AN ALTERNATE PROOF OF THE DE MOIVRE'S THEOREM

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1 AN ALTERNATE PROOF OF THE DE MOIVRE'S THEOREM K.V. Vidysgr Deprtment of Mthemtics, Government Degree College, Nrsiptnm-536, Viskhptnm * Author for Correspondence ABSTRACT Trnsformtion is method for solving the system, where we fce difficult. As the well known trnsformtions like Fourier trnsformtion, Lplce trnsformtion, Z-trns formtions helps us to solve different equtions like differentil lgebric problems. In this pper I propose new trnsformtion clled C-trnsformtion, from the set of complex numbers onto the set of ll mtrices of the form where, b re rel numbers. I defined new set M, which is nothing but the set of ll mtrices of the form where, b re rel numbers. The C-trnsformtion is trnsformtion from the set of complex numbers onto M. I further show tht the C-trnsformtion is isomorphism from C onto M. The existence nd uniqueness of such trnsformtion is proved. By using this trnsformtion, I try to give the geometricl interprettion for the mtrix of the form which is nothing but point in the rgnd plne. I gve n lternte proof for the well known Demovier s theorem. I further try to solve conforml mpping from z-plne to w-plne using this C-trnsformtion provided the ngle preserving, ie ( the ngle between the curves in z-plne is equl to the ngle between the curves in w-plne with n exmple wf(z z t the point z+i. INTRODUCTION It is known tht the set of rel number system R is ring with respect to generl ddition, generl multipliction i.e., ( R,+,. is ring. In this ring (R,+,. is the identity element nd - is the dditive inverse of. Now define new set M, which is nothing but, the set of ll mtrices of the form where, b re rel numbers. Hence the set M is ring with respect to mtrix ddition, mtrix multipliction. In 0 0 this ring I (unit mtrix is I 0 the identity element nd -I, I 0 is the 0 dditive inverse of I, I 0 (unit mtrix. When we define complex number Z, we define Z (+ib, where, b re rel numbers. And i, (dditive inverse of. I suppose tht there my be 0 0 unique mtrix J in the set M such tht J -I, I 0 dditive inverse of I, I 0 When we suppose J -I, we should get J 0 i i 0, since J 0 i 0 i i 0 i 0 -I {Of course, we my get 4 mtrix in the form, stisfying the condition J -I

2 i 0 i 0 i 0 When J 0 i, J 0 i 0 i -I When J 0 0, J I 3 When J 0 0 ; J I 4 When J 0 i i 0, J 0 i 0 i i 0 i 0 -I } mong the four the forth mtrix is only the mtrix in M Every complex number z+ib cn be represented s mtrix m I+bJ,,b re rel numbers I identity mtrix nd J 0 i i 0 Proposition: There is isomorphism from The set of complex number C into the set M of ll mtrices of the form where, b re rel numbers. Proof Suppose tht f is trnsformtion from The set of complex number C into the set M of ll z z z z mtrices of the form where, b re rel numbers nd Defined s f(z for z z z z z C, z z z z When z+ib, ; bi then f(z f (+ib As per the definition of the set M, M { M r / M r is the mtrix in the form } ib Proof Prt I:- Result We know tht the set of ll complex number system C is field with respect to generl ddition nd multipliction. Clim The set M is ring with respect to mtrix ddition nd mtrix multipliction For proving this, we lredy Know tht M is belin group with respect to mtrix ddition nd M is group with respect to mtrix ddition i.e. (M,+ is n belin group (M,. is group And since ib ib ib ib (verified ib ib ib ib Hence (M,. is n belin group.

3 Therefore (M,+ is n belin group (M,. is semi group 3Distributive properties holds good Therefore (M,+,. is ring under mtrix ddition nd mtrix multipliction. Clim The set M is commuttive ring with unit element, here unit element is the unit mtrix I. Clim3 The ring (M,+,. is skew field. Clim4 The ring (M,+,. is field since, every non zero element is invertble under mtrix multipliction. Clim 5 The ring (M,+,. is withoutzero divisors, Hence (M,+,. is Integrl domin.. Proof PART II Clim z z z z The defined function f: C z C is n isomorphism M such tht f(z Proof f is one- one nd onto nd well defined Proof f ( z z f ( z f ( z Proof 3 f ( zz f ( z f ( z for z z z z Note when z z, hence f ( zz f ( z f ( z ( f ( z Similrly, f ( zz z3z4... zn f ( z f ( z... f ( zn If z z. z n f ( z z z z... z f ( z f ( z... f ( z ( f ( z n Hence Definition C-TRANSFORMATION z z z z If Z+ib then the c trnsformtion for Z+ib is ib z z z z Tht is C-T(z C-T(+ib z z z z And the inverse trnsformtion for ib is +ib z z z z z ib 3

