mso202a: introduction to complex analysis - iitkhome.iitk.ac.in/~gp/mso202lect1.pdf · mso202a:...
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MSO202A:IntroductionToComplexAnalysisInstructor:Dr.G.P.KapoorFB565,DepartmentofMathematicsandStatisticsTel.7609,Email:[email protected]
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Lecture1
TextBooks: E.Kreyszig,AdvancedEngineeringMathematics,8thEd.,JohnWiley&Sons.
RuelV.Churchill,etal:ComplexVariablesandApplications,McGrawHill.
JohnB.Conway:FunctionsofOneComplexVariable,IIEd.,SpringerInternationalStudentAddition.
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ReferenceBooks:
JanG.Krzyz:ProblemsinComplexVariableTheory,AmericanElsevierPublishingCompany.
LarsV.Ahlfors:ComplexAnalysis,McGrawHill.
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SupplementaryCourseMaterial
LectureNotes,AssignmentsandCoursePlanwillbeavailableonthiscourseatthewebpagehttp://home.iitk.ac.in/~gpthroughthelinkMSO202A.In the lecturenotes, someproofsaremarked (*).Suchproofswillnotbeaskedintheexams.TutorialClasses
The assignment problems marked (T) on the assignmentsheetswillbediscussedinthetutorialclasses.
The solutions/hints to the assignment problemsmarked (D)willbemadeavailableonthecourseweb‐site.
Theexercisesgiveninthetextbooksareusuallynotdiscussedin the tutorial classesand the studentsareexpected to solvethese problems on their own. However, the students canapproach the tutor if theyhaveanydifficulty in solving suchproblems.
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Evaluationplan
There will be 2 pre‐announced Quizzes of 40‐minutesdurationandaweightageof20%marksforeach.
TheEnd‐CourseExaminationwillbeof2‐hoursdurationwithaweightageof60%marks.
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ReviewofComplexNumberSystemComplex numbers were introduced to have solutions ofequationslike 2 1 0x whichdonotpossessasolutionintherealnumbersystem.Acomplexnumber z isanorderedpair( , )x y ofrealnumbers.If 1 1 1 2 2 2( , ), ( , )z x y z x y , the elementary operations aredefinedas
1 2 1 2 1 2( , )z z x x y y
1 2 1 2 1 2,z z if x x y y
1 1 1( , )z x y
1 2 1 2 1 2 1 2 2 1( , )z z x x y y x y x y
( , )z x y , 2 2z x y
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Notations:
Throughout in the sequel, denote Complexnumber( ,0) , (0,1)a a i . With these notations, RC,where R is set of all real numbers and C is set of allcomplexnumbers.
TheEuclideandistancebetweenanytwopoints 1 2,z z Cisdefinedas 1 2z z andissometimesdenotedby 1 2( , )d z z .
Notethat 2,| | 1, 1.i i i i Thus,thecomplexnumberi isthesolutionoftheequation 2 1 0x .Further,writing ( ,0), ( ,0), (0,1)x x y y i ,itiseasilyseenbyusing the definitions of addition and product of complexnumbers that x + i y = z. This is called the cartesianrepresentation of the complex number z.
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Proposition1. 2
zz z
Proof: 22 2 2 2( , ).( , ) ( ,0)x y x y x y x y z
Proposition2.2 2 2 2
1( , ), 0
x yif z
z x y x y
Proof:2 2 2 2 2
1( , )
z z x y
z z x y x yz
.
Proposition3.Re ,Im2 2
z z z zz z
i
Proof.Wegiveheretheproofofthesecondpartoftheproposition.Thefirstpartfollowssimilarly. (0,2 )(0, 1) (2 ,0)
Im2 2 2
z z i y yy z
.
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PolarrepresentationofComplexNumbers
With cos , sin , (cos sin ) ( cos , sin )x r y r z r i r r iscalledthepolarrepresentationofthecomplexnumberz.
,r z 0
arg
angle between the line segment from to z and
positive real axis
z
Itfollowsimmediatelythat 1 2 1 2 1 2 2z z r r and k .Further,if 1 1 1 1 2 2 2 2(cos sin ), (cos sin )z r i z r i ,itfollowsthat
1 2 1 2 1 2 1 2(cos( ) sin( ))z z r r i .Thus, 1 2 1 2arg( ) arg( ) arg( )z z z z .
