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ANALYTIC FUNCTION S.KARTHIK ASSISTANT PROFESSOR MATHEMATICS, MSEC

ANALYTIC FUNCTION S.KARTHIK ASSISTANT PROFESSOR MATHEMATICS, MSEC. ANALYTIC FUNCTION. A single valued function which is defined and differentiable at each point of a domain D is said to be analytic in that domain.

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ANALYTIC FUNCTION S.KARTHIK ASSISTANT PROFESSOR MATHEMATICS, MSEC

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  1. ANALYTIC FUNCTIONS.KARTHIKASSISTANT PROFESSORMATHEMATICS, MSEC

  2. ANALYTICFUNCTION • Asinglevaluedfunctionwhichisdefinedanddifferentiable at each point of a domain D is said to be analytic in that domain. • A function is said to be analytic at a point if its derivative existsnotonlyatthatpointbutinsomeneighbourhoodof thatpoint. • Thepointwherethefunctionisnotanalyticiscalleda singular point of thefunction.

  3. Notes • Letf(z)andg(z) beanalyticfunctionsinsomedomainD; • f (z)g(z) and f(z).g(z) are analytic functions inD. (ii) f(z) isanalyticinDexceptwhereg(z)=0. g(z) (iii) f[g(z)]andg[f(z)]areanalyticfunctionsinD.

  4. 2z is notanalytic? Example-1:Statewherethefunctionf(z) z21 Solution:-f(z) isnotanalyticatziasthefunctionisnotdefinedatthese points. Hence, the points z i are singularpoints. z2 1 Example-2: State where the function f (z) (z2)(z2 isnot 2z2) analytic? Solution:-Thefunctionisnotanalyticatthepoints z2 and 2z20 z20 and z 2 z 2 481i 2

  5. CAUCHY-RIEMANN PARTIAL DIFFERENTIAL EQUATIONS Theorem:If afunctionf(z)=u(x,y) +iv(x,y)isanalyticatanypointz= x+iy, then the partial derivatives ux , uy , vx , vy should exist and satisfythe or u v , u v equations ux =vy , uy =-vx x y y x Proof:Letf(z)beananalyticfunctionatanypointz. lim z 0 f(zz)f(z) z exist f '(z) lim z 0 (uu)i(vv)(uiv) z Now f '(z) lim u iv z 0 z …..(1) 

  6. CAUCHY-RIEMANN PARTIAL DIFFERENTIAL EQUATIONS Forequation(1)limitexistalonganypathasz0.Considerthepathalongreal axis(z=x). z0x0 from(1), lim u iv x0 x f '(z) u iv x x ….(2) Now consider the path along y axis i.e.z=iy z0y0 from(1), lim u ivi u v f '(z) ….(3) y0iy yy

  7. CAUCHY-RIEMANN PARTIAL DIFFERENTIAL EQUATIONS Itisobviousthatthelimits(2)and (3)aresame. uiviuv x x y y ux vy uy vx and

  8. Example 3 : Use C-R equation concept to find derivative of f (z) z 2. Solution:-Wehave f (z) z2 x2 y2 i2xy u(x,y)x2y2,v(x,y)2xy ux 2x vy and uy 2 y vx So, C-R equations are satisfied everywhere inz-plane. f '(z) exists everywhere. Thus weget f'(z)uxivxvyiuy2xi2y2z Example 4 :Show that neither f (z) z nor f (z) z is an analyticfunction. Solution:-We have f (z) z x iy u(x,y)x,v(x,y)y ux 1, uy 0, vx 0, vy 1 ux vy So, C-R equation is notsatisfied. f (z)z is notanalytic.

  9. Continue… Now f (z) zx2 y2 u(x, y) x2y2,v(x,y)0 x u ,v 0 u v x y x y x2 y2 So, C-R equation is notsatisfied. is notanalytic. f (z)z x iy dw isananalyticfunction,find .  Example 5 : Show that w x2 x2 y2 y2 dz Solution:-Wehave, u x y ,v x2 y2 x2 y2 2xy y 2 x2 v   x (x2 y2)2 ux (x2 y 2 )2, y 2 x2 2xy uy (x2 y2 )2, vy (x2 y 2)2 uxvyand . So, w isanalytic. uy vx

  10. Continue… Now, dw ux ivx dz y2x2 2xy   (x2 y2)2 i (x2 y2)2 Example 6 :Check whether the following functions are analytic or not at anypoint: (a)f(z)ez(b)f(z)2xixy2. Solution:- (a) we have f (z) ez exiy ex(cos yisiny) u excosy, v ex siny, u ex cosy, v excosy x y ux vy f (z) ez is not analyticanywhere. (b)Wehave f(z)2xixy2 v xy2 u 2x, ux 2, isnot ux vy f(z)2xixy2 v 2xy analyticanywhere. y

