12MC Term 1 Complex Numbers - Weebly...We can also graph a Complex number using the Modulus and...
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1 Year 12 Complex Numbers – Term 1 Chapter 1A - Operations with Complex Numbers Revision - = −1 - ! = −1 - Standard form = + - Conjugate = − You need to be able to Add, Subtract, scalar multiply and multiply. ( Can you recall the difference between multiplication and Scalar multiply? ) Recall from Term 1 Year 11 the definition for the inverse of a number: A number multiplied by its Inverse will result in the Identity Element. For complex numbers that means: × !! = 1 so clearly, it follows that; !! = 1 TASK: Set = + and use algebra to find !! in its simplest form. And you will find … Inverse of ‘z’ What are the pro’s and con’s of remembering this as a rule? ( Hint: Don’t remember the rule, but remember how to find the Inverse of a Complex number) z −1 = 1 z = a a 2 + b 2 − b a 2 + b 2 i
Transcript of 12MC Term 1 Complex Numbers - Weebly...We can also graph a Complex number using the Modulus and...
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Year12ComplexNumbers–Term1
Chapter1A- OperationswithComplexNumbers
Revision
- 𝑖 = −1- 𝑖! = −1- Standardform 𝑧 = 𝑎 + 𝑏𝑖- Conjugate 𝑧 = 𝑎 − 𝑏𝑖
YouneedtobeabletoAdd,Subtract,scalarmultiplyandmultiply.
(CanyourecallthedifferencebetweenmultiplicationandScalarmultiply?)
RecallfromTerm1Year11thedefinitionfortheinverseofanumber:AnumbermultipliedbyitsInversewillresultintheIdentityElement.
Forcomplexnumbersthatmeans:
𝑧 × 𝑧!! = 1
soclearly,itfollowsthat;
𝑧!! =1𝑧
TASK: Set𝑧 = 𝑎 + 𝑏𝑖andusealgebratofind𝑧!!initssimplestform.
Andyouwillfind…Inverseof‘z’
Whatarethepro’sandcon’sofrememberingthisasarule?
(Hint:Don’tremembertherule,butrememberhowtofindtheInverseofaComplexnumber)
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z−1 =1z
=a
a2 + b2−
ba2 + b2
i
2
PolarForm:
𝑧 = 𝑟 cos𝜃 + 𝑖 𝑟 sin𝜃 = 𝑟 cis𝜃
requirestheModulusandArgumentofthecomplexnumber…where;
𝑀𝑜𝑑 𝑧 = 𝑧 = 𝑟 = 𝑎! + 𝑏!
Arg 𝑧 = 𝜃 = tan!!𝑏𝑎
WecanalsographaComplexnumberusingtheModulusandArgumentonanArganddiagram…horizontalaxis=Realandverticalaxis=Imaginary
ToconvertfromPolarFormbacktogeneralform:
-𝑎 = 𝑧 cos𝜃 = 𝑟 cos𝜃
-𝑏 = 𝑧 sin𝜃 = 𝑟 sin𝜃
TASK:Google“HowtopronounceDeMoivre”
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DeMoivres’sTheorem:
𝑧! = 𝑟! cos 𝑛𝜃 + 𝑟! sin(𝑛𝜃) 𝑖
or
𝑧! = 𝑟! cis 𝑛𝜃
SpecificallyDeMoivre’sTheoremreferstocalculatingPowersofcomplexnumbers,howeverbasedonDeMoivre’sTheorem,itfollowsthat;
𝑧!× 𝑧! = 𝑟!𝑟! cos 𝜃! + 𝜃! + 𝑟!𝑟! sin 𝜃! + 𝜃! 𝑖
or
𝑧!× 𝑧! = 𝑟!𝑟! cis 𝜃! + 𝜃!
and
𝑧!𝑧!=𝑟!𝑟!cos 𝜃! − 𝜃! +
𝑟!𝑟!sin 𝜃! − 𝜃! 𝑖
or
𝑧!𝑧!=𝑟!𝑟!cis 𝜃! − 𝜃!
SometimesoperationsoncomplexnumberswillbeeasierinPolarform,ratherthanGeneralform.Soyouneedtobecomfortableconvertingfromonetotheother.
TASK: Theaboverevisionpointsmustberoutineworkforyou.Wherenecessary,Practice!ItmaybeworthwhiletorevisityourYear11MathsCComplexNumbersworkbooks.
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Chapter1B- FactorisationofPolynomialsin‘C’
HerearesomeYr10PracticequestionswithREALrootsonly.Trythisfirstifyouneed,andthenheadintoquestionsthatinvolveComplexRoots.
