Exercise 15.3.2

Proof. (a) By the definition of $\phi$ on $\mathrm{\Omega }=\{z\in ℂ\mid z\ne (m+\mathit{in})\frac{\varpi }{2},m\equiv n\equiv 1(\mathit{mod}2)\}$, for all $z=x+\mathit{iy}\in \mathrm{\Omega }$,

where

If we write $d=1-{\phi }^2(x){\phi }^2(y)$ the denominator, then $d\ne 0$ on $\mathrm{\Omega }$.

Using ${\phi }^{″}(x)=-2{\phi }^3(x)$ (see Exercise 15.2.3), and ${\phi {}^{\prime }}^2(x)=1-{\phi }^4(x)$, we obtain

and, using $v(x,y)=u(y,x)$,

Therefore, using $d\ne 0$ on $\mathrm{\Omega }$,

so that $\phi$ satisfies the Cauchy-Riemann equations on $\mathrm{\Omega }$. Thus $\phi$ is analytic on $\mathrm{\Omega }$. (b) For $s\in [0,\frac{\varpi }{2}]$, and $r\in [0,1]$,

Since $0={\int }_0^r\frac{1}{\sqrt{1-t^4}}dt$, $\phi (0)=0$, and since $\frac{\varpi }{2}={\int }_0^1\frac{1}{\sqrt{1-t^4}}dt$, $\phi \left(\frac{\varpi }{2}\right)=1$. Using ${\varphi }^{\prime }(x)=\sqrt{1-{\phi }^4(x)}$ for $0\le x\le \frac{\varpi }{2}$ (see Section 15.2), we obtain ${\phi }^{\prime }(0)=1,{\phi }^{\prime }\left(\frac{\varpi }{2}\right)=0$.

By (15.9), for all real $s$, $\phi (\varpi -s)=\phi (s)$, which gives $-{\phi }^{\prime }(\varpi -s)={\phi }^{\prime }(s)$, thus $\phi (\varpi )=0$, and ${\phi }^{\prime }(\varpi )=-{\phi }^{\prime }(0)=-1$.

Moreover $\phi$ is odd, thus $\phi \left(-\frac{\varpi }{2}\right)=-1,{\phi }^{\prime }\left(\frac{\varpi }{2}\right)=0$. Since $\phi$ has period $2\varpi$, $\phi \left(\frac{3\varpi }{2}\right)=-1,{\phi }^{\prime }\left(\frac{3\varpi }{2}\right)=0$.

We have proved (15.23):

By definition of $\phi$ on $\mathrm{\Omega }\subset ℂ$, for all $z=x+\mathit{iy}\in \mathrm{\Omega }$,

Therefore, since $\phi$ is odd and ${\phi }^{\prime }$ is even,

Using the Chain Rule (see Exercise 1), $i{\phi }^{\prime }(\mathit{iz})=i{\phi }^{\prime }(z)$, thus ${\phi }^{\prime }(\mathit{iz})={\phi }^{\prime }(z)$, for all $z\in \mathrm{\Omega }$. This proves (15.24):

Since $\phi$ and ${\phi }^{\prime }$ have period $2\omega$ on $ℝ$,if $k\in \mathbb{Z}$, $\phi (2\mathit{k\varpi })=\phi (0)=0$, and $\phi ((2k+1)\varpi )=\phi (\varpi )=0$. Similarly, ${\phi }^{\prime }(2\mathit{k\varpi })=\phi (0)=1,{\phi }^{\prime }((2k+1)\varpi )={\phi }^{\prime }(\varpi )=-1$. Using (15.24), $\phi (\mathit{m\varpi i})=\mathit{i\phi }(\mathit{m\varpi }),{\phi }^{\prime }(\mathit{m\varpi i})={\phi }^{\prime }(m\varpi )$.This shows (15.25):

Using the Addition Law, for all $z\in \mathrm{\Omega }$ (then $z+\mathit{m\varpi }+\mathit{n\varpi i}\in \mathrm{\Omega }$ for $m,n\in \mathbb{Z}$),

This proves (15.26), and

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