dx f y dy f z dz f 9 z d z 1 2 4 f x i f y 5 dx idy 1 2 4 f x i f y 5 dx idy 1

Dx f y dy f z dz f 9 z d z 1 2 4 f x i f y 5 dx idy 1

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dx + @f @y dy @f @z dz + @f @ 9 z d 0 z = 1 2 4 @f @x " i @f @y 5 ( dx + idy ) + 1 2 4 @f @x + i @f @y 5 ( dx " idy ) = 1 2 @f @x dx + 1 2 @f @y dy + 1 2 @f @x dx + 1 2 @f @y dy + i 2 h @f @x dy " @f @y dx " @f @x dy + @f @y dx i = df 49
IV.8.6 1 2 3 P L K Show that if D is a domain with smooth boundary, and if f ( z ) and g ( z ) are analytic on D [ @D , then Z @D f ( z ) g ( z ) dz = 2 i ZZ D f ( z ) g 0 ( z ) dx dy: Compare this formula with Exercise 1.4. Solution By (8.4), R @D f ( z ) g ( z ) dz = 2 i RR D @ @E ( f 0 g ) dz = 2 i RR D fgdz Using the Leibnitz rule, obtain @ @ 9 z ( f 0 g ) = f @ 9 g @ 9 z + 0 g @f @ 9 z = f @g @z = f g 0 To get Ex1.4, take f = 1 , g ( z ) = z , get R @D 0 z dz = 2 i RR D dxdt = 2 i Area 50
IV.8.7 1 2 3 P L K LLL Show that the Taylor series expansion at z 0 = 0 of a smooth function f ( z ) , through the quadratic terms, is given by f ( z ) = f (0)+ @f @z (0) z + @f @ 0 z (0) 0 z + 1 2 # @ 2 f @z 2 (0) z 2 + 2 @ 2 f 0 z (0) j z j 2 \$ + o % j z j 3 & : Solution Since a smooth function f is a linear combination of the functions 1 ; z; 0 z; z 2 ; 0 z 2 ; j z j 2 and a reminder term O % j z j 3 & , it su¢ces to check the formula for these six functions. The formula holds fro each of these. For instance, for j z j 2 = z 0 z are harmonic. @f @z (0) = 0 , @f @ 9 z (0) = 0 , @ 2 f @ 9 z 2 (0) = 0 , f (0) = 0 , @ 2 f 9 z (0) = 1 . The term O % j z j 3 & has Taylor series 0 + O % j z j 3 & . That ????? it. 51
IV.8.8 1 2 3 P L K LLL Establish the following version of the chain rule for smooth complex- valued functions w = w ( z ) and h = h ( w ) . @ @ 0 z ( h 7 w ) = @h @w @w @ 0 z + @h @ 0 w @ 0 w @ 0 z ; @ @z ( h 7 w ) = @h @w @w @z + @h @ 0 w @ 0 w @z : Solution Suppose w (0) = 0 , h (0) = 0 , write h ( w ) = aw + b 0 w + O % j w j 2 & , a = @h @w (0) , b = @h @ 9 w (0) and w ( z ) = Yw + Z 0 w + O % j z j 2 & , Y = @w @z (0) , b = @w @ 9 z (0) then h ( w ( z )) = a % Yz + Z 0 z + O % j z j 2 && + b % 0 Y 0 z + 0 Zz + O % j z j 2 && + O % j w j 2 & = aYz + aZ 0 z + b 0 Y 0 z + b 0 Zz + O % j z j 2 & = % aY + b 0 Z & z + ( aZ + b 0 Y ) 0 z + O % j z j 2 & This gives, @h , w @ 9 z (0) = aZ + b 0 Y = @h @w @w @ 9 z + @h @ 9 w @w @z @ 9 [email protected] 9 z = @h @w @w @ 9 z + @h @ 9 w @ 9 w @ 9 z @h , w @z (0) = aY + b 0 Z = @h @w @w @z + @h @ 9 w @w @ 0 z |{z} @ 9 [email protected] = @h @w @w @z + @h @ 9 w @ 9 w @z 52
IV.8.9 1 2 3 P L K LLL Show with the aid of the preceding exercise that if both h ( w ) and w ( z ) are analytic, then ( h 7 w ) ( z ) and ( h 7 w ) 0 ( z ) = h 0 ( w ( z )) w 0 ( z ) . Solution If the functions h and w are analytic, then @h , w @ 9 z = 0 + 0 = 0 so h 7 w is analytic. ( h 7 w ) 0 ( z ) = @ @z h 7 w = @h @w @w @z + 0 = h 0 ( w ( z )) w 0 ( z ) 53
IV.8.10 1 2 3 P L K Let g ( z ) be a continuously di§erentiable function on the complex plane that is zero outside of som compact set. Show that g ( w ) = " 1 ; ZZ C @g @ 0 z 1 z " w dx dy; w 2 C : Remark. If we integrate this formally by parts, we obtain g ( w ) = 1 ; ZZ C g ( z ) @ @ 0 z 6 1 z " w 7 dx dy: Thus the "distribution derivative" of 1 = ( ; ( z " w )) with respect to z is the point mass w ("Dirac delta-function"), in the sense that it is equal to 0 away from w , and it is inÖnite at w in such a way that its integral (total mass) is equal to 1. Solution Apply Pompeeiuës formula to a large disk fj z j < R g . 54
V 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 4 5 6 7 8 1
V.1.1 1 2 3 P L K (Harmonic Series) Show that n X k =1 1 k ! log n: Deduce that the series P 1 k =1 1 k does not converge. Hint . Use the estimate 1 k ! Z k +1 k 1 x dx: Solution Let k be a nonnegative integer. Then since the function x 7! 1 =x is decreasing on the positive reals, we have for all x 2 [ k; k + 1] that 1 =x % 1 =k . Hence locking at the oversum in the interval [ k; k + 1] 1 k (( k + 1) & k ) = 1 k !

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• Fall '07
• Lim
• Complex number, Logarithm, -2, Complex Plane, Branch point, Z Öxed