Factorise 484412)( 23 −+−= xxxxP .
(Hint:x=2isonesolution)
Solution:
0
48244824
20104410
2
24104844122
2
2
23
2
23
−
−
+−
+−
−
+−−+−−
xx
xxxx
xx
xxxxxx
Quotient:x2−10x+24=(x−6)(x−4)
P(x)=(x−2)(x−4)(x−6)
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Now,forthosequadraticswherethediscriminantis“negative’,wecannowfindComplexsolutions…!
TASK: UsetheQuadraticformulatosolvefor𝑧.
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z2 + 8z + 25
Whatdoyounoticeaboutthetworootsoftheaboveexpression?Whattermcanwegivethem?
Whendealingwithasecondorderpolynomial,weareexpectingtwosolutions.Howmanysolutionsareweexpectingforanorder4polynomial?
TASK: Trythisone.
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z4 − 81 = 0 Howareyougoingtoget4solutions?
WhatdoyounoticeaboutthetwoComplexrootsoftheaboveexpression?Whattermcanwegivethem?
Factorisingacubic…Trialanderrorforthefirstfactor,then,
PolynomialDivision…canyourecallhowto…?
TASK: Manuallyfactorise (hint:set𝑧 = −2)
WhatdoyounoticeaboutthetwoComplexrootsoftheaboveexpression?Whattermcanwegivethem?
Andyes,wecangotoaregularquarticlike𝑎𝑥! + 𝑏𝑥! + 𝑐𝑥! + 𝑑𝑥 + 𝑒 = 0,butletsnotgettoocarriedawaywithanythinghigherthanthat.Wemayalsodosomethingalittlemoreinterestinglikesolving;
TASK: Solvefor𝑧,given;𝑧 = 𝑧! + 25𝑧! + 144 = 0
(Makesureyouworkouttheshortcut!)
Youmaynoticetwo“couples”ofsolutions.Whatdoyounoticeaboutthem?
Whatisyourhypothesisaboutcomplexroots?
Yes…it’scalledtheconjugateroottheorem.Basically,if𝑎 + 𝑏𝑖isaroot,then,𝑎 − 𝑏𝑖isalsoaroot!
TASK:Createyourownorder4polynomialequationthathasatleast1complexrootandonerealroot,andthenprepareyourfullyworkedsolution.
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z3 − z2 − z +10 = 0
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SinceweareonthesubjectofConjugates,(andthischapterisabiteasy)letsdigressandtrythese:
1.Showthat𝑧×𝑧 =aRealNumber
2.Prove𝑧×𝑤 = 𝑧×𝑧
3.Find𝑧suchthat;𝑧! + 𝑧 ! = 0
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Solutions:
1.
Set𝑧 = 𝑎 + 𝑏𝑖
Nowwehave;
𝑎 − 𝑏𝑖 𝑎 + 𝑏𝑖
= 𝑎! − 𝑏𝑖 !
= 𝑎! − 𝑏!𝑖!
= 𝑎! + 𝑏!
Since𝑎&𝑏 ∈ ℝ,thenclearly𝑎! + 𝑏! ∈ ℝ
QED.
2.
Set𝑧 = 𝑎 + 𝑏𝑖and𝑤 = 𝑐 + 𝑑𝑖
𝑅𝐻𝑆 = 𝑎 − 𝑏𝑖 𝑐 − 𝑑𝑖
= 𝑎𝑐 − 𝑎𝑑𝑖 − 𝑐𝑏𝑖 + 𝑏𝑑𝑖!
= 𝑎𝑐 − 𝑎𝑑 + 𝑏𝑐 𝑖 − 𝑏𝑑
= 𝑎𝑐 − 𝑏𝑑 − 𝑎𝑑 + 𝑏𝑐 𝑖
𝐿𝐻𝑆 = 𝑎 + 𝑏𝚤 𝑐 + 𝑑𝚤
= 𝑎𝑐 + 𝑎𝑑𝚤 + 𝑏𝑐𝚤 + 𝑏𝑑𝚤!
= 𝑎𝑐 + 𝑎𝑑 + 𝑏𝑐 𝚤 − 𝑏𝑑
= 𝑎𝑐 − 𝑏𝑑 + 𝑎𝑑 + 𝑏𝑐 𝚤
= 𝑎𝑐 − 𝑏𝑑 − 𝑎𝑑 + 𝑏𝑐 𝑖
= 𝑅𝐻𝑆
QED.
