Question 10.42: Prove that (a) sn(2K + 2iK′) = 0, (b) cn(2K + 2iK′) = 1, (c)...
Prove that \quad (a) \operatorname{sn}\left(2 K+2 i K^{\prime}\right)=0, \quad (b) \operatorname{cn}\left(2 K+2 i K^{\prime}\right)=1, \quad (c) \operatorname{dn}\left(2 K+2 i K^{\prime}\right)=-1.
From the addition formulas with z_{1}=z_{2}=K+i K^{\prime}, we have
(a) \operatorname{sn}\left(2 K+2 i K^{\prime}\right)=\frac{2 \operatorname{sn}\left(K+i K^{\prime}\right) \operatorname{cn}\left(K+i K^{\prime}\right) \operatorname{dn}\left(K+i K^{\prime}\right)}{1-k^{2} \operatorname{sn}^{4}\left(K+i K^{\prime}\right)}=0
(b) \operatorname{cn}\left(2 K+2 i K^{\prime}\right)=\frac{\mathrm{cn}^{2}\left(K+i K^{\prime}\right)-\operatorname{sn}^{2}\left(K+i K^{\prime}\right) \mathrm{dn}^{2}\left(K+i K^{\prime}\right)}{1-k^{2} \operatorname{sn}^{4}\left(K+i K^{\prime}\right)}=1
(c) \operatorname{dn}\left(2 K+2 i K^{\prime}\right)=\frac{\operatorname{dn}^{2}\left(K+i K^{\prime}\right)-k^{2} \operatorname{sn}^{2}\left(K+i K^{\prime}\right) \operatorname{cn}^{2}\left(K+i K^{\prime}\right)}{1-k^{2} \operatorname{sn}^{4}\left(K+i K^{\prime}\right)}=-1