Question 10.41: Prove that (a) sn(K + iK′) = 1/k, (b) cn(K + iK′) = ik′/k, (...

(a) We have

K^{\prime}=\int_{0}^{1} \frac{d t}{\sqrt{\left(1-t^{2}\right)\left(1-k^{\prime 2} t^{2}\right)}}

where k^{\prime}=\sqrt{1-k^{2}}.

Let u=1 / \sqrt{1-k^{\prime 2} t^{2}}. When t=0, u=1; when t=1, u=1 / k. Thus as t varies from 0 to 1, u varies from 1 to 1 / k. By Problem 2.43, page 69, with p=1 / k, it follows that \sqrt{1-t^{2}}=-i k^{\prime} u / \sqrt{1-k^{\prime 2} u^{2}}. Thus, by substitution, we have

K^{\prime}=-i \int\limits_{1}^{1 / k} \frac{d u}{\sqrt{\left(1-u^{2}\right)\left(1-k^{2} u^{2}\right)}}

from which

K+i K^{\prime}=\int\limits_{0}^{1} \frac{d u}{\sqrt{\left(1-u^{2}\right)\left(1-k^{2} u^{2}\right)}}+\int\limits_{1}^{1 / k} \frac{d u}{\sqrt{\left(1-u^{2}\right)\left(1-k^{2} u^{2}\right)}}=\int\limits_{0}^{1 / k} \frac{d u}{\sqrt{\left(1-u^{2}\right)\left(1-k^{2} u^{2}\right)}}

i.e., \operatorname{sn}\left(K+i K^{\prime}\right)=1 / k.

(b) From part (a),

\operatorname{cn}\left(K+i K^{\prime}\right)=\sqrt{1-\operatorname{sn}^{2}\left(K+i K^{\prime}\right)}=\sqrt{1-1 / k^{2}}=-i \sqrt{1-k^{2}} / k=-i k^{\prime} / k

(c) \operatorname{dn}\left(K+i K^{\prime}\right)=\sqrt{1-k^{2} \operatorname{sn}^{2}\left(K+i K^{\prime}\right)}=0 by part (a).