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3.
Set𝑧 = 𝑎 + 𝑏𝑖
Nowwehave;
𝑎 + 𝑏𝑖 ! + 𝑎 − 𝑏𝑖 ! = 0
𝑎! + 2𝑎𝑏𝑖 + 𝑏𝑖 ! + 𝑎! − 2𝑎𝑏𝑖 + 𝑏𝑖 ! = 0
2𝑎! + 2 𝑏𝑖 ! = 0
𝑎! + 𝑏𝑖 ! = 0
𝑎! + 𝑏!𝑖! = 0
𝑎! − 𝑏! = 0
𝑎 + 𝑏 𝑎 − 𝑏 = 0
byNFL,wehave
𝑎 = −𝑏, 𝑜𝑟 𝑎 = 𝑏
similarlywecouldsay
𝑏 = −𝑎, 𝑜𝑟 𝑏 = 𝑎
Wecannowseeoursolutionfor𝑧;
𝑧 = 𝑎 − 𝑎𝑖 𝑜𝑟 𝑧 = 𝑎 + 𝑎𝑖
where;𝑎 ∈ ℝ
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Chapter1C- Solvingequationsin‘C’
Afollowonfromlastchapter…IfwecanFactorise,thenwecanfindsolutionsbyusingtheNullFactorLaw…!
Letsfindouthowtocalculate… 𝑧
Lookforeither 𝑎 + 𝑏𝑖 or 𝑧! = 𝑎 + 𝑏𝑖 (samequestion)
TASK: Find𝑧,given𝑧! = 1+ 3𝑖
(Hint…set𝑧 = 𝑎 + 𝑏𝑖andsubintoequation…then‘correlate’coefficients…andnotethatboth𝑎and𝑏arerealnumbers…andrememberwearelookingfor2solutions)
Tryingthatstrategytofindrootsdoesgetcomplicated,butDeMoivretotherescue!
𝑧! = 𝑧!! = 𝑟
!! cis
𝜃 + 2𝑘𝜋𝑛
So,thetrickybitinthebracketistheprocesstogiveus𝑛solutions.Thatisacubicrootwillhave3solutions,aquarticrootgives4solutions,etcetc.
Important… 𝑘 = 0,1,2,3,… ,𝑛 − 1.
Eg.If𝑛 = 5,then𝑘 = 0, 1, 2, 3, 4
TASK: Find𝑧,given𝑧! = 1+ 3𝑖
(ThistimeuseDeMoivre)
Clearly,forlargevaluesof𝑛,DeMoivreistheweaponofchoice…J
TASK: Plotasolutiontoacubedroot(orlarger)onanarganddiagram.Infact,plotafewofyoursolutions,andseeifyoucancomeupwithahypothesisaboutageometricrelationshipbetweenthesolutions,thatisforacubicrootyouhavethreesolutions,foraquarticrootyouhavefoursolutions.***MakesureyouknowtheanswertothisTask!
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Questions:
1.Writetheexpression𝑧! + 1astheproductofthreeREALfactors.
Solution:
Set
𝑧! + 1 = 0
𝑧! = −1
byDeMoivre,𝑧!! = 𝑟
!! cis !!!!"
!
𝑧! = 1 cis !!= !
!+ !
!,givesafactorof 𝑧 − !
!− !
!
𝑧! = 1 cis !!= 0+ 𝑖,givesafactorof 𝑧 − 𝑖
𝑧! = 1 cis !!!= ! !
!+ !
!,givesafactorof 𝑧 + !
!− !
!
𝑧! = 1 cis !!!= ! !
!− !
!,givesafactorof 𝑧 + !
!+ !
!
𝑧! = 1 cis !!!= 0− 𝑖,givesafactorof 𝑧 + 𝑖
𝑧! = 1 cis !! !!= !
!− !
!,givesafactorof 𝑧 − !
!+ !
!
∴ 𝑧! + 1
= 𝑧 −32 −
𝑖2 𝑧 − 𝑖 𝑧 +
32 −
𝑖2 𝑧 +
32 +
𝑖2 𝑧 + 𝑖 𝑧 −
32 +
𝑖2
= 𝑧 − 𝑖 𝑧 + 𝑖 𝑧 −32 +
𝑖2 𝑧 −
32 −
𝑖2 𝑧 +
32 −
𝑖2 𝑧 +
32 +
𝑖2
= 𝑧! + 1 𝑧! − 3 𝑧 + 1 𝑧! + 3 𝑧 + 1
