Exercise
1. Find
for
the following values of z.
1 (g).
.
Solution 1 (g).
See text and/or instructor's solution manual.
One solution is obtained by simplifying the
quotient
as follows:
![]()
then
We are done.
Another solution is obtained by using the
formula
, where
n is an integer.
Since
and
and
and
, an
argument of
is easily found to be
If we choose
, then
lies
in the interval
,
so the principal value of the argument is found to be
![]()
We are done.
Aside. We can let Mathematica double check our work.
![[Graphics:../Images/ComplexGeometryContinuedModHome_gr_119.gif]](../Images/ComplexGeometryContinuedModHome_gr_119.gif)
This solution is complements of the authors.
(c) 2008 John H. Mathews, Russell W. Howell