Question 10.40: Prove that (a) sn(z + 4K) = sn z, (b) cn(z + 4K) = cn z, (c)...
Prove that (a) \operatorname{sn}(z+4 K)=\operatorname{sn} z, (b) \operatorname{cn}(z+4 K)=\operatorname{cn} z, (c) \operatorname{dn}(z+2 K)=\operatorname{dn} z.
From Problem 10.39
(a) \operatorname{sn}(z+4 K)=-\operatorname{sn}(z+2 K)=\operatorname{sn} z
(b) \operatorname{cn}(z+4 K)=-\operatorname{cn}(z+2 K)=\operatorname{cn} z
(c) \operatorname{dn}(z+2 K)=\sqrt{1-k^{2} \operatorname{sn}^{2}(z+2 K)}=\sqrt{1-k^{2} \operatorname{sn}^{2} z}=\operatorname{dn} z
Another Method. The integrand 1 / \sqrt{\left(1-t^{2}\right)\left(1-k^{2} t^{2}\right)} has branch points at t= \pm 1 and t= \pm 1 / k in the t plane [Fig. 10-10]. Consider the integral from 0 to w along two paths C_{1} and C_{2}. We can deform C_{2} into the path A B D E F G H J A+C_{1}, where B D E and G H J are circles of radius \epsilon while J A B and E F G, drawn separately for visual purposes, are actually coincident with the x axis.
We then have
\begin{aligned} \int\limits_{\underset{C_2}{0}}^{w} \frac{d t}{\sqrt{\left(1-t^{2}\right)\left(1-k^{2} t^{2}\right)}}= & \int\limits_{0}^{1-\epsilon} \frac{d x}{\sqrt{\left(1-x^{2}\right)\left(1-k^{2} x^{2}\right)}}+\int\limits_{B D E} \frac{d t}{\sqrt{\left(1-t^{2}\right)\left(1-k^{2} t^{2}\right)}} \\ & +\int\limits_{1-\epsilon}^{0} \frac{d x}{-\sqrt{\left(1-x^{2}\right)\left(1-k^{2} x^{2}\right)}}+\int\limits_{0}^{-1+\epsilon} \frac{d x}{-\sqrt{\left(1-x^{2}\right)\left(1-k^{2} x^{2}\right)}} \\ & +\int\limits_{G H J} \frac{d t}{-\sqrt{\left(1-t^{2}\right)\left(1-k^{2} t^{2}\right)}}+\int\limits_{-1+\epsilon}^{\theta} \frac{d x}{\sqrt{\left(1-x^{2}\right)\left(1-k^{2} x^{2}\right)}} \\ & +\int\limits_{\underset{C_1}{0}}^{w} \frac{d t}{\sqrt{\left(1-t^{2}\right)\left(1-k^{2} t^{2}\right)}} \\ = & 4 \int\limits_{0}^{1-\epsilon} \frac{d x}{\sqrt{\left(1-x^{2}\right)\left(1-k^{2} x^{2}\right)}}+\int\limits_{\overset{0}{c_1}}^{w} \frac{d t}{\sqrt{\left(1-t^{2}\right)\left(1-k^{2} t^{2}\right)}} \\ & +\int\limits_{B D E} \frac{d t}{\sqrt{\left(1-t^{2}\right)\left(1-k^{2} t^{2}\right)}}+\int\limits_{G H J} \frac{d t}{-\sqrt{\left(1-t^{2}\right)\left(1-k^{2} t^{2}\right)}} \end{aligned}
where we have used the fact that in encircling a branch point, the sign of the radical is changed.
On B D E and G H J, we have t=1-\epsilon e^{i \theta} and t=-1+\epsilon e^{i \theta}, respectively. Then, the corresponding integrals equal
\begin{aligned} & \int\limits_{0}^{2 \pi} \frac{-i \epsilon e^{i \theta} d \theta}{\sqrt{\left(2-\epsilon e^{i \theta}\right)\left(\epsilon e^{i \theta}\right)\left\{1-k^{2}\left(1-\epsilon e^{i \theta}\right)^{2}\right\}}}=-i \sqrt{\epsilon} \int\limits_{0}^{2 \pi} \frac{e^{i \theta / 2} d \theta}{\sqrt{\left(2-\epsilon e^{i \theta}\right)\left\{1-k^{2}\left(1-\epsilon e^{i \theta}\right)^{2}\right\}}} \\ & \int\limits_{0}^{2 \pi} \frac{i \epsilon e^{i \theta} d \theta}{\sqrt{\left(\epsilon e^{i \theta}\right)\left(2-\epsilon e^{i \theta}\right)\left\{1-k^{2}\left(-1+\epsilon e^{i \theta}\right)^{2}\right\}}}=i \sqrt{\epsilon} \int\limits_{0}^{2 \pi} \frac{e^{i \theta / 2} d \theta}{\sqrt{\left(2-\epsilon e^{i \theta}\right)\left\{1-k^{2}\left(-1+\epsilon e^{i \theta}\right)^{2}\right\}}} \end{aligned}
As \epsilon \rightarrow 0, these integrals approach zero and we obtain
\int\limits_{\underset{C_2}{0}}^{w} \frac{d t}{\sqrt{\left(1-t^{2}\right)\left(1-k^{2} t^{2}\right)}}=4 \int\limits_{0}^{1} \frac{d x}{\sqrt{\left(1-x^{2}\right)\left(1-k^{2} x^{2}\right)}}+\int\limits_{\underset{C_1}{0}}^{w} \frac{d t}{\sqrt{\left(1-t^{2}\right)\left(1-k^{2} t^{2}\right)}}
Now, if we write
z=\int\limits_{\substack{0 \\ C_{1}}}^{w} \frac{d t}{\sqrt{\left(1-t^{2}\right)\left(1-k^{2} t^{2}\right)}}, \quad \text { i.e., } w=\operatorname{sn} z
then
z+4 K=\int\limits_{\substack{0 \\ C_{2}}}^{w} \frac{d t}{\sqrt{\left(1-t^{2}\right)\left(1-k^{2} t^{2}\right)}}, \quad \text { i.e., } w=\operatorname{sn}(z+4 K)
and since the value of w is the same in both cases, \operatorname{sn}(z+4 K)=\operatorname{sn} z.
Similarly, we can establish the other results.
