Question 10.43: Prove that (a) sn(z + 2iK′) = sn z, (b) cn(z + 2K + 2iK′) = ...
Prove that (a) \operatorname{sn}\left(z+2 i K^{\prime}\right)=\operatorname{sn} z, \quad (b) \operatorname{cn}\left(z+2 K+2 i K^{\prime}\right)=\operatorname{cn} z, \quad (c) \operatorname{dn}\left(z+4 i K^{\prime}\right)=\operatorname{dn} z.
Using Problems 10.39, 10.42, 10.170, and the addition formulas, we have
(a) \operatorname{sn}\left(z+2 i K^{\prime}\right)=\operatorname{sn}\left(z-2 K+2 K+2 i K^{\prime}\right)
\begin{aligned} & =\frac{\operatorname{sn}(z-2 K) \operatorname{cn}\left(2 K+2 i K^{\prime}\right) \operatorname{dn}\left(2 K+2 i K^{\prime}\right)+\operatorname{sn}\left(2 K+2 i K^{\prime}\right) \operatorname{cn}(z-2 K) \operatorname{dn}(z-2 K)}{1-k^{2} \operatorname{sn}^{2}(z-2 K) \operatorname{sn}^{2}\left(2 K+2 i K^{\prime}\right)} \\ & =\operatorname{sn} z \end{aligned}
(b) \quad \operatorname{cn}\left(z+2 K+2 i K^{\prime}\right)=\frac{\operatorname{cn} z \operatorname{cn}\left(2 K+2 i K^{\prime}\right)-\operatorname{sn} z \operatorname{sn}\left(2 K+2 i K^{\prime}\right) \operatorname{dn} z \operatorname{dn}\left(2 K+2 i K^{\prime}\right)}{1-k^{2} \operatorname{sn}^{2} z \operatorname{sn}^{2}\left(2 K+2 i K^{\prime}\right)}=\operatorname{cn} z
(c) \operatorname{dn}\left(z+4 i K^{\prime}\right)=\operatorname{dn}\left(z-4 K+4 K+4 i K^{\prime}\right)
\begin{aligned} & =\frac{\operatorname{dn}(z-4 K) \operatorname{dn}\left(4 K+4 i K^{\prime}\right)-k^{2} \operatorname{sn}(z-4 K) \operatorname{sn}\left(4 K+4 i K^{\prime}\right) \operatorname{cn}(z-4 K) \operatorname{cn}\left(4 K+4 i K^{\prime}\right)}{1-k^{2} \operatorname{sn}^{2}(z-4 K) \operatorname{sn}^{2}\left(4 K+4 i K^{\prime}\right)} \\ & =\operatorname{dn} z \end{aligned}