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Let $\operatorname{cn}(u\mid m)$ be Gudermann's notation for the Jacobi elliptic function $\operatorname{cn}$. It is well known to be doubly periodic. For $0<m<1$ the two periods, $4 K(m)$ and $4 i K(1-m)$, are purely real and purely imaginary respectively. For any $n_1, n_2 \in \mathbb{Z}$: $$ \operatorname{cn}\left(u + 4 n_1 K(m) + 4 n_2 i K(1-m) \mid m \right) = \operatorname{cn}\left(u \mid m \right) $$ and where $K(m)$ denotes the complete elliptic integral of the first kind (in that link $m=k^2$).

For $m=\frac{1}{2}$, $f(x,y) = \operatorname{cn}(x+i y \mid m)$ is doubly periodic in a square with sidelength $4K\left(\frac{1}{2}\right) \approx 7.4163$.

By looking at Peirce quincuncial projection I noticed empirically that, for $x\in \mathbb{R}$, $f(x,x)$ is of unit magnitude: $$ \left| f(x,x) \right| = \left| \operatorname{cn}\left(x+i x \mid \frac{1}{2}\right) \right| = 1 $$ enter image description here

I am looking for either a proof, or a reference to one.

Thank you.

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Jacobi's imaginary transformation $$ \mathrm{cn}(iu|m)=\frac{1}{\mathrm{cn}(u|m')},$$ and parity of $\mathrm{cn}(u)$ will do the job. $\blacksquare$

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  • $\begingroup$ Thank you! This is exactly it. I will post my solution shortly $\endgroup$ – Sasha Apr 29 '13 at 2:19
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    $\begingroup$ Yours is quite a lot neater! $\endgroup$ – Sasha Apr 29 '13 at 2:47
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Using the addition theorem: $$ \operatorname{cn}\left(x+ i x \mid m\right) = \frac{\operatorname{cn}(x \mid m) \operatorname{cn}(i x \mid m) - \operatorname{sn}(x \mid m) \operatorname{sn}(i x \mid m) \operatorname{dn}(x \mid m) \operatorname{dn}(i x \mid m) }{1-m \operatorname{sn}^2(x \mid m) \operatorname{sn}^2(i x \mid m)} $$ Using the Jacobi's imaginary transformations: $$ \operatorname{cn}(i x \mid m) = \frac{1}{\operatorname{cn}(x \mid 1-m)} \quad \operatorname{sn}(i x \mid m) = i \frac{\operatorname{sn}(x \mid 1-m)}{\operatorname{cn}(x \mid 1-m)} \quad \operatorname{dn}(i x \mid m) = \frac{\operatorname{dn}(x \mid 1-m)}{\operatorname{cn}(x \mid 1-m)} $$ we get $$ \operatorname{cn}\left(x+ i x \mid m\right) = \frac{\operatorname{cn}(x \mid 1-m) \operatorname{cn}(x\mid m)-i \operatorname{dn}(x \mid 1-m) \operatorname{dn}(x\mid m) \operatorname{sn}(x\mid 1-m) \operatorname{sn}(x \mid m)}{\operatorname{cn}^2(x\mid 1-m) + m \operatorname{sn}^2(x \mid 1-m) \operatorname{sn}^2(x \mid m)} $$ Substituting $m={1 \over 2}$ we get $$ \operatorname{cn}\left(x+ i x \mid {1 \over 2}\right) = \frac{\operatorname{cn}^2(x\mid {1\over 2}) - i \operatorname{sn}^2(x\mid {1\over 2}) \operatorname{dn}^2(x\mid {1\over 2}) }{\operatorname{cn}^2(x\mid {1\over 2}) + {1\over 2} \operatorname{sn}^4(x\mid {1\over 2})} $$ The absolute value now follows to be 1: $$ \left| \operatorname{cn}\left(x+ i x \mid {1 \over 2}\right) \right|^2 = \frac{\operatorname{cn}^4(x\mid {1\over 2}) + \operatorname{sn}^4(x\mid {1\over 2}) \operatorname{dn}^4(x\mid {1\over 2}) }{\left( \operatorname{cn}^2(x\mid {1\over 2}) + {1\over 2} \operatorname{sn}^4(x\mid {1\over 2}) \right)^2} $$ Using defining relations $\operatorname{cn}^2(x \mid 1/2) + \operatorname{sn}^2(x \mid 1/2) = 1$ and $\operatorname{dn}^2(x \mid 1/2) + {1 \over 2} \operatorname{sn}^2(x \mid 1/2) = 1$ the right hand side equals one.

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