| ia
If you represent x as re , where r and a are real numbers, you get
x
x = -1 --> x ln x = (2N+1)*pi*i
--> (r cos a + i r sin a) (ln r + a i) = (2N+1)*pi*i
note that, if we don't restrict a to lie in (-pi,pi],
we don't have to worry about the multiple values of ln x
(this is opinion, not fact - I passed Complex Variables
in college by the skin of my teeth, because I didn't
find the subject very interesting - oh, well.....)
Setting real and imaginary parts of the two sides equal, we get:
a) r(ln r)(cos a) - ra(sin a)= 0
--> ln r = a(tan a) (or r = 0 which we know is not
a solution)
b) r(ln r)(sin a) + ra(cos a) = (2N+1)pi
substituting (a) into (b) we get:
ra[(sin a)(tan a) + (cos a)] = (2N+1)pi
ra/(cos a) = (2N+1)pi
a(tan a)
ae
------- = (2N+1)pi
cos a
Solving the members of this family of equations for a gives the possible
"radials" on which solutions may lie. It is pretty easy to substitute various
values for a to see what "radials" have solutions: e.g. a = pi is a solution
of this equation when N=0 and yields the "degenerate" solution of x = -1
Radials that don't have any solutions include: a = 0 [positive reals],
a = pi/2 [LHS is not defined, but equation (a) yields r=0 which cannot be
a solution of equation (b)], a = pi/4, etc. Note that any rational multiple of
pi (except pi itself) is very unlikely to be a root of any equation in the
family because the equation will simplify to:
exp(something)/cos(something else) = rational.
Anyone have a rootsolving program they would like to try out?
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| If we look at the equation in .4,
a = (2n+1)pi*cos(a)/exp(a*tan(a))
The right hand side is an even funtion with periodic singularities
at odd multiples of pi/2 (essential singularities, in fact), with
zeroes at even multiples of pi.
This implies that it will have solutions at approximately
odd multiples of pi/2 since the right hand side of the equation
varies monotonically in each interval from k*pi/2 to (k+1)*pi/2.
I don't know if there is a nice closed form expression for the
roots though.
- Jim
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