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31 Unit_III Comple Nmbes: In the sstem of eal nmbes R we can sole all qadatic eqations of the fom a + b + c 0, a 0, and the disciminant b 4ac 0. When the disciminant b 4ac < 0, the soltion of this qadatic eqation do not belong to the sstem of. In fact, a simple qadatic eqation of the fom + 0, does possesses soltion in eal. This difficlt was oecame b intodcing the imagina pat nit i, whee i. Ths the set of comple nmbes defined as. C {( + i) :, R and i }. Some Basic Reslts:. If +i is a comple nmbe, then the comple nmbe i is called the comple conjgate of, and ( + i)( i) +.If +i is a comple nmbe, then the modls of, denoted b + 3. A comple nmbe + i is epesented b a point p(,) in the Catesian plane with abscissa and odinate. Then the -ais is called eal ais and the -ais is called the imagina ais.the point p(, ) is efeed to as the point. Let OP and XOP θ. Then cosθ, sinθ Ee Comple nmbe can +i be epessed in the fom as gien below (cosθ + isinθ ) pola fom iθ e eponential fom We obsee that + the modls of and it epesents the distance of the point fom the oigin. Also θ tan - Skps Media the angle θ is called the ag ment of. + i - ( ) + i( 4. Let then ) ( ) ( ) Now - 0 ma be epesented as

32 - 0 R (cosθ + i sinθ ) and 0 R sees as the comple eqation of the cicle C with (, 0 0 ) and adis R. In paticla epesents the cicle with cente at the oigin and adis eqal to. Fnctions of a comple aiable: Let C be a set comple nmbes. If to each comple nmbe in C thee coesponds a niqe comple nmbe w., then w is called a comple fnction of defined on C, and we wite w f(). Hence,w has a eal pat, sa and an imagina pat, sa. Then, w has the epesentation W f() (,) + i (,) (Catesian fom) W (,θ) + i (,θ) (pola fom) Continit : A comple aled fnction f() is said to be continos at a point 0 if f() lim f() is defined at 0 and. 0 0 Note:. If a comple aled fnction f() is diffeentiable at a point 0, then it is continos at 0.. The conese of the aboe eslt is not alwas te. The continit of a comple fnction need not impl its diffeentiabilit. Deiatie of a comple fnction: The deiatie of a comple fnction f at a point 0, denoted b f ( 0 ), f(0 + ) f(0 ) is defined as f ( 0 ) lim, poided this limit eists. 0 Sbstitting -,we hae 0 Skps Media f() f(0 ) f ( 0 ) lim 0 We shold emembe that b the definition of limit f() is defined in a neighbohood of 0 and ma appoach 0 fom an diection in the comple plane. The deiatie of a fnction at a point is niqe if it eists. Analtic Fnctions. Cach-Riemann eqations. In comple analsis we ae inteested in the fnctions, which ae diffeentiable in some domain, called the analtic fnctions. A lage aiet of fnctions of comple aiables which ae sefl fo applications ppose ae analtic.

33 A Fnction f() is said to be analtic at a point 0, if it is diffeentiable at 0 and, in addition, it is diffeentiable thoghot some neighbohood of 0. Fthe a fnction f() is said to be analtic in a domain D if f() is defined and diffeentiable at all points of D. In fact, analticit is a global popet while diffeentiabilit is a local popet. The tems egla and holomophic ae also sed in place of analtic. Cach-Riemann Eqations : Cach Riemann eqations poide a citeion fo the analticit of a comple fnction W f() (,) + i (,). Statement: Necessa conditions fo a fnction to be analtic.: If f() (,) + i (,) is continos in some neighbohood of a point + i and is diffeentiable at, then the fist ode patial deiaties of (,) and (,) eist and satisf the Cach-Riemann eqations and. At the point + i. Poof: Since f() is diffeentiable at, we hae f( + ) f() f ( ) lim 0 { ( +, + ) + i ( +, + ) } { (.) + i (, ) } f ( ) lim 0 + i Skps Media Let s assme to wholl eal and wholl imagina. Case I: When wholl eal, then 0, so that the ight side of eqation (I) becomes, (I).The limit on { ( +, ) - (.) } { ( +, ) - (.) } f ( ) lim + i lim i Case II: When wholl imagina, then 0, so that limit on the ight side of eqation (I) becomes, (II) i.the