  11. Solution:-Let z0 be any point in thedomain. Example 7 :Examine the analyticity of sinhz. lim z z0 lim z z0 f (z)f (z0 )  sinhzsinhz0  z z0 z z0 z z z z    2cosh sinh 0 0     lim z z0 lim 2 2     z z0 z z   sinh 0   lim z z 2   cosh0 z z0 z z0 2 z z0   2   coshz0 f '(z) exist at any pointz. isanalytic. f(z)

  12. f(z)sinhzsinh(xiy) Continue…. OR Wehave cosysinhxisinycoshx u(x,y)cosysinhx, uxcosycoshx, uy sin y sinhx, v(x,y)sinycoshx vycosycoshx vxsinysinhx, ux vy uy vx and So,C-Requationissatisfiedforanypoint. So, f(z) is analyticfunction.

  13. POLAR FORM OF C.R.EQUATIONS xrcos,yrsin Wehave rx2 y2 ,tan1y x  u 1 v , 1 u v r rr are C-R equations in polarform. r and f'(z)eiuiv r r  

  14. Harmonic Function & Conjugate harmonic function Harmonic Function:-Afunction (x,y)issaidtobeharmonicina domain Dif (1) (x, y) Satisfy Laplace’s equation xx yy 0 and arecontinuousfunctionsofxandyinD. (2)xx,xy,yy Conjugate harmonic function:-If f(z)=u+iv is an analytic functionof z, then v is called a conjugate harmonic function of u and u in its turn is termed a conjugate harmonic function of v. Or u and v are called conjugate harmonicfunctions.

  15. Example8 :Isthefunctionu=xsinxcoshy-cosxsinhyharmonic? Solution:-Wehave uxsinxcoshyycosxsinhy uxsinxcoshyxcosxcoshyysinxsinhy uxxcosxcoshycosxcoshyxsinxcoshyycosxsinhy 2cosxcoshyxsinxcoshyycosxsinhy uyxsinxsinhycosxsinhyycosxcoshy uyyxsinxcoshycosxcoshycosxcoshyycosxsinhy xsinxcoshy2cosxcoshyycosxsinhy uxx uyy 0 also uxx,uxy,uyyarecontinuousfunctions. So, u is a harmonicfunction. And

  16. x Example 9 : Showthat isharmonic. u x2 • y2 x Solution:- Wehave u x2 • y2 y2 (x2 y2 )(1) x(2x) • x2 ux  (x2 y2)2 (x2 y2 )2 (x2y2)2(2x)(y2x2)2(x2y2)(2x) uxx (x2 y2)4 (x y)[(x y )(2x) 4x(y x)] 2 2 2 2 2 2 (x2 y2)4 2x3 2xy2 4xy2 4x3 2x3 6xy2 (x2 y2)3 (x2 y2)3 ….(1)

  17. Continue… 2xy uy (x2 y2)2 (x2y2)2(2x)(2xy)2(x2y2)2y uyy   (x2 y2)4  (x y )[(x y )(2x)8xy(x y)] 2 2 2 2 2 2 2 2 (x2 y2)4 [2x32xy28xy2]2x36xy2 …(2) (x2 y2)3 So, From equation (1) &(2), uxx uyy 0 (x2 y2)3 Alsouxx,uxy,uyyarecontinuousfunctions. So, u is a harmonicfunction.

  18. Example 10 : Determine a and b such that u ax3 bxy its conjugateharmonic. is harmonic andfind Solution:-Wehave u ax3 bxy ux3ax byu 6ax 2 xx uy bx uyy 0 since u is harmonicfunction, uxx uyy 0 6ax0a0 and b assumes anyvalue u bxy ux by, uy bx Now, dvvxdxvydy uydxuxdy bxdxbydy

  19. x2 y2 Continue… v b • b c 2 2 bx2 by 2 2   2 f(z)uivbxyic 

  20. Method of constructing a regularfunction Ifonlytherealpartofananalyticfunctionf(z)isgiventhen f(z)2uz,zu(0,0)ci 2 2i  wherecisarealconstant. z z 2 2i uz , z Replace xby and yby tofind 2 2i and put x=y=0 to find u(0,0)   inu(x,y)

  21. Example 11 :Find an analytic function f (z) uiv if u x3 3xy. Solution:-Wehave u x3 3xy z3 z z 3 z3 z z u, 3 z2i   2 2i 8 4 222i andu(0,0)=0 f(z)2uz,zu(0,0)ci 2 2i   z3 3 4 2 2 f (z) z i ci

  22. Example12 :Show that the function u x2 y2 x corresponding analyticfunction. is harmonic and findthe u x2 y2 x Solution:-Wehave ux 2x 1uxx 2 uy2yuyy2 uxx uyy 0 So, u isharmonic. z2 z2 z2 z z 22i44222 z z Now, u,  u(0,0) 0 f(z)2uz,zu(0,0)ciz2zci 2 2i  

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