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

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

  1. Luna Moon

    Luna Moon Guest

    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
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  2. 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
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  3. 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
    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

    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. 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. There are those two, together with the trivial third solution: z = 0.
    Luna Moon: I suspect that Stephen Montgomery-Smith and Robert have already
    answered your question adequately for your needs. But if not, you might
    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

    David W. Cantrell, Jan 7, 2008
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