34 { } + i (.) - ), ( ) ( f lim0 { } + + i (.) - ), ( lim i 0 i + i (III) Since f() is diffeentiable the ale of the limits obtained fom (II) and (III) mst be eqal. i + i Compaing the eal and imagina pats, we get and at the (,). These ae known as the Cach-Riemann eqations. Satisfaction of these eqations is necessa fo diffeentiabilit and analticit of the fnction f() at a gien point. Ths, if a fnction f() does not satisf the Cach- Riemann eqations at a point, it is not diffeentiable and hence not analtic at that point. E : If w log, find d dw, and detemine whee w is not analtic. Let s conside in eponential fom, ) sin (cos e i θ θ θ + i ( ) + tan, θ i) log( i w + + ( ) + + tan i log - Eqating eal and imagina pats ( ) + tan : log - : + + : + + Skps Media

35 Now fom- C-R eqations and Ths w log the C-R eqations holds good fo ( + ) 0 Fthe, dw d dw d d + i d - i i ( + ) + + ( )( ) Ths ee point othe than oigin ( i.e. 0 ) + wlog is diffeentiable and the fnction log is analtic ee whee ecept at oigin. E: Show that the fnction w sin is analtic and find the deiatie. w + i sin( + i) sin cosi + cos sini () i i i -i e e e e Now sin and cos i sini isinh : cosi cosh Using these in eqation () W ( + i) sin cosh + icos sinh Eqating eal and imagina pats, we get sin cosh : cos sinh cos cosh, sin sinh () sin sinh, cos cosh The C-R eqations ae satisfied and f() sin is analtic. d f ( ) + i cos cosh + i( - sin) sinh d cos cosi sin sin i Skps Media

36 ( + i) cos cos Conseqences of C-R Eqations: ). If f() + i is an analtic fnction then and both satisf the two dimensional Laplace eqation. φ φ + 0 This eqation is also witten as φ 0. Hee is the two- dimensional Laplacian. Since f() is analtic we hae Cach-Riemann eqations ( I) and ( II) Diffeentiating (I) w,,t. and (II) w..t patiall we get and Bt is alwas te and hence we hae o + 0,this implies is hamonic. Similal Diffeentiating (I) w..t. and (II) w..t. patiall we get o + 0, this implies hamonic. If f() + i is an analtic fnction, then and ae hamonic fnctions. Hee, and ae called hamonic conjgates of each othe. Conseqence II: If f() + i is an analtic fnction, then the eqations (, ) c And (, ) c epesent othogonal families of ces. Soln:, c (i) ( ) (, ) c (ii) Diffeentiating eqn (i) patiall w..t Skps Media

37 + d d 0 ( ) d o m (I) d Diffeentiating eqn (ii) patiall w..t. ( ) d d + 0 o m (II) d d The two families ae othogonal to each othe, then m m, And sing C-R eqations ( )( ) m m ( )( ) ( )( ) ( )( ) Hence the ces intesect othogonall at ee point of intesection. Note: The conese of the aboe eslt is not te. The following eample eeals the popet. : + d d c : + c ( ) ( ) m fo ce c (- ) ( ) ( ) Skps Media d m fo ce c d ( 4) m m. The intesect othogonall. Bt C-R Eqations ae not satisfied

38 and. Some diffeent foms of C-R Eqations: If w f() + i, is analtic, then the following eslts follows.. ( ) f i + i + i i i + + i i w i w. ( ) f + ( ) f sing C-R Eqations. Based on the eslts aboe mentioned the following eslts ae alid, a) ( ) f ( ) f + ( ) f b) ψ is an diffeential fnction of and then + ψ ψ + ψ ψ ( ) f. c) + ( ) [ ] f Re ( ) f Skps Media

39 d) + f ( ) 4 f ( ) Constction of An Analtic Fnction When eal o Imagina pat is Gien (Ptting in Eact diffeential M d + N d 0) The Cach-Riemann eqations poide a method of constcting ± is gien. an analtic fnction f() +i when o o Sppose is gien, we detemine the diffeential d, since (,), d d + d sing C-R eqations,this becomes d d + d M.d + N d And it is clea that N M + 0 Becase is hamonic. This shows that M d + N d is an eact diffeential. Conseqentl, can be obtained b integating M w..t. b teating as a constant and integating w..t. onl those tems in N that do not contain, and adding the eslts. Similal, if is gien then b sing Skps Media d d + d d d. Following the pocede eplained aboe we find, and hence f() + i can be obtained. Analogos pocede is adopted to find +i when ± is gien. Milne-Thomson Method: An altenatie method of finding is gien. Sppose we ae eqied to find an analtic fnction f() + i when is gien. We ecall that ± i when o o ±

