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I'm looking for a continuous function, polynomial or transcendental, whatever, that takes real values when its argument is real, but has at least one pair of complex valued roots with non-zero imaginary part. The Hardy Z function would fit this description, but it isn't known to have any complex roots. Does anyone have any examples?

Thanks

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    Would a quadratic with negative discriminant satisfy your needs? – 2017-02-26
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    Like $f(z) = z^2+1$ – 2017-02-26
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    that also works.. yes, thanks! – 2017-02-26

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Consider $f = e^{x}- e^{-x}$ Observe that

$$f(i \pi ) = e^{i \pi} - e^{- i \pi} = -1 - (-1) = 0 $$ So there is your complex root.

Also observe that $f$ is real on all real arguments.

(curious fact i realized after that answer)

$$ f = 2 \sinh(x) $$

I realized you said "pair" of roots.

So in that case observe that $k i \pi \ \forall \ k \in \mathbb{N}$ is a root here (i.e. $...-2i\pi,-i \pi, 0, i \pi, 2 i \pi ... $)

Thus we have infinitely many such pairs of roots.