F19MC2 Solutions 9 Complex Analysis

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1 F9MC Solutions 9 Complex Analysis. (i) Let f(z) = eaz +z. Then f is ifferentiable except at z = ±i an so by Cauchy s Resiue Theorem e az z = πi[res(f,i)+res(f, i)]. +z C(,) Since + has zeros of orer at z = ±i, f has poles of orer at z = ±i. Also we have Thus C(,) an res(f,i) z i (z i)e az + res(f, i) z i (z +i)e az + e az + z = πi i [eia e ia ] = πisin(a). e az z i z +i = eia i e az z i z i = e ia i. (ii) Let f(z) = ( +). Then f is ifferentiable except at z = ±i an so by Cauchy s Resiue Theorem z = πi[res(f,i)+res(f, i)]. ( +) C(,) Since ( +) has zeros of orer at z = ±i, f has poles of orer at z = ±i. Also we have res(f,i) z i z [(z i) ( +) ] z i z (z +i) Moreover Thus C(,) i(z +i) (z +i) = 4ie 4ie z i (z +i) 4 6 res(f, i) z i z [(z +i) ( +) ] z i z = i e. (z i) i(z i) (z i) = 4ie +4ie z i (z i) 4 6 i z = πi( z e ) = π e. (iii) Let f(z) = ( +) = (z +i) (z i). Then f has poles of orer at z = ±i. Only i is insie. Therefore by Cauchy s resiue theorem (z i) f(z)z = πires(f,i) = πilim z i z ( +) = πilim z i z ( (z +i) ) (z +i) z (z +i) = πilim = πi 8i+4i = π z i (z +i) 4 6. =.

2 (iv) Let f(z) = ( +)( +4) = (z +i)(z i)(z +i)(z i). Then f has poles of orer at z = ±i, z = ±i. Only i an i are insie. (z i) res(f,i) z i ( +)( +4) z i (z +i)( +4) = i. = i 6 ; res(f,i) z i (z i) ( +)( +4) z i Therefore by Cauchy s resiue theorem ( +)(z +i) = 4 ( ).4i = i. f(z)z = πi[ i 6 i ] = π. (v) Let f(z) = eiz +z. Then f has poles of orer at z = ±i. Only i is insie. Therefore by Cauchy s resiue theorem (z i) f(z)z = πires(f,i) = πi lim = πi lim z i + z i z +i = π e. (vi) Let f(z) = z ( +). Then f has poles of orer at z = ±i. Only i is insie. Therefore by Cauchy s resiue theorem f(z)z = πires(f,i). res(f,i) z i z [(z i) z ] ( +) z i z ( z (z +i) ) z i (z +i) [iz + ] (z +i)z (z +i) 4 = ( 4).[e e ] 4e 6 = 4e f(z)z = πi 4e = πi e.. In this question we shall integrate roun the same curve given in the previous question. This was efine as follows. Let : [,R] C be efine by (t) = t, let = S(,R) an let =. (i) Let f(z) = ( +). By the previous question ( +) z = π. Now πrsup{ f(z) : z } = πrsup{ ( +) : z }. πr R as R. (R ) R Also f(z)z = f(x) x. Therefore

3 (ii) Since g(x) = x R x x (x +) (x +) x ( f(z)z + f(z)z) f(z)z = π. x (x +)(x +4) x x (x +)(x +4) is an even function, R Let f(z) =. By the previous question ( +)( +4) Now πrsup{ f(z) : z } = πrsup{ πr R as R. (R )(R 4) R Also f(z)z = f(x) x. Therefore x x (x +)(x +4) = R lim x x (x +)(x +4). ( +)( +4) z = π. ( +)( +4) : z }. R x x (x +)(x +4) (x +)(x +4) x ( f(z)z + f(z)z) f(z)z = π. x (x +)(x +4) x = x (x +)(x +4) x = π 6. (iii) Let f(z) = eiz +z. By the previous question +z z = π e. Now πrsup{ f(z) : z } = πrsup{ + : z }. as = e i(x+iy) = e y+ix = e y when z πr as R. R R Also f(z)z = f(x) x. Therefore x

4 e ix R e ix x +x +x x ( f(z)z + f(z)z) f(z)z = π e. Taking the real part of both sies of the last equality we get (iv) Since g(x) = x sin(x) is an even function, (x +) R cos(x) +x x = π e. x sin(x) (x +) x sin(x) (x +) x = lim x sin(x) (x +) x. Let f(z) = z z πi ( +). By the previous question z = ( +) e. By Joran s lemma π sup{ z ( +) : z R } π (R ). as R. R Also f(z)z = f(x) x. Therefore R R x e ix x e ix x (x +) (x +) x ( f(z)z + f(z)z) f(z)z = πi e. Taking the imaginary part of both sies of the last equality we get x sin(x) (x +) x = π x sin(x) an so e (x +) x = π 4e. Let r,r be real numbers such that < r < < R an let : [r,r] C be efine by (t) = t, let = S(,R) let : [, r] C be efine by (t) = t an let 4 = S(,r). Let = 4 so that is a close path. e iπz Let f(z) =. Then f has poles of orer at z = an z = ±i. z( +) Only i is insie an so by Cauchy s resiue theorem f(z)z = πires(f,i). Now f(z)z = (z i)e iπz res(f,i) z i z(+ ) e iπz z i z(z +i) = e π so R r f(x)x an f(z)z = r f(x)x. Thus f(z)z = πie π.