40 f ( ) i (I) φ () φ (, ) i (, Let s we set (, ) and (, ) (II) φ Then ) (III) f φ Replacing b and b 0, this becomes (IV) f () φ (,0) iφ (,0) Fom which the eqied analtic fnction f() can be got. Similal, if is gien we can find the analtic fnction f() + i b stating with f ( ) + i Analogos pocede is sed when ± is gien. Applications to flow poblems: As the eal and imagina pats of an analtic fnction ae the soltions of the Laplace s eqation in two aiable. The conjgate fnctions poide soltions to a nmbe of field and flow poblems. Let be the elocit of a two dimensional incompessible flid with iigational motion, V i + j () Since the motion is iotational cl V 0. Hence V can be witten as φ φ φ i + j (II) Theefoe, φ is the elocit component which is called the elocit potential. Fom (I) and (II) we hae Skps Media

41 φ φ, (III) Since the flid is incompessible di V 0. φ φ (IV) φ φ + 0 This indicates that φ is hamonic. The fnction φ (, ) is called the elocit potential, and the ces, ae known as eqi -potential lines. φ ( ) c ψ so that Note : The eistence of conjgate hamonic fnction (, ) (, ) iψ (, ) w() φ + is Analtic. d ψ The slope is Gien b d ψ φ φ This shows that the elocit of the flid paticle is along the tangent to the ce ψ (, ) c, the paticle moes along the ce. ψ (, ) c - is called steam lines φ(, ) c - called eqipotential lines. As the eqipotential lines and steam lines ct othogonall. w() φ + (, ) iψ (, ) Skps Media dw d φ ψ + i φ φ i

42 The magnitde of the flid elocit ( + ) dw d The flow patten is epesented b fnction w() known as comple potential. Comple potential w() can be taken to epesent othe two-dimensional poblems. (stead flow). In electostatics (, ) c (, ) c φ --- intepeted as eqipotential lines. ψ --- intepeted as Lines of foce f. In heat flow poblems: φ(, ) c --- Intepeted as Isothemal lines ψ, c --- intepeted as heat flow lines. ( ) Cach Riemann eqations in pola fom: iθ Let f() f(e ) (, θ ) + i(, θ ) be analtic at a point, then thee eists fo continos fist ode patial deiaties,,,, θ θ : θ θ,. and satisf the eqations Poof: The fnction is analtic at a point ( ) lim0 f( + ) f() iθ e. eists and it is niqe. θ + θ Now f() (, ) i(, ). Let be the incement in, coesponding incements ae, θ in andθ. Skps Media

43 f ( ) lim f ( 0 { ( +, θ + θ ) + i ( +, θ + θ ) } { (. θ ) + i (, θ ) } { ( +, θ + θ ) - (. θ ) } { ( +, θ + θ )- (. θ) } lim ) Now lim 0 + i (I) iθ e and is a fnction two aiables and θ, then we hae + θ. θ θ iθ iθ ( e ) + ( e ) θ iθ e + i e When (I). Let iθ θ tends to eo, we hae the two following possibilities. θ 0, so that e And Z 0, implies 0 f ( iθ { ( +, θ ) - (. θ ) } { ( +, θ ) - (. θ ) } ) i lim lim 0 e iθ + 0 e iθ The limit eists, f ( ) e iθ + i (I) iθ. Let 0, so that i e And 0, impl 0 θ { (, θ + θ ) - (. θ ) } { (, θ + θ ) - (. θ ) } f ( ) lim θ 0 Skps Media i e iθ θ + i 0 lim iθ i e θ

44 + θ θ θ i e i i + θ θ θ i e i θ θ θ i e () f -i (II) Fom (I) and(ii) we hae θ θ o θ θ, Which ae the C-R Eqations in pola fom. Hamonic Fnction: A fnction φ -is said to be hamonic fnction if it satisfies Laplace s eqation 0 φ Let ) i(, ) (, ) f(e f() i θ θ θ + be analtic. We shall show that and satisf Laplace s eqation in pola fom θ φ φ φ The C-R eqations in pola fom ae gien b, θ (I) θ (II) Diffeentiating (I) w..t and (II) w..θ, patiall, we get + θ : θ θ And we hae θ θ Skps Media

