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On the Complex Differentiability of the Hyperbolic Secant

Date: 09/24/2010 at 06:14:41
From: Matthew
Subject: is sech(z) complex differentiable

Is sech(z) complex differentiable?

To show complex differentiability, I typically use the Cauchy-Riemann
equations. However, they require you to break the function into real and
imaginary parts, which I can't seem to do with the hyperbolic secant of z.

Date: 09/24/2010 at 15:59:32
From: Doctor Jordan
Subject: Re: is sech(z) complex differentiable

Hi Matthew,

Indeed, I don't see how to break sech(z) into real and imaginary parts.
Maybe there is a clever way to do it -- but in any case, we can solve the
problem differently.

Start with the hyperbolic trigonometric identity

   sech(z) = 1/cosh(z)

In terms of the exponential function, the formula for cosh(z) is complex
differentiable on the entire complex plane. So cosh(z) is complex
differentiable, or holomorphic, on the entire complex plane.

It is true in general that if f(z) is a holomorphic function, then 1/f(z)
is holomorphic wherever f(z) is not equal to 0. You can prove this using
the Cauchy-Riemann equations you're already familiar with, letting

   f(z) = u(x,y) + iv(x,y)


   1/f(z) = u/(u^2 + v^2) - iv/(u^2 + v^2)
Therefore sech(z) is holomorphic wherever cosh(z) is not equal to 0. 

Now, by definition,

   cosh(z) = (e^z + e^(-z))/2

   cosh(z) = 0     if and only if    e^z = -e^(-z)
This is true if and only if

   e^(2z) = -1

Write z = x + iy. Now,

   e^(2z) = -1     if and only if    e^(2x + 2iy) = -1
This holds if and only if

   e^(2x)(cos(2y) + isin(2y)) = -1

Comparing the real and imaginary parts of this equation, we have

   e^(2x)cos(y) = -1 
   e^(2x)sin(y) = 0.

The second of these equations is true if and only if y = pi*k for some
integer k. For y = pi*k,

   cos(y) = (-1)^k, 
so the first equation becomes

   e^(2x)*(-1)^k = -1,
which holds if and only if 

   e^(2x) = (-1)^(k + 1).
This, in turn, holds if and only if x = 0 and k is odd, since the
exponential of a real number cannot be -1.

Therefore, for x = 0 and y = k*pi for k odd,

   cosh(z) = 0    if and only if    z = x + iy
Therefore sech(z) is holomorphic for all z *not* of that form, i.e.,
sech(z) is holomorphic for all z not of the form 0 + i*k*pi for k odd.

The two ideas we used to determine where sech(z) is complex differentiable
were 1) for f(z) complex differentiable, 1/f(z) is complex differentiable
wherever f(z) is not equal to 0; and 2) to find the z that solve the
equation e^(2z) = -1, write z in the form x + iy and compare the real and
imaginary parts of each side of the equation.

- Doctor Jordan, The Math Forum 
Associated Topics:
College Calculus
College Imaginary/Complex Numbers
College Trigonometry

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