So for every complex number z, f(z)≠0 and all derivatives fⁿ(z)≠=0.
This is a followup to another question:
http://answers.yahoo.com/question/index?qid=20100824143346AAZYTHU
Hi KB. The condition that all derivatives are also non-zero restricts things somewhat. The second derivative of e^(e^z) is zero at z=πi.
kb
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Update: Here's a proof in progress (I hope)...
Since C is open (in itself), every zero-free analytic function f(z) has an analytic logarithm.
Thus, f(z) = e^(g(z)) for some analytic g(z) on C.
Note that f' = g' e^g.
Since we want f'(z) ≠ 0 for all z in C, this forces
g'(z) ≠ 0 for all z in C.
Thus, g'(z) = e^(h(z)) for some analytic h(z) on C.
Next, we look at the second derivative.
f'' = (g'' + (g')^2) e^g
= (h' e^h + (e^h)^2) e^g
= (h' + e^h) e^(g+h).
Since we also want f''(z) ≠ 0 for all z in C, we require that
h' + e^h ≠ 0.
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Major case:
This is certainly satisfied if h(z) is a nonzero constant
Then, g'(z) = k for some nonzero k in C.
==> g(z) = kz + r for some r in C.
==> f(z) = e^(kz + r) = Ce^(kz) for some nonzero constants C and k.
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Note that f^(n)(z) = Ck^n e^(kz) is never 0 for all z in C.
Hence, we've found one such family of functions.
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Still to be addressed:
Are there other functions which satisfy h' + e^h ≠ 0 for all z in C which will fit the criteria for this posted question?
(This is certainly false if h(z) = az + b with a ≠ 0.)