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4 Pages

### soln5

Course: MATH 116, Fall 2009
School: Stanford
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Word Count: 1169

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Analysis Complex HW 5: Solutions 9. Suppose not. Then there exists a positive integer N such that for all {zn } converging to P we have |(zn P )N f (zn )| &lt; N. By the Riemann Removable Singularities theorem, limn (zn P )N f (z) exists and is nite. Dene (z P )N f (z) if z = P g(z) = . limn (zn P )N f (z) if z = P The map g is analytic in D(P, r). This implies that f has a pole of order N at P , a...

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Analysis Complex HW 5: Solutions 9. Suppose not. Then there exists a positive integer N such that for all {zn } converging to P we have |(zn P )N f (zn )| < N. By the Riemann Removable Singularities theorem, limn (zn P )N f (z) exists and is nite. Dene (z P )N f (z) if z = P g(z) = . limn (zn P )N f (z) if z = P The map g is analytic in D(P, r). This implies that f has a pole of order N at P , a contradiction. Let P1 = (x1 , y1 , z1 ) be a point in S 2 . Then the line through P1 and 1 (0, 0, 1) has equation xx1 = yy1 = zz1 . The image of P1 under the stereographic x1 y1 1z projection is then 31. x1 y1 , ) . 1 z1 1 z1 Now consider the antipodal point, P2 = (x1 , y1 , z1 ) S 2 . The line through P2 and (0, 0, 1) has equation y + y1 z + z1 x + x1 = = , x1 y1 1 + z1 (x1 , y1 , z1 ) = ( (x1 , y1 , z1 ) = ( y1 x1 , ) w. 1 + z1 1 + z1 2 2 2 which implies that x1 y1 y1 x1 y x1 y x1 Then w = ( 1+z1 , 1+z1 ), and so w = ( 1z2 1z2 , 1z1 1z1 ) = ( z1 1 , 0) = 2 2 2 1z1 1 1 1 1 (1, 0) = 1. 34. f) We have the formula n f= j=1 Rj 1 , z Pj 1 where the Pj are the poles (here 0, i, 1) of f enclosed by , and Rj = (k1)! ( z )k1 ((z 1 1 Pj )k f (z))|z=Pj is the residue of f at Pj . Now, z+i dz = 0, z dz = 2i, and 1 dz = 0. Also, R0 = z ( z2cos z )|z=0 = 2i + 1. Thus (z+i) z+1 f = (2i + 1) 2i. 2 cos 2i+3i sin 2i . 16 g) We use the same formulae as in part (f). In this case R0 = 0, R2i = 1 Also, z+2i dz = 4i. Thus we obtain f= 2 cos 2i + 3i sin 2i (4i). 16 Let g(z) = + an z n , R < |z| < . Let C0 be a curve going counn= terclockwise along a circle of radius R0 > R. Note that for a point lying outside 41. 1 2 1 of C0 this is a clockwise orientation. Thus an = 2i 2ia1 = C0 g(z)dz. On the other hand, g(z) dz. C0 z n+1 In particular, 1 an2 1 1 an g( ) = = (0 < |z| < ), 2 n+2 n z z z z R n= n= 1 which gives Res [g(z)] = Resz=0 [ z12 g( z )] = a1 , which is the desired result. To see that the denition of residue of g at remains unchanged if the origin is replaced by some other point in the nite plane, apply the transformation z zP. 1 1 1 Then Resz=P [ (zP )2 g( zP )] = Resz=0 [ z12 g( z )]. P (z) 42. Let f (z) = Q(z) . Let Q(z) = (z P1 )1 . . . (z Pm )m be a factorization of Q. Let be the circle |z| = R such that P1 , . . . , Pm are in the interior of |z| R. By the residue formula m f= j=1 Rj 2i. Also, f = 2ia1 , where a1 is the coecient of z 1 in the Laurent expansion of f. By the previous problem, a1 is the residue of f at . Thus we nally get m Rj + R = a1 a1 = 0. j=1 Precisely the same proof will go through for any meromorphic function f with nitely many poles (and whose poles are hence contained in a compact disk). 47. Let be the path C1 going from R to R along the real axis, followed by C2 , the semi-circle of radius R above the y-axis in the counterclockwise direction. Then eiz dz 1 + z4 = C1 R eiz + 1 + z4 ix C2 eiz 1 + z4 eiz dz. 1 + z4 = R e dx + 1 + x4 C2 The second summand goes to zero as R : | C2 eiz dz| 1 + z4 1 1 dz dz |1 + z 4 | |z 4 | 1 C2 C2 0 R as R . R4 1 eie 4e i 4 By the Cauchy integral formula: 1 dz 1 + z4 = 2i 3i 4 + R eie 4e 3i 4 9i 4 = limR R cos x + i sin x dx 1 + x4 = cos x dx, 1 + x4 3 where the last equality follows from the fact that sin x is an odd function. 1 Let be the path going rst from R to R along the real axis, C1 , then along ui the upper semi-circle of radius R counterclockwise up to the point P = Re 3 , C2 , and then from P along a straight line connecting P with 0 to where it intersects the 1 circle of radius R around the origin, C3 , and from this point along the semi-ricle of 1 1 radius R in clockwise direction until the point R R,C4 . Then 48. z4 dz = 1 + z3 1 4 Ci i=1 z4 dz. 1 + z3 1 By Cauchy's integral formula: z4 dz = 2i 1 + z3 1 e 12 3e i 2i 3 . Now, (1) z4 dz = 1 + z3 1 R 1 R x4 dx + 1 + x3 1 + C2 C4 z4 dz + 1 + z3 1 1 1 R R x 4 e 12 2i e 3 dt. 1 + x3 1 2i The second summand converges to zero as we let R go to innity: | C2 z4 dz| 1 + z3 1 C2 | z4 |dz 1 + z3 1 4 1 C2 |z| 4 dz |z|3 1 = R R 0 as R , R3 1 while the length of C4 gets short and the integrand goes to 0, so that the intgral over C4 also approaches 0. Hence the limit of (1) as R is 0 5i x4 dx + e 6 3 1+x 1 0 5i x4 dx = (1 + e 6 ) 3 1+x 1 0 x4 dx, 1 + x3 1 so that nally 0 49. x4 dx = 2i 1 + x3 1 e 12 3e i 2i 3 (1 e 5i 6 )1 . Let be the curve in gure 4.7, page 136. Then log z dz z3 + z + 1 log z log z dz + dz z3 + z + 1 z3 + z + 1 1 2 log z log z + dz + dz z3 + z + 1 z3 + z + 1 3 4 0 as R . = By the calculation on page 136, We also have 2 log(x + i/ 2R) log(x i/ 2R) 2i. log z dz 0 as R . z3 + z + 1 log z dz 0 as R , z3 + z + 1 4 4 Thus, 1 z3 log z dz + +z+1 3 z3 log z dz 2i +z+1 0 1 dx. x3 + x + 1 Let P1 , P2 , P3 be the roots of f (z) = z 3 + ...

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