§&+ππ§+π, where π,π,ππππββ. Two of the roots of π(π§)are β2 and (β3+2π). Find the values of π,πand π(ans a=16, b=50, c=52)Example 1:Solve π§&+4π§+8=0π§=5)Β±:)"5)(,)(*)&(,)=5)Β±β5,7&=5)Β±)1&=β2Β±2πExample 2:Completely factorise π(π₯)=π₯)β2π₯-+2π₯&β2π₯+1π(1)=0so dividing by (π₯β1)gives π(π₯)=π₯-βπ₯&+1π₯β1π(1)=0so divide again by π₯β1which gives π₯&+1π₯)β2π₯-+2π₯&β2π₯+1=(π₯β1)(π₯β1)(π₯&+1)=(π₯β1)&(π₯βπ)(π₯+π)Example 3:Find all complex numbers z, such that π§)βπ§-+6π§&βπ§+15=0andπ§=1+2πis a solution to the equation1β2πmust be another root since roots occur in conjugate pairsπ§&β(π π’πππππ‘π )π§+πππππ’ππ‘ππππ‘π =π§&β2π§+5(π§)βπ§-+6π§&βπ§+15)Γ·(π§&β2π§+5)=π§&+π§+3Solve π§&+π§+3π§=5,Β±:,"5)(,)(-)&(,)=5,Β±β,,1&=β,&Β±β,,&πLociβ’|π§|=πβcircle center origin and radius kβ’|π§|<πβinside of circle, centered at origin and radius kβ’|π§|β₯πβoutside of circle including circumference, centered at origin and radius kβ’|π§βπ§>|=πβis a circle of radius πcentered at π§>β’|π§βπ§>|<πβinside of circle centered π§>and radius kβ’|π§βπ§>|β₯πβoutside of circle including circumference, centered at π§>and radius kβ’|π§βπ§>|=|π§βπ§,|To deal with this we let π§=π₯+ππ¦and the take the modulus of each side and see which equation you get. Might be a straight line or circle etc.β’πππ(π§βπ§>)=πis a line from π₯=π§>on the π₯axis with angle πfrom the positive π₯axisDescribe clearly the locus in the complex plane defined by the equation |π§+2π|=|2ππ§β1||π₯+ππ¦+2π|=|2π(π₯+ππ¦)β1||π₯+π(π¦+2)|=|(β1β2π¦)+2π₯π|Gπ₯&+(π¦+2)&=G(β1β2π¦)&+4π₯&π₯&+(π¦+2)&=(β1β2π¦)&+4π₯&3π₯&+3π¦&=3π₯&+π¦&=1Unit circle i.e. circle centre (0,0)radius 1Inequalitiesβ’|π
π(π§)|β€|π§|and |πΌπ(π§)|β€|π§|β’|π§+π€|β€|π§|+|π€|β’|π§+π€|β₯|π§|β|π€|β’|π?@/A|=π?@/A
3 FormsCartesian Formπ§=π₯+ππ¦,with π₯,π¦ββAll above methods and properties have dealt with this formModulus Argument Form (aka Polar form)π§=π(cosπ+πsinπ)where, |π§|=πand argπ§=πBasic Properties:β’π§β=π§#=π(cosπβπsinπ)β’π§"=π"(cosππ+πsinππ) by De Moivreβs Theoremβ’π§#"=π#"(cosππβπsinππ)by De Moivreβs TheoremMultiplying and Dividing Properties:Mod Resultsβ’|π§$π§%|=|π§$||π§%|β’3&#&"3=|&#||&"|Arg Resultsβ’arg(π§$π§%)=arg(π§$)+arg(π§%). This works like indices rules. When we multiply, we add the powers.So [π$(cosπ$+πsinπ$)][π%(cosπ%+πsinπ%)]can be multiplied quickly and is π$π%[cos(π$+π%)+πsin(π$+π%)]=π$π%π((*#+*")β’πππ<&#&"==arg(π§$)βarg(π§%)This works like indices rules. When we divide. we subtract the powers.So -#(./0*#+(012*#)-"(./0*"+(012*")can be divided quickly and is -#-"[cos(π$βπ%)+πsin(π$βπ%)]Useful Results: β’π§+$&=2cosπβ’π§β$&=2πsinπβ’π§"+$&$=π§"+π§#"=2cosππrearranging