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Similarly,usinginduction,if(cos sin ), 1,2,..., ,j j j jz r i j n then
1 2 1 2 1 1... ... (cos( ... ) sin( ... ))n n n nz z z r r r i Thus, 1 2 1 2arg( ... ) arg arg ... argn nz z z z z z .
Inparticular, (cos sin ), 0.n nz r n i n n Toprovethisidentityforn<0,wehave1 1 cos sin 1
(cos sin( )).(cos sin )
ii
z r i r r
Sothat,forn<0,
1 1( ) [ (cos sin( )]
[cos sin ].
n n n
n
z z ir
r n i n
Thus, [cos sin ]n nz r n i n forallintegersn.Takingr=1inthisidentity,(cos sin ) cos sin ,ni n i n for all integersn whichiscalledDe‐Moivre’sTheorem.
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Aspecialwordaboutargumentofacomplexnumberargzisnotafunction,sincefor ,iz re arg z has all the values
, 2 , 4 ,... soitisnotsinglevalued.Theidentity 1 2 1 2arg( ) arg argz z z z hastobeinterpretedinthesensethatforsomevalueofargonLHS, suitable values of 1arg z and 2arg z on RHS so thatequalityholds.Conversely, forgivenvaluesof 1 2arg argz and z on RHS, suitable values of 1 2arg( )z z on RHS so thatequalityholds.For example, if 1 2z z i and the values of their arguments
are given as 1 23 3
arg , arg2 2
z z
, then 1 2 1z z andout of
all the values 3 2 , 0,1,2,...k k . of 1 2arg( )z z , we mustchoose 1 2arg( ) 3z z ,sothat 1 2 1 2arg arg arg( )z z z z holds.
1 2
3arg arg
2z z
i
1
1 2arg( ) 3z z
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Conversely,if 1 2arg( ) 5z z isgiven,thenwecantake
1 23 11 3
arg 2 , arg 42 2 2 2
z z .
Tomakeargzafunctionofzinthestrictsenseofthedefinitionofafunction,werestricttherangeofargzas ( , ] (orwithanother convention, some authors restrict this range as[0,2 ) ). Once the range of arg z is so restricted, arg z isdenotedbyArgz.Thus, Arg z (or,0 2Arg z withtheotherconvention).
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Remark:If z x iy ,theprincipalvalueof 1tany
x ,denoedas
1Tany
x ,satisfies 1
2 2
yTan
x
,while Arg z .The
relationbetweenArg z and 1 yTan
x isthereforegivenbythe
following:
1
1
1
, 0
, 0, 0
, 0, 0
yTan if x
xy
Arg z Tan if x yxy
Tan if x yx
Graphof 1Tan x
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Remark:Notethat,ingeneral, 1 2 1 2( )Arg z z Arg z Arg z (1)Forexample,withtheconvention Arg z
if 1 2 1 21, , ,2
z z i then Arg z Arg z ,sothat
1 23
2Arg z Arg z
.
But, 1 2( )2
Arg z z
.This illustrates (1),when theconvention
is Arg z .Similarly,withtheconvention0 2Arg z ,if
1 2 1 23
1, , ,2
z z i then Arg z Arg z ,sothat
1 25
2Arg z Arg z
.
But, 1 2( )2
Arg z z
.Thisillustrates(1),whentheconventionis
0 2Arg z .
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Solutionoftheequation nz c :Let
0 0 0(cos sin )
(cos sin )
c r i
and
z r i
Theequation nz c gives 0 0 0(cos sin ) (cos( 2 ) sin( 2 ))nr n i n r k i k
0 0
1/ 00
2
2, 0,1,..., 1
n
n
r r and n k
kr r and k n
n
Therefore,thensolutionsoftheequation nz c are
1/ 0 00
2 2[cos( ) sin( )], 0,1,..., 1n
kk k
z r i k nn n
.
