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 + ...