Download Complex Analysis 9, Exercises Solution - Mathematics and more Exercises Complex Numbers Theory in PDF only on Docsity!
Solution of Homework Assigned on February 26, 2009
due March 3, 2009
(numbering of problems continued from
the last assignment with the same due date)
Problem 6. Verify that
|z|=
z
15
(z^2 + 1)
2 (z^4 + 2)
3
dz = 2πi
by using the change of variables z =
1 w
Solution of Problem 6. Use the change of variables z =
1 w
to transform
the integral to
|w|= 1 4
1 w^15 ( 1 w^2
1 w^4
dw
w
2
The negative sign in front of the integral comes from the change of orientation
when the parametrization z = 4e
iθ for 0 ≤ θ ≤ 2 π is transformed to the
parametrization w =
1 4
e
−iθ for 0 ≤ θ ≤ 2 π. This new integral can be
rewritten as (^) ∮
|w|= 1 4
dw
w (1 + w^2 )
2 (1 + 2w^4 )
which is equal to 2πi, because the only pole of the meromorphic function
w (1 + w^2 )
2 (1 + 2w^4 )
3
inside the circle |w| <
1 4
is the simple pole at the origin whose residue is equal
to the value of
1
(1 + w
2 )
2 (1 + 2w
4 )
3
at w = 0, namely 1.
Problem 7. Evaluate the integral
∞
0
x
1 3
1 + x
2
dx
by applying the method of residues to a branch of the function
f (z) =
z
1 3
1 + z
2
defined on C − [0, ∞).
Solution of Problem 7. Define a branch of the holomorphic function
f (z) =
x
1 3
1 + z^2
on C minus the nonnegative real axis by specifying
z
1 (^3) = e
1 3 log^ z^ with log z = log |z| + i arg z and 0 < arg < 2 π.
We are going use as contour of integration the boundary of the set in C which
is equal to the disk of radius R center at the origin minus the union of the
closed disk of radius r centered at the origin and the strip { |Im z| ≤ r, Re z ≥ 0 }
and the we will take limit as R → ∞ and r → 0.
The points i and −i are simple poles for f (z) and the residues resi f and
res−if are computed as follows.
resi f = (z − i)
z
1 3
1 + z^2
z=i
e
1 3 log^ z
z + i
z=i
e
πi 6
2 i
res−i f = (z + i)
z
1 3
1 + z
2
z=−i
e
1 3 log^ z
z − i
z=−i
e
πi 2
− 2 i
Thus
1 − e
2 πi 3
) ∫^ ∞
x=
x
1 (^3) dx
1 + x^2
= 2πi (resi f + res−i f ) = 2πi
e
πi 6
2 i
e
πi 2
− 2 i
and
∫ (^) ∞
x=
x
1 (^3) dx
1 + x
2
= π
e
πi (^6) − e
πi 2
1 − e
2 πi 3
= π
e
−πi 3
e
πi (^6) − e
πi 2
e
−πi 3
1 − e
2 πi 3
) (^) = π
e
−πi (^6) − e
πi 6
e
−πi (^3) − e
πi 3
on C minus the nonnegative real axis by specifying
z
1 −α = e
(1−α) log z with log z = log |z| + i arg z and 0 < arg < 2 π.
We are going use as contour of integration the boundary of the set in C which
is equal to the disk of radius R center at the origin minus the union of the
closed disk of radius r centered at the origin and the strip { |Im z| ≤ r, Re z ≥ 0 }
and the we will take limit as R → ∞ and r → 0.
The points i and −i are simple poles for f (z) and the residues resi f and
res−if are computed as follows.
resi f = (z − i)
z
1 −α
1 + z^2
z=i
e
(1−α) log z
z + i
z=i
e
(1−α)πi 2
2 i
res−i f = (z + i)
z
1 −α
1 + z^2
z=−i
e
(1−α) log z
z − i
z=−i
e
3(1−α)πi 2
− 2 i
Thus
1 − e
2(1−α)πi
x=
x
1 −α dx
1 + x^2
= 2πi (resi f + res−i f ) = 2πi
e
(1−α)πi 2
2 i
e
3(1−α)πi 2
− 2 i
and
x=
x
1 −α dx
1 + x^2
= π
e
(1−α)πi (^2) − e
3(1−α)πi 2
1 − e2(1−α)πi
= π
e
−(1−α)πi
e
(1−α)πi (^2) − e
3(1−α)πi 2
e−(1−α)πi^ (1 − e2(1−α)πi)
= π
e
−
(1−α)πi (^2) − e
(1−α)πi 2
e−(1−α)πi^ − e(1−α)πi
= π
sin
(1−α)π 2
sin ((1 − α)π)
π
2 cos
(1−α)π 2
π
2 sin
απ 2
The integral of f (z) dz over the curve CR which is the portion of the upper
half-circle of radius R centered at 0 minus outside of the strip { |Im z| ≤ r, Re z ≥ 0 }
approaches 0 as R → ∞ and r → 0, because
CR
f (z) dz
R
1 −α
R^2 − 1
πR → 0 as R → ∞.
The integral of f (z) dz over the left half-circle Γr of radius r centered at 0
approaches 0 as r → 0, because
Γr
f (z) dz
r
1 −α
1 − r^2
πr → 0 as r → 0.
The final answer is
∫ (^) ∞
0
log (1 + x
2 ) dx
x
1+α
π
α sin
απ 2