45 + θ Diiding b we, get θ Hence -satisfies Laplace s eqation in pola fom. The fnction is hamonic. Similal - is hamonic. Othogonal Sstem: dθ Let f( θ ) and tanφ, φ being the angle between d the adis ecto and tangent. The angle between the tangents at the point of intesection of the ces is φ φ. Tanφ Tanφ,is the condition fo othogonal. Conside (, θ ) c. Diffeentiating w..t θ, teating as a fnction of θ. d + 0 dθ θ Ths Tan d θ dθ dθ φ ( ) d ( ) θ (, θ ) c Similal fo the ce ( ) ( ) (I) θ - ( (II) ) θ Tanφ ( ) Skps Media

46 T anφ T anφ θ B C-R Eqations ( ) ( ) ( ) ( ) θ θ, θ The eqation edces to T anφtanφ ( ) (- ) θ θ ( ) ( ) θ θ Hence the pola famil of ces θ and θ, intesect othogonall. (, ) c (, ) c Constction of An Analtic Fnction When eal o Imagina pat is Gien(Pola fom.) The method de to Eact diffeential and Milne-Thomson is eplained in ealie section. E: Veif that ( cosθ ) fnction. Soln: is hamonic. find also an analtic Skps Media cosθ : sin θ θ 6 cosθ : 4 cosθ 4. θ Then the Laplace eqation in pola fom is gien b, + + θ 6 cos θ 4 cos 4 cos θ 0 4 θ

47 Hence -satisfies the laplace eqation and hence is hamonic. Let s find eqied analtic fnction f() +i. We note that fom the theo of diffeentials, d d + θ dθ Using C-R eqations - d + dθ θ sinθ d cosθ dθ 3 d - sinθ Fom this - sinθ + c θ, f() + i cosθ + i - sinθ + c θ [ cosθ - isin ] ic + Skps Media -iθ e + ic + ic iθ ( e ) f() ic. + E :Find an analtic fnction f() +i gien that - sinθ 0 θ

48 Soln: + sinθ : cos θ θ To find sing the diffeentials d d + θ dθ Using C-R eqations, θ θ d + - dθ θ - cosθ d - + sinθ dθ cosθ d - sinθ dθ + d + cosθ + cos θ + c f() +i Skps Media + cosθ + c + i - sinθ (cos θ + isinθ + cosθ - isinθ + iθ -iθ f() e + e + + c ( ) c E: Constction an analtic fnction gien (Milne Thomson Method) cosθ.

49 cosθ (I) cosθ f ( ) e iθ + i Using C-R eqations f θ θ, sinθ -iθ ( ) e cosθ + i (- sinθ ) - [ cosθ i ( sinθ )] θ e -i + e -iθ [ cosθ + isinθ ] Now pt, and θ 0. ( ) ( ) + c f on integating f. COMPLETION OF UNIT-I Skps Media θ

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55 Unit.3 Analtic Fnctions UNIT III ANALYTIC FUNCTIONS Pat-A Poblem State Cach Riemann eqation in Catesian and Pola coodinates. Soltion: Catesian fom:, Pola fom:, Poblem State the sfficient condition fo the fnction f() to be analtic. Soltion: The sfficient conditions fo a fnction f i to be analtic at all the points in a egion R ae (), (),,, ae continos fnctions of and in egion R. Poblem 3 Show that f e Soltion: f i e i e i e e e cos isin e cos, e sin e cos, e sin e sin, e cos i.e.,, Hence C-R eqations ae satisfied. f e is analtic. is an analtic Fnction. Skps Media Poblem 4 Find whethe f ( ) Soltion: Gien f i i.e.,, is analtic o not.

56 Unit.3 Analtic Fnctions, 0 0, C-R eqations ae not satisfied anwhee. Hence f is not analtic. Poblem 5 State an two popeties of analtic fnctions Soltion: (i) Both eal and imagina pats of an analtic fnction satisf Laplace eqation. i.e., 0o 0. (ii) If w i is an analtic fnction, then the ces of the famil, c. othogonall the ces of the famil Poblem 6 Show that Soltion: f ( ) f (0) Lim Lim Lim Lim f is diffeentiable at 0. Let i i f i0 i i, 0, 0, 0, c, ct f is diffeentiable at 0 bt not analtic at 0. The C-R eqation and ae not satisfied at points othe than 0. Theefoe f is not analtic at points othe than 0. Bt a fnction can not be analtic at a single point onl. Theefoe Skps Media f is not analtic at 0 also. Poblem 7 Detemine whethe the fnction Soltion: Gien f i i.e.,, i( ) is analtic.

