# how do I find all the roots of tanh(z)=z/b?

Discussion in 'Maple' started by Luna Moon, Jan 4, 2008.

1. ### Luna MoonGuest

Hi all,

How do I find out all the roots of tanh(z)=z/b in the complex domain,
where b is a real number and z is complex valued?

Thanks a lot!

Luna Moon, Jan 4, 2008

2. ### Dave L. RenfroGuest

Methods for actually finding such solutions are
a little outside my interests (but I have, at
times, been interested in the algebraic transcendence
of the solutions to equations such as this for "most"
algebraic values of b), but perhaps something in
these searches will help:

Dave L. Renfro

Dave L. Renfro, Jan 4, 2008

3. ### Stephen Montgomery-SmithGuest

Solve the equation (*):
X''(x) = z^2 X(x)
X(0) = 0
X'(1) = b X(1)

All solution must satisfy
X(x) = c_1 cosh(z x) + c_2 sinh(z x). Then
c_1=0
and c_2!=0 if and only if z cosh(z) = b sinh(z).

Hence the solutions to tanh(z)=z/b are exactly those z for which (*) has
a non-zero solution.

Next, if X(x) is a solution, then let bX(x) be its complex conjugate. Then

z^2 int_0^1 |X(x)|^2 dx =
int_0^1 X''(x) bX(x) dx =
int_0^1 |X'(x)|^2 dx

where the last equality follows by integration by parts (being careful
to show that the cross terms really are zero).

Hence z^2 is a real number.

Hence z is either real (graphically has 1 or three solutions) or pure
imaginary (graphically has infinitely many solutions, since z=iy, and
tan(y)=y/b).

I have to admit that I only came to this solution because I am familiar
with Robin boundary conditions. Surely there is a more direct way to
prove this.

Stephen Montgomery-Smith, Jan 4, 2008
4. ### Robert IsraelGuest

Considering that there are infinitely many, what exactly do you
mean by "find"?

(I'm removing the removable singularity at 0, writing tanh(z)/z = 1 at z=0)

You might note that tanh(z)/z is real if and only if z either real or imaginary.
This is because if z = x + i y with x and y real,
Im(tanh(z)/z) = (sin(2y) x - sinh(2x) y)/(2 (x^2+y^2)(sinh(x)^2 + cos(y)^2))
and sin(2y)/y < 2 < sinh(2x)/x for x, y <> 0.
For real z, tanh(z)/z is an even function, decreasing on [0,infty) and
always in the interval (0,1]. So if b > 1, there are two real solutions
to tanh(z) = z/b (one the negative of the other).
For imaginary z = it, tanh(z)/z = tan(t)/t. If 0 < b < 1, there is one
solution to tan(t)/t = 1/b for 0 < t < pi/2 and its negative in -pi/2 < t < 0.
Then for any real nonzero b, there is one solution to tan(t)/t = 1/b in each
of the intervals (n-1/2) pi < t < (n+1/2) pi for nonzero integer n.

Robert Israel, Jan 4, 2008
5. ### David W. CantrellGuest

There are those two, together with the trivial third solution: z = 0.
Luna Moon: I suspect that Stephen Montgomery-Smith and Robert have already
look at my response

in the sci.math thread "roots of trancendental equation" (2007 Aug. 29).
There, I considered the equation cot(x) = c*x for c > 0. I could probably
write up something similar for your problem; if you would like that, let me
know.

David

David W. Cantrell, Jan 7, 2008