5 f(z)z = = r R = i r R r f(x)x+ e iπx R x(+x ) x+ sin(πx) x(+x ) x r f(x)x R r e iπx x(+x ) x As = e i(x+iy) = e y+ix = e y when z πr as R. R(R ) As f has a simple pole at it follows that res(f,) i Therefore ze iπz z(+ ) lim f(z)z = lim f(z)z = πi res(f,) = πi. r r 4 S(,r) sin(πx) R x x(+x ) limi sin(πx) r r x(+x ) x lim( f(z)z + f(z)z) r lim( f(z)z f(z)z r 4. Let f(z) = πcos(πz) sin(πz) lim r = πi( e π ). sin(πx) x(+x ) x = π ( e π ). f(z)z lim 4 f(z)z) e iπz (+ ) f(z)z lim f(z)z r 4 =. So. Then f has singularities at z = n where n Z (z +)(z +) an at z = an z =. If g(z) = cos(πz) then g (z) = πsin(πz) so the orer of g at is. Therefore f has a removable singularity at. Exactly the same argument as in the lecture notes shows that res(f,n) = f has a pole of orer at an res(f, ) (z + )πcos(πz) z sin(πz) πcos(πz) z sin(πz)(z +) πcos( π = ) = π sin( π).( ) = π. (n+)(n+). (z +)(z +)

6 Let n, = L((n+ )+i(n+ ), (n+ )+i(n+ )), n, = L( (n+ )+i(n+ ), (n+ ) i(n+ )), n, = L( (n+ ) i(n+ ),(n+ ) i(n+ )) an n,4 = L((n+ ) i(n+ ),(n+ )+i(n+ )). Let n = n, n, n, n,4. Then n is the bounary of the square with vertices at ±(n+ )±(n+ )i. By Cauchy s Resiue Theorem [ ] [ ] f(z)z = πi π k=n + res(f, k) = πi π k=n + ( ). n (k +)(k +) We now show that lim n =. From the result in the lecture notes cos(πz) n sin(πz) for z n. lim n f(z)z n sup{ πcos(πz) n sin(πz) (z +)(z +) : z n} 4(n+) π sup{ (z +)(z +) : z n} 4 π(n+) (n )(n ) ( as z n for z n) Therefore lim n =. Thus from equation (*) n [ ] lim π k=n + =. n (k +)(k +) n= n= (n+)(n+) = π. 5. Let f(z) = πcos(πz) sin(πz) z. Then f has poles at z = n for all n Z. If n, f has a pole of orer at z = n an res(f,n) = n. Since Or( sin(πz),) = Or(,)+ Or(sin(πz),) = + =, it follows that f has a pole of orer at. Also res(f,) =! lim [πz cos(πz) sin(πz) ] =! lim πzcos(πz) sin(πz) = π! lim z (g h )(z) where g(z) = cos(πz) an h(z) = sin(πz). z Now z (g h ) = g h gh an h (g h ) = h [g h gh ] hh [g h gh ]. h 4 Now g (z) = πsin(πz), g (z) = π cos(πz) an so g () = an g () = π. Since h(z) = (πz) [πz +...] = π π +..., z! 6 h() = π, h () = an h () = π. res(f,) = π π 5 + π5. π 4 = π.

7 Let n = n, n, n, n,4 be the same path as in the last question so that n is the bounary of the square with vertices at ±(n+ )±(n+ )i. By Cauchy s Resiue Theorem f(z)z = πi[ n π k= + k + k=n k= k] = πi[ π k=n + We now show that lim n =. From the result in the lecture notes cos(πz) n sin(πz) for z n. lim n f(z)z n sup{ πcos(πz) n sin(πz) : z n} 4(n+) π sup{ : z n} 4 π(n+) n ( as z n for z n) Therefore lim n =.Thus by (*) n k=n lim n k= k = π ; that is, n= k=n lim n πi[ π + k] =. n = π Recall that sin(x+iy) = sinh (y)+sin (x). When z = ±(n+ )π, sin (z) sin (n+ )π =. When z = ±(n+ )πi, sin (z) sinh (n+ )π sinh π (π ). sin(πz) for all z n. π Let f(z) =. Then f has poles of orer at z = n for n = ±,±,... sin(πz) Since Or( sin(πz),) = Or(,)+ Or(,) =, f has a pole of orer at. Also if z = n, where n = ±,±,..., then Also res(f,n) z n π(z n) sin(πz) z n res(f,) =! lim k= z lim π(z n) z n sin(πz) = n lim z n [ πz sin(πz) ] =! lim k= k ] ( ) π cos(πz) = n cos(nπ) = ( )n n. [ πz sin(πz) ] =! lim [ h (z)] where h(z) = sin(πz). πz Now z ( h ) = h an h ( h ) = h h +hh. h 4 Since h(z) = sin(πz) = [ (πz ) +...], h() =, h () = an h () = π πz 6. res(f,) = π = π 6. Thus, by Cauchy s Resiue Theorem,

8 f(z)z = πi[ n π + sum of resiues of f at ±,...±n] 6 = πi[ π k= 6 + ( ) k k + k=n ( ) k k ] k= = πi[ π k=n 6 + ( ) k k ] ( ) k= But f(z)z n sup{ f(z) : z } n π = 4(n+)sup{ z sin(πz) : z } 4(n+) π n (as z n an sin(z) for z n). f(z)z as n. n Letting n in (*) we obtain lim[ π k=n n 6 + ( ) k k] =. k=n lim ( ) k+ n k = π an so ( ) n+ n = π. k= n= k=