57 Unit.3 Analtic Fnctions,, and C-R eqations ae not satisfied. Hence f is not analtic fnction. Poblem 8 Show that sinh cos is hamonic Soltion: sinhcos coshcos, sinhsin sinhcos, sinhcos Hence is a hamonic fnction. sinhcos sinhcos 0 Poblem 9 Constct the analtic fnction f ( ) Soltion: e cos e cos Assme,,0,0 e e sin Assme,,0,0 0 ',0,0 e d i 0. f f d d i d f e C fo which the eal pat is e cos. Skps Media 3

58 Unit.3 Analtic Fnctions Poblem 0 Poe that an analtic fnction whose eal pat is constant mst itself be a constant. Soltion: Let f i be an analtic fnction,...() Gien c a constant 0, 0 f i 0 & 0 b We know that f i f 0 i0 f 0 Integating with espect to, f C Hence an analtic fnction with constant eal pat is constant. Poblem Define confomal mapping Soltion: A tansfomation that pesees angle between ee pai of ces thogh a point both in magnitde and sense is said to be confomal at that point. Poblem If w f ( ) is analtic poe that dw w i w d w poe that 0 Soltion: w, i, is an analtic fnction of. As Skps Media f is analtic we hae, dw f i i i i d i i i i i i w w i ' Now whee w i and 4

59 Unit.3 Analtic Fnctions w W.K.T. 0 w 0 Poblem 3 Define bilinea tansfomation. What is the condition fo this to be confomal? Soltion: The tansfomation a b w, a d bc 0 c d whee a, b, c, d ae comple nmbes is called a bilinea tansfomation. dw The condition fo the fnction to be confomal is 0 d. Poblem 4 Find the inaiant points o fied points of the tansfomation w. Soltion: The inaiant points ae gien b i.e., i i The inaiant points ae i, i Poblem 5 Find the citical points of (i) Soltion: (i). Gien w dw Fo citical point 0 d dw 0 d i ae the citical points Skps Media 3 w (ii) w. 5

60 Unit.3 Analtic Fnctions 3 (ii). Gien w dw 3 0 d 0 0 is the citical point. Pat-B Poblem Detemine the analtic fnction whose eal pat is Soltion: 3 Gien 3 3 3, 3 3 6,0 3 6, 6 6,0 0 B Milne Thomason method f,0 d i,0 d 3 6 d C C Poblem Find the egla fnction e cos sin Soltion: e cos sin, cos cos sin,0 e e e e e, e sin cos sin e e e, B Milne s Thomson Method f,0 d i,0 d f whose imagina pat is Skps Media 6

61 Unit.3 Analtic Fnctions 0 d i e d e e i C i e e C i e e e C i e C Poblem 3 Detemine the analtic fnction whose eal pat is Soltion: sin Gien cosh cos cosh cos cos sin sin, cosh cos,0 cos cos cos sin cos cos cos cos cos cos cos cos cos cos cos cos cos cos cos cos cos cos ec sin cosh cos 0 sin sinh, cosh cos sin sinh cosh cos,0 0 B Milne s Thomson method f,0 d i,0 d Skps Media cosec d 0 cot C 7 sin. cosh cos

62 Unit.3 Analtic Fnctions Poblem 4 Poe that the eal and imagina pats of an analtic fnction w i satisf Laplace eqation in two dimensions i 0 and 0. Soltion: Let f w i be analtic To Poe: and satisf the Laplace eqation. i.e., To poe: 0and 0 Gien: f is analtic and satisf C-R eqations.. () and.. () Diff. () p.w. to we get.. (3) Diff. () p.w.. to we get.. (4) The second ode mied patial deiaties ae eqal i.e., (3) + (4) satisfies Laplace eqation Diff. () p.w. to we get Diff. () p.w.. to we get (5) + (6) 0 (5) (6) 0 i.e., 0 Satisfies Laplace eqation Skps Media Poblem 5 If f() is analtic, poe that Soltion: Let f() + i be analtic. f () 8 4.f()

