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Would the complex function sech(z) be holomorphic because cosh(z) defined by its power series is holomorphic? I am not sure why sech (z) is even (complex) differentiable everywhere since surely it is not a "flat map" and the "gradient" along a contour line ($=0$) and that along a perpendicular of a contour line ($\neq 0$) are different! Please explain!

Also is there a effective way of telling whether a complex function is differentiable and/or holomorphic?

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1 Answer 1

up vote 1 down vote accepted

It cannot be holomorphic because $\cosh i\pi/2=0$ therefore ${\rm sech} ~ i\pi/2 =\infty$. It is meromorphic instead because the reciprocal of a nonzero holomorphic function ($\cosh$ in this case) is always meromorphic.

If a function is given by its power series, you can often calculate its radius of convergence: the radius of the greatest $0$-centered open disk where it converges, which is in this case the same as being holomorphic there. See http://en.wikipedia.org/wiki/Radius_of_convergence

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Thanks, Tib! Is holomorphic the same as being analytic/ –  James C Jan 30 '12 at 12:22
Thanks again, Tib –  James C Jan 30 '12 at 14:50

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