On the Complex Differentiability of the Hyperbolic SecantDate: 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) and 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 So 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 http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/