I'm working on Example $4$, page $262$, of Harris Hancock's book Lectures on the Theory of Elliptic Functions which reads:

Prove that $\dfrac{1}{\operatorname{sn}(iu,k)^2} + \dfrac{1}{\operatorname{sn}(u,k)^2} = 1$.

I can only get this result if $\operatorname{sn}(u,k) = \operatorname{sn}(u,k')$, where $k'$ is the complimentary modulus of $k$. But that does not make sense.

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    $\begingroup$ You might want to specify what the function $\operatorname{sn}$ is. $\endgroup$ Jul 28, 2016 at 19:02
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    $\begingroup$ It holds like this: $\dfrac{1}{\operatorname{sn}(iu,k)^2} + \dfrac{1}{\operatorname{sn}(u,k')^2} = 1$. There's probably a misprint in the book. $\endgroup$ Jul 28, 2016 at 19:34

1 Answer 1


Numerical computations quickly show that the identity doesn't hold.

Apparently all the editions of the book after 1910 are reprints, so it's not unusual that the typo has survived this long.

The correct identity is the one that მამუკა ჯიბლაძე mentiones in the comments,

$$\dfrac{1}{\operatorname{sn}(iu,k)^2} + \dfrac{1}{\operatorname{sn}(u,k')^2} = 1$$


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