βcosππ=&$+&%$%β’π§"β$&$=π§"βπ§#"=2πsinππrearranging βsinππ=&$#&%$%(Eulerβs Formππ$,Basic Properties:β’π§=ππ(*=π(cosπ+πsinπ)β’π§β=π§#=ππ#(*=π(cosπβπsinπ)β’π§"=π"π("*=π"(cosππ+πsinππ)β’π§#"=π#"π#("*=π#"(cosππβπsinππ)Converting Between The 3 FormsModulus ArgumentβCartesian π(ππππ½+πππππ½)βπ+ππ:Work out values of the trig function and this will immediately be in cartesian formExample: Convert 2<πππ 34+ππ ππ34=into cartesian form=2M<$%=+π<β4%=N=1+β3πCartesianβModulus Argumentπ+ππβπ(ππππ½+πππππ½)Need to find πππππ. To find πand πwe use the fomrulaπ+ππβTπ=βπ%+π%π=π‘ππ#$<3673=and then draw the angle πin the quadrant where the complex number π+ππliesRead off πby starting on the positive π₯axis (like when you solve for trig using the CAST diagram)(remember: βπβ€π<π)Or use the formula π=β©βͺβ¨βͺβ§tan#$<67=(ππππ’ππππππ‘1ππ4)tan#$<67=+π(ππππ’ππππππ‘2)tan#$<67=βπ(ππππ’ππππππ‘3)Sub this r and πfound into π(πππ π+ππ πππ)Example: Convert π§=1ββ3πinto modulus argument formπ=b(1)&+(ββ3)&=β4=2π=π‘ππ5,'β-,(=B-Complex number is in 4thquadrant so angle is βB-Plugging this into the form we get2'cos(βB-)+ππ ππ(βB-()=2fcos'B-(βππ ππ'B-(gModulus ArgumentβEuler π(ππ¨π¬π½+ππ¬π’π§π½)βππππ½Get into Modulus Argument form if given cartesian form and then you can read Eulerβs form straight off by locating r and πThis is easy since we already know πand πfrom modulus argument formExample:Convert 2<πππ 34+ππ ππ34=into Euler form=2π(34CartesianβEuler π+ππβππππ½Turn this into modulus argument form using second column and then see third columnExample:Convert β2β2πinto Euler Formπ=G(β2)&+(β2)&=β8=2β2π=π‘ππ5,'&&(=B)π=βπβ-B)Complex number is in 3rd quadrant so angle is β-B)Plugging this into the form we get2β2'cos'β-B)(+ππ ππ'β-B)((=2β2π51-B)We commonly use mod/arg form to find the roots of complex numbers. Finding square roots is easy since we can compare coefficients instead like in example 3 under solving with complex numbers section, but when we get to third roots and above the algebra becomes messy and it is easier to put into modulus argument form and use De Moivreβs Example:Find the cube roots of 8ππ§4=8πβΊπ§=(8π)#&Letβs turn 8πinto modulus argument form: r=β8%=8, π=π‘ππ#$<:;==3%π§=M8kπππ <3%+2ππ=+ππ ππ<3%+2ππ=lN#&=8#&kπππ <3<+%"34=+ππ ππ<3<+%"34=lπ§=2kπππ <3<+%"34=+ππ ππ<3<+%"34=lChoose 3 values for πLet π=β1:2kπππ <3<β%34=+ππ ππ<3<β%34=l=2kπππ <#43<=+ππ ππ<#43<=l=2kπππ <3%=βππ ππ<3%=l=2π#'"Let π=1: 2kπππ <3<+%34=+ππ ππ<3<+%34=l=2kπππ <>3<=+ππ ππ<>3<=l=2π(')Let π=0: 2kπππ <3<=+ππ ππ<3<=l=2π')These roots are equispaced around a circle of radius r (see the diagram on the right)