Since, 1/
0n
kz r , all the roots of nz c lie on the circle1/ 1/
0 0(0, ) { : }.n nC r z z r Further, since the angles
0 2, 0,1,..., 1
kk n
n
,dividethiscircleintonequalsectors,
alltheserootsareequispacedon 1/0(0, ).nC r
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Example:Alltherootsof 1nz ,callednthrootsofunity,canbewrittenas
2 2 2( 1) 2( 1)cos0 sin0, cos sin ,...,cos sin
n ni i i
n n n n
or
2 1 2 21, , ,..., ; cos sin .n
n n n nw w where w in n
Notethatif 0z isanyrootoftheequation
nz c ,thenalltherootsofthisequationaregivenby 2 1
0 0 0 0, , ,..., nn n nz z z w z w
since, 1/ 0 02 2[cos( ) sin( )], 0,1,..., 1
nk
k kz c i k n
n n
givesthat
1/ 0 02 2 2 2
[cos( ) sin( )],
0 1
nlk n
k l k lz c i
n nl n
,
whosedistinctvaluesareobtainedfor 1.k k l k n
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VectorRepresentationofComplexNumbersAny complex number ( , )z x y can be represented as thevector , ( )z x i y j r say
.
Thisrepresentationhelpsingeometricallyvisualizingadditionandsubtractionofcomplexnumbersasvectors.However,itdoesnothelpinvisualizingtheproductofcomplexnumbers as this is different from the vector product ofcorrespondingvectors.
1 1 1 1 2 2 2 2
1 2 1 2 1 2 1 2 2 1
1 2
1 2 1 1 1 2 1 2 1 2
2 2
( ,
( , )
, , ,
0 ( ) ( )
0
, if z x i y j r and z x i y j r then
z z x x y y x y x y is in xy plane itself
while for the corresponding vectors r r
i j k
r r x y x y y x k z z k
x y
since
.)is perpendicular to xy plane
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RepresentationofPoints,CurvesandRegionsbyComplexNumbersRepresentationofPoints
z
-
r
1/r 1/z
z
-z z
z
1z
1 2z z
2z
ziz
90
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EquationofacircleanddiskintermsofComplexNumbersEquationofacirclewithcenter 0z andradiusr :
0z z r Equationofanopendiskwithcenter 0z andradiusr :
0z z r Equationofacloseddiskwithcenter 0z andradiusr :
0z z r EquationofaLineintermsofComplexNumbersEquationofalineLpassingthrougha
andparalleltovectorb
is r a t b
, t
or,intermsofnotationofacomplexvariablesz,aandb,thisequationis z a t b
z a
tb
Im( ) 0
z a
b
.
Thus,equationofthelineLisgivenby
:Im( ) 0z a
L zb
.
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*AlgebraicStructureofComplexNumbersField:( , , .)X isafieldif
(i) (X,+)isanabeliangroup.(ii) (X–{0},.)isanabeliangroup.(iii) ‘.’isdistributiveover‘+’.
Itiseasilyverifiedthat(C,+,.)isafieldthatcontainsthefield(R,+,.).OrderedSet:(X,<),where,‘<’isarelation,iscalledanorderedsetif
(i) Oneandonlyoneofthestatementsx<y,x=y,y<xholdsforanyxandy.
(ii) ‘<’istransitive.OrderedField:AnorderedsetXiscalledanorderedfieldif
(i) Xisafield(ii) Xisanorderedset(iii) Ify<z,thenx+y<x+zforallx,yandzX(iv) Ifx>0,y>0,thenxy>0.
Itiseasilyverifiedthat(C,+,.)isafieldaswellasanorderedsetwithrespecttodictionaryordering(dictionaryorderon isdefinedby
1 1 2 2 1 2 1 2 1 2( , ) ( , )x y x y if either x x or if x x then y y .
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However, (C, +, .) is not an ordered fieldwith any order,sinceineveryorderedfield1isalwayspositive(for,either1ispositiveor‐1ispositiveand,if‐1ispositive,then(‐1)(‐1)=1ispositive,whichisacontraction),sothat‐1=(‐1,0)isalwaysnegative.Now,either(0,1)>0or‐(0,1)>0
If(0,1) 0 (0,1).(0,1) ( 1,0)then (0,0) , whichimplies(R,+,.)cannotbeanorderedfield.If (0,1) 0 (0,1). (0,1) ( 1,0)then (0,0) ,whichimplies( ,+,.)cannotbeanorderedfield.
Alternatively,ineveryorderedfield,squareofeveryelementispositive. This gives ‐1 is positive being square of (0,1), acontradiction since ‐1 is always negative as in the abovearguments.