Exercise 9. Describe a Riemann surface for the domain of definition of the multivalued function
9
(a).
.
Solution 9 (a).
See text and/or instructor's solution manual.
Answer. For
we
have
,
as the three branches of the cube root with
domains
.
As in the text, form three domains
which are copies of the complex plane slit along the negative real
axis, and stack
directly
above each other.
![[Graphics:../Images/ComplexFunBranchModHome_gr_376.gif]](../Images/ComplexFunBranchModHome_gr_376.gif)
The
image of
.
![[Graphics:../Images/ComplexFunBranchModHome_gr_381.gif]](../Images/ComplexFunBranchModHome_gr_381.gif)
The
image of
.
![[Graphics:../Images/ComplexFunBranchModHome_gr_385.gif]](../Images/ComplexFunBranchModHome_gr_385.gif)
![[Graphics:../Images/ComplexFunBranchModHome_gr_386.gif]](../Images/ComplexFunBranchModHome_gr_386.gif)
The
image of
.
Join
the edge of
in the upper half plane to the edge of
in the lower half plane.
Join
the edge of
in the upper half plane to the edge of
in the lower half plane.
It is possible to make above construction, it would look something like a spiral parking ramp.
![[Graphics:../Images/ComplexFunBranchModHome_gr_391.gif]](../Images/ComplexFunBranchModHome_gr_391.gif)
The
three sheets
of
the Riemann surface domain for the cube root
function
.
Finally, join the edge of
in the upper half plane to the edge of
in the lower half plane.
![[Graphics:../Images/ComplexFunBranchModHome_gr_395.gif]](../Images/ComplexFunBranchModHome_gr_395.gif)
The
Riemann surface domain for the cube root
function
.
Background for the
Solution. In Exercise
5, Exercise 6,
and Exercise 7 we
investigated three branches of the cube root
function
.
For each branch we used the coordinate notation
in the same domain set
.
Thus we obtained three branches of
:
![]()
with
, for
.
The images of D under
is
the sector
, for
.
Important note. In
Exercise 5,
Exercise 6, and
Exercise 7 we used
the same domain set
and three different formulas
for the functions.
In the construction of the Riemann surface for
we will use three different domain sets
and one formula
for
the function.
As in the text, each domain
is
a copy of the z-plane slit along the
negative real axis.
Solution for the Riemann
Surface.
To construct the Riemann surface
for
we
need to consider extending Equation
(2-30) for the case of the cube
root.
For each branch
we
must use it's own special coordinate notation
for
in
the z-plane, i.e. use the special special domain set
with
.
The image of the domain set
is
the sector
, for
.
Notice that in the
w-plane the angles for the three sectors
are
,
, and ![]()
and taken all together these angles cover the
interval
.
So when we glue together the image sectors
they
cover the entire complex w-plane.
The domain set will be a
Riemann surface, and it is formed as follows:
Join the edge of
in the upper half plane to the edge of
in the lower half plane.
Join the edge of
in the upper half plane to the edge of
in the lower half plane.
It is possible to make above
construction, it would look something like a spiral parking ramp.
![[Graphics:../Images/ComplexFunBranchModHome_gr_376.gif]](../Images/ComplexFunBranchModHome_gr_376.gif)
The
image of
under
is
the sector
.
![[Graphics:../Images/ComplexFunBranchModHome_gr_381.gif]](../Images/ComplexFunBranchModHome_gr_381.gif)
The
image of
under
is
the sector
.
![[Graphics:../Images/ComplexFunBranchModHome_gr_385.gif]](../Images/ComplexFunBranchModHome_gr_385.gif)
![[Graphics:../Images/ComplexFunBranchModHome_gr_386.gif]](../Images/ComplexFunBranchModHome_gr_386.gif)
The
image of
under
is
the sector
.
![[Graphics:../Images/ComplexFunBranchModHome_gr_391.gif]](../Images/ComplexFunBranchModHome_gr_391.gif)
The
three sheets
of
the Riemann surface domain for the cube root
function
.
Finally, join the edge of
in the upper half plane to the edge of
in the lower half plane.
This final connection makes the a construction in the real world of
3D space difficult.
![[Graphics:../Images/ComplexFunBranchModHome_gr_395.gif]](../Images/ComplexFunBranchModHome_gr_395.gif)
The
Riemann surface domain for the cube root
function
.
This solution is complements of the authors.
(c) 2008 John H. Mathews, Russell W. Howell