4 Tht is C T ( z z z z z z z z C T ( ib ib +ib z Trnsformtion on Addition of complex Numbers Let z +ib nd z nd z + z is lso complex number then C-trnsformtion on z + z is C-T( z + z C-T {( +ib +( +ib } ( b b i C-T{( + +i( b + b } ( b b i bi bi + bi bi And the Inverse C-trnsformtion C T bi bi ( + C T bi ( + C T bi ( bi bi bi bi ( +( z + z Let z +ib nd z then the C-Trnsformtion on z z ( ( is C-T ( z z C-T( z.c-t( z b i b i b i b i bb i( b b i( b b bb And the inverse C-Trnformtion for bb i( b b C T ( bb i( b b i( b b bb i( b b bb ( b b +i( b b. By the bove definition of ddition nd multipliction of complex numbers, we cn esily verify ll others xioms. 4

5 Multiplictive Inverse of non-zero complex number Let Z+ib be non zero complex number, then C-Trnsformtion for Z is C-T( Z bi bi Since the multipliction inverse of Z is Z And hence the C-trnsformtion for Z C T ( Z C T ( Z bi bi bi And hence the C-trnsformtion for Z ( b bi bi ib And the inverse C-Trnsformtion for ( b ( b b i ( b ( b ib Hence the Inverse of +ib. ( b bi bi bi bi bi bi bi ( b Note: by the bove definition of ddition nd multipliction of complex number cos If Z cos + then C-T(Z A cos (sy nd cos the inverse trnsformtion C T ( cos C T ( A cos + Z Now we shll prove The De-Movier s theorem by pplying C-trnsformtion De-Moiver s Theorem (Specil Proof: Sttement: Let n be positive integer.then (cos n (cos n n cos Proof : Since C-T(cos +i sin C-T(Z A cos (sy 5

6 n cos n n We know tht A n cos n for nn nd the proof s follows n cos n n For n, A n cos n cos m sin m Let it be true for nm, so tht ( A m sin m cos m cos sin cos m sin m ( A m sin cos sin m cos m cos cos m sin sin m cos sin m sin cos m sin cos m cos sin m cos cos m sin sin m cos( m sin( m sin( m cos( m Thus the result is true for nm then it is lso true for n (m+.hence, by the principle of mthemticl induction,it is true for ll nturl numbers. n cos n sin n A sin n cos n And hence Inverse C-Trnsformtion C T ( n A C T ( cos n sin n sin n cos n (cos n n And hence (cos n (cos n n for nn. Cor: If n is negtive integer, then (cos n (cos n n Proof: Let n-p, where p is positive integer, then (cos n (cos p (cos p (cos p p (cos p p (cos p p (cos p p (cos p p (cos( p ( p (cos n n Hence the proof. Note (: (cos n (cos n n. 6

7 Note (: (cos n (cos n n. Note (3: (cos n (cos n n. Scope: Mtrix opertion is quite esy to ll for solving; I hope this trnsformtion my help for solving the equtions involving complex vribles. I wnt to develop this trnsformtion nd I wish to pply this in the conforml mppings. REFERENCES Jmes wrd brown, Ruel V Churchill (004. Complex vrible nd pplictions/.-7 th ed.p.cm-(brown- Churchill series Interntionl Series in Pure nd Applied Mthemtics. Churchill, Ruel Vnce (899. Functions of complex vrible I. -II. Title. III. series. IV, series: Interntionl series in pure nd pplied mthemtics. John B. Convy Functions of one complex vrible - nd edition, (Springer interntionl student edition, (nros publictions Wlter rudin Rel nd Complex nlysis-3 rd edition (Pushp print series, New Delhi (Tt-MCGrw Hill Publishing compny Ltd. H Elton. Lcey The Isometric theory of Clssicl Bnch Spces-primry edition Springer verlg New York, Heidelberg Berlin. Giovnni snsone-john gerretsen (969. Lectures on the Theory of functions of S Complex Vrible (Wolters Noordhoff publishing Groningen 969 the Netherlnds. WK Hymn. Multivlent Functions (Cmbridge AT the University Press 958 Dr B S Grewl. Higher engineering mthemtics th edition (Hindustn offset press (Khnn Publictions AR vsist, Dr R K Gupt (009. Integrl Trnsforms -8 th edition (Krishn prkhsm medi (P Ltd.meerut Book code 4-8 Kenneth Hoffmn Ry Kunze (00. Liner lgebr - nd edition (Printed hll Indi