We know the identity $\tanh(ix)=i \tanh(x)$.
My question: is it true that $\tanh^{-1}(ix) = i \tanh^{-1}(x)$ ?
If not then is there a similar identity for arctangents? I think there might not be but would like to know how to find one.
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Mhenni Benghorbal
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Milan Amrut Joshi
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1Note that the "known" identity is wrong. We have $\tanh (ix) = i \tan x$, not $i\tanh x$. – Daniel Fischer Oct 02 '13 at 13:56
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please check it – Milan Amrut Joshi Oct 02 '13 at 14:03
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$$\tanh (ix) = \frac{\sinh (ix)}{\cosh (ix)} = \frac{\frac12(e^{ix} - e^{-ix})}{\frac12(e^{ix}+e^{-ix})} = \frac{i\sin x}{\cos x} = i \tan x$$ – Daniel Fischer Oct 02 '13 at 14:23
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Here is a start. Assume $y=\tanh^{-1}(ix)$ and then apply $\tanh$ to both sides and do some manipulations and see what you get.
Note: As in the comment
$$ \tanh(ix)=i\tan(x). $$
Mhenni Benghorbal
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1Also note that 1/i = -i, and that will get you a similar but not quite the same relationship that you are looking for. – DanielV Oct 02 '13 at 13:25
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Well, to follow this line of reasoning, you actually need that tan(ix) = i tanh(x) , rather than the above, but you could get that by setting z = ix and knowing that tangent is an odd function. – DanielV Oct 03 '13 at 06:19