63 Unit.3 Analtic Fnctions Then and - () Also + 0 and + 0 () Now f() + and f() + i f ().. and f ().. Similal f ().. Adding (3) and (4) f () 4 4.f () ( (0) ) Poblem 6 Poe that Re f f Soltion: Let f i Re f 0 f (0) ( Skps Media ) (3) (4) 9

64 Unit.3 Analtic Fnctions Poblem 7 Find the analtic fnction e cos sin Soltion: e cos sin Gien f i... if i... f i...(3) i f i F U iv U e cos sin f i gien that 3 ( )...(4) U, e cos e sin o e,, e sin e cos o e, B Milne Thomson method F, o i, o F ie C F d e d i e d (5) Fom (4) & (5) i e C i f i C f e i i 3i C f e 5 i Poblem 8 Find the Bilinea tansfomation that maps the points i, i, i of the -plane into the points 0,, i of the w-plane. Soltion: Gien i, w 0 i, w 3 i, w3 i Coss-atio w w w w3 3 w w w w 3 3 Skps Media 0

65 Unit.3 Analtic Fnctions w 0 i i i i 0 w i ii i i w i i i w i i w i i i w i i w i i i w i w i i i i w i w i i i i w i w i 3 i i w i 3 i i w i i 3 i i w i i 3 i i w i i 3 i i w i i i 3 i i w i w i i i 3 i i i i w i 3 i i i i i w i i i w i 6 3i 3 i i i i w. 5 3i i Skps Media Poblem 9 Poe that an analtic fnction with constant modls is constant. Soltion: Let f i be analtic

66 Unit.3 Analtic Fnctions B C.R eqations satisfied i.e.,, f i f C f C C...() Diff () with espect to () Diff () with espect to (3) () (3) 0 0 () (3) 0 0 W.K.T f i 0 f 0 Integate w..to f C Poblem 0 When the fnction f i is analtic show that, C, C ae Othogonal. Soltion: If f i is an analtic fnction of, then it satisfies C-R eqations,, C...() Gien C,...() B total diffeentiation d d d d d d Skps Media and

67 Unit.3 Analtic Fnctions Diffeentiate eqation () & () we get d 0, d 0 d d 0 d d 0 d / m ( sa) d / d / m ( sa) d / / / m m ( ) / / m m The ces, C and, C ct othogonall. Poblem Show that the fnction conjgate. Soltion: Gien log.. 0 Hence is hamonic fnction To find conjgate of,, o log is hamonic and detemine its Skps Media 3

68 Unit.3 Analtic Fnctions,, o 0 B Milne Thomson Methods f, o i, o f d d 0 log c i f log e f i log i log, log, tan tan Conjgate of is tan. Poblem Find the image of the infinite stips nde the 4 tansfomation w. Soltion: w w i i...()...() Gien stip is when 4 4 ( b ) 4 ( ) 4...(3) which is a cicle whose cente is at 0, in the w -plane and adis. When ( b ) 0 ( )...(4) Skps Media 4

69 Unit.3 Analtic Fnctions which is a cicle whose cente is at 0, and adis is in the w -plane. Hence the infinite stip is tansfomed into the egion between cicles 4 ( ) and ( ) 4 in the w -plane. Poblem 3 Obtain the bilinea tansfomation which maps the points, i, into the points w 0,,. Soltion: We know that w w w w3 3 w w w3 w 3 w 0 i 0 w i w. i i i w. i w i plane 4 0, 0, Skps Media w plane Poblem 4 Find the image of i nde the tansfom Soltion: Gien w w w Now w i w 5

70 Unit.3 Analtic Fnctions i i w i i i i i.e., i...()...() Gien i i i i( ) (3) Sb () and () in (3) ( 0) 4 0 This is staight line in w -plane. 4 Skps Media 4 0, 0 plane 4 w plane 6

71 Unit.3 Analtic Fnctions Poblem 5 Poe that of the w-plane. Soltion: w w( ) w w w w w ( w ) w w Pt i, w i i i i i i i i i i i Eqating eal and imagina pats, w maps the ppe half of the -plane onto the ppe half Skps Media Ths the ppe half of the plane is mapped onto the ppe half of the w plane. 7

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Unit_III Complex Numbers: Some Basic Results: 1. If z = x +iy is a complex number, then the complex number z = x iy is

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