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I have written a monograph called Elliptic Integrals and Functions, which contains a sizeable collection of formulas and results relating to the named mathematical topics. At the time of this question, however, the monograph was missing a result on the following problem.

Assume $0<m<1$. For an arbitrary complex number $\gamma$, decompose $F(\gamma,m)$ into real and imaginary parts.

(The argument convention, as with all my other questions and answers on the MSE, follows Mathematica/mpmath convention; $m$ is the parameter.)

Because of the addition formulas – section 2.5 in the monograph – as soon as I can write $$F(\gamma,m)=F(\alpha,m)+F(i\beta,m)$$ where $\alpha$ and $\beta$ are real, I solve the above problem as well as the corresponding problems for the other two kinds of elliptic integrals because the extra parts in the addition formulas can themselves be decomposed (though writing out the full formula would be tedious).

Without loss of generality we can take $|\operatorname{Re}(\gamma)|<\frac\pi2$ (if it is exactly $\frac\pi2$, the formula below doesn't work and my answer here applies). Abramowitz and Stegun 17.4.11 gives the following solution: suppose $\gamma=a+bi$, then letting $$B=(1-m)-\cot^2a-m\sinh^2b\csc^2a\qquad C=(m-1)\cot^2a$$ we have $$\alpha=\tan^{-1}\sqrt{\frac2{-B+\sqrt{B^2-4C}}}\qquad\beta=\sinh^{-1}\sqrt{\frac{\tan^2a\cot^2\alpha-1}m}$$ and the signs of $\alpha$ and $\beta$ (which sign for the outer square roots) match those of $a$ and $b$ respectively.

Like the rest of A&S, no proof is given inside. The references, though numbered, are actually in alphabetical order and give no further leads. How can the above solution be shown and can it be simplified in any way?

Parcly Taxel
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  • K. Oldham, J. Myland, & J. Spanier, An Atlas of Functions, Springer, 2009 (Chapter 62 covers the incomplete elliptic functions) contains specific solutions for the cases of imaginary amplitude $m$ and modulus $\gamma$ and point to additional imaginary transformations in G&R, Table 8.127. – Cye Waldman Dec 20 '20 at 16:33
  • @CyeWaldman All those cases are already covered in my EIF monograph. – Parcly Taxel Dec 20 '20 at 18:16

1 Answers1

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Let $c=\sin^2a$, $d=\sinh^2b$, $A=\sin^2\alpha$ and $B=\sinh^2\beta$. The expression for $\sin\gamma$ in the addition formula yields the following two equations in real numbers, corresponding to real and imaginary parts: $$c(1+d)(1+mAB)^2=A(1+B)(1+mB)$$ $$(1-c)d(1+mAB)^2=B(1-A)(1-mA)$$ A resultant computation (eliminating $A$ and $B$ individually) shows that $A$ and $B$ are the smaller roots of $$(d+1)mA^2-(1+(c+d)m)A+c$$ $$(c-1)mB^2-(1-(c+d)m)B+d$$ Still the solution requires solving a quadratic, but I find this more symmetric than the A&S solution. This has been put into the EIF update.

Parcly Taxel
  • 103,344