Separate elliptic integrals into real and imaginary parts

This is somewhat of a follow-up question to ‘Elliptic integrals with parameter outside 0<m<1’.

I have an equation that I'm attempting to simplify that has terms that look something like this:

$\Re\bigl( -i (c-1) \mathrm K(c) -2i \mathrm E(c) \bigr)$

where $c>1$. I.e., they're being evaluated outside their standard parameter range of $0\le m\le 1$. I know the equation has been derived with Mathematica, and I'm suspicious that the $\Re(i\mathrm K(c))$ terms, etc., are more complex than they need to be.

But I've taken a look at Legendre's Relation which seems to be related to splitting up these integrals into real and imaginary parts, and it doesn't seem to click—I suspect that the expression above is about as simple as I can get it (i.e., even if I manage to replace the integrals by alternate versions that have only real components, the resulting expression will be more complex than that above). Is there anything I'm missing?

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What's the admissible range of $c$? $E$ (the complete version) and $K$ for real parameters only become complex for parameters greater than 1. The elliptic integral of the third kind is... quite complicated; that might require separate analysis. You might want to give example values for all those variables your expression uses. – J. M. Nov 4 '10 at 3:49
@J.M. Oh, of course that's required information here — sorry for the omission. I've already pruned out the terms that have zero value in the parameter ranges of the equation. (Fixed in the question.) – Will Robertson Nov 4 '10 at 4:24
Well at least the $-b K(1-c)$ term is real (though some transformation might be needed if this is intended for routines that are restricted only to unit interval parameters). Is the imaginary part of your expression actually nonzero? – J. M. Nov 4 '10 at 4:39
@J.M. Yes, there are non-zero imaginary components; consider K(2)=2.5+1.53i (approx). But you're right; those last two terms are always real (I think), so it's just the Re(iK()) and Re(iE()) terms to worry about. – Will Robertson Nov 4 '10 at 4:59
Nah... as I said, that last term you had is a complicated beastie (the characteristic $n$ in $\Pi(n|m)$ complicates a lot of things). But I'll have a look at those other terms... – J. M. Nov 4 '10 at 5:04

Well, for the simplified case that you now presented in your current version of the question (again, as I said, the complete elliptic integral of the third kind requires a different sort of analysis), the identities in formula 19.7.3 in the DLMF can be rewritten in parameter form as

$$K(m)=\frac1{\sqrt{m}}\left(K\left(\frac1{m}\right)-i K\left(1-\frac1{m}\right)\right)$$

and

$$E(m)=\sqrt{m}\left(E\left(\frac1{m}\right)-\left(1-\frac1{m}\right)K\left(\frac1{m}\right)+i\left(E\left(1-\frac1{m}\right)-\frac1{m} K\left(1-\frac1{m}\right)\right)\right)$$

If we apply both to your expression above, it can be simplified to

$$2\sqrt{c}E\left(1-\frac1{c}\right)-\frac{1+c}{\sqrt{c}}K\left(1-\frac1{c}\right)$$

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Thanks! I can now see how this comes together. So by assuming m>0, we know that K(1/m) will be real so will K(1-1/m). I guess what I was missing before was a good understanding of the correspondence between the input ranges and the real/imaginary components of the output, and hence where/when those identities could be applied to my situation. (It's always obvious in hindsight :)) – Will Robertson Nov 4 '10 at 13:25
@Will, Teeny nitpick: you probably intended $m > 1$; there's no reason to reciprocate when your parameter is already within the unit interval. :) – J. M. Nov 4 '10 at 15:29
Yes, that is what I meant :) – Will Robertson Nov 4 '10 at 23:25