Exercise 15.3.4

Proof. Note that, since $\phi (z+\mathit{k\varpi }+\mathit{l\varpi i})={(-1)}^{k+l}\phi (z)$, we obtain by differentiation

$${\phi }^{\prime }(z+\mathit{k\varpi }+\mathit{l\varpi i})={(-1)}^{k+l}{\phi }^{\prime }(z),(z\in \mathrm{\Omega },k,l\in \mathbb{Z}).$$

Suppose that $m+n$ is odd, where $m,n\in \mathbb{Z}$. $\text{∙}$ If $m$ is odd, and $n$ even, then $m=2k+1,n=2l$, where $k,l$ are integers. Then

$$\begin{array}{llll}\hfill {\phi }^{\prime }\left((m+\mathit{in})\frac{\varpi }{2}\right)& ={\phi }^{\prime }\left(\frac{\varpi }{2}+\mathit{k\varpi }+\mathit{l\varpi }\right)& \hfill & \\ \hfill & ={(-1)}^{k+l}{\phi }^{\prime }\left(\frac{\varpi }{2}\right)=0.& \hfill & \end{array}$$ $\text{∙}$ If $m$ is even , and $n$ odd, then $m=2k,n=2l+1$, where $k,l$ are integers. Then, using (15.24), $$\begin{array}{llll}\hfill {\phi }^{\prime }\left((m+\mathit{in})\frac{\varpi }{2}\right)& ={\phi }^{\prime }\left(\frac{\varpi }{2}i+\mathit{k\varpi }+\mathit{l\varpi }\right)& \hfill & \\ \hfill & ={(-1)}^{k+l}{\phi }^{\prime }\left(\frac{\varpi }{2}i\right)& \hfill & \\ \hfill & ={(-1)}^{k+l}{\phi }^{\prime }\left(\frac{\varpi }{2}\right)=0.& \hfill & \end{array}$$

Thus ${\phi }^{\prime }(z)$ vanishes at all points of form $(m+\mathit{in})\frac{\varpi }{2}$, $m+n$ odd.

But are these points the only zeros of ${\phi }^{\prime }(z)$ ? In Exercise 6, we need also the converse, which will prove now.

Suppose that ${\phi }^{\prime }(z)=0$. As in the proof of Theorem 15.3.2, for all $z\in \mathrm{\Omega }$ such that $\phi (z)\ne \pm i$,

$$\phi \left(z+\frac{\varpi }{2}\right)=\frac{\phi (z){\phi }^{\prime }\left(\frac{\varpi }{2}\right)+\phi \left(\frac{\varpi }{2}\right){\phi }^{\prime }(z)}{1+{\phi }^2(z){\phi }^2\left(\frac{\varpi }{2}\right)}=\frac{{\phi }^{\prime }(z)}{1+{\phi }^2(z)},$$

thus

$${\phi }^{\prime }(z)=(1+{\phi }^2(z))\phi \left(z+\frac{\varpi }{2}\right).$$

By the Principle of Analytic Continuation (see Exercise 5), since both members are analytic, this formula, which is true for all $z\in ℝ$, is true for all $z\in \mathrm{\Omega }$ such that $z+\frac{\varpi }{2}$ is not a pole of $\phi$.

Therefore, for all $z\in \omega$

$${\phi }^{\prime }(z)=0\Rightarrow \phi \left(z+\frac{\varpi }{2}\right)=0,\text{ or }\phi (z)=\pm i,\text{ or }z+\frac{\varpi }{2}\text{ is a pole}.$$

$\text{∙}$ If $\phi \left(z+\frac{\varpi }{2}\right)=0$, by Proposition 15.3.2,

$$z+\frac{\varpi }{2}=(p+\mathit{iq})\varpi ,p,q\in \mathbb{Z},$$

thus $z=(2p-1+i2q)\frac{\varpi }{2}=(m+\mathit{in})\frac{\varpi }{2}$, where $m=2p-1,n=2q$, and $m+n$ is odd.

$\text{∙}$ If $\phi (z)=-i$, then $\phi (\mathit{iz})=1=\phi \left(\frac{\varpi }{2}\right)$, thus, using ${\phi }^{\prime }(\mathit{iz})={\phi }^{\prime }(z)$, and the addition formula,

$$\phi \left(\mathit{iz}-\frac{\varpi }{2}\right)=\frac{\phi (\mathit{iz}){\phi }^{\prime }\left(\frac{\varpi }{2}\right)-\phi \left(\frac{\varpi }{2}\right){\phi }^{\prime }(\mathit{iz})}{1+{\phi }^2(\mathit{iz}){\phi }^2\left(\frac{\varpi }{2}\right)}=-\frac{{\phi }^{\prime }(\mathit{iz})}{2}=-\frac{{\phi }^{\prime }(z)}{2}=0.$$

Therefore, by Proposition 15.3.2(a),

$$\mathit{iz}-\frac{\varpi }{2}=(p+\mathit{iq})\varpi ,p,q\in \mathbb{Z},$$

thus, multiplying by $-i$,

$$z+i\frac{\varpi }{2}=(-\mathit{ip}+q)\varpi ,$$

and

$$z=\left[2q+(-2p-1)i\right]\frac{\varpi }{2}=(m+\mathit{ni})\frac{\varpi }{2},\text{where }m=2q,n=-2p-1\in \mathbb{Z},m+n\text{ odd}.$$

$\text{∙}$ If $\phi (z)=i$, then $\phi (\mathit{iz})=-1=\phi \left(-\frac{\varpi }{2}\right)$, thus

$$\phi \left(\mathit{iz}+\frac{\varpi }{2}\right)=\frac{\phi (\mathit{iz}){\phi }^{\prime }\left(\frac{\varpi }{2}\right)+\phi \left(\frac{\varpi }{2}\right){\phi }^{\prime }(\mathit{iz})}{1+{\phi }^2(\mathit{iz}){\phi }^2\left(\frac{\varpi }{2}\right)}=\frac{{\phi }^{\prime }(\mathit{iz})}{2}=\frac{{\phi }^{\prime }(z)}{2}=0.$$

Therefore

$$\mathit{iz}+\frac{\varpi }{2}=(p+\mathit{iq})\varpi ,p,q\in \mathbb{Z},$$

thus

$$z=i\frac{\varpi }{2}+(-\mathit{ip}+q)\varpi ,$$

and

$$z=\left[2q+(-2p+1)i\right]\frac{\varpi }{2}=(m+\mathit{ni})\frac{\varpi }{2},\text{where }m=2q,n=-2p+1\in \mathbb{Z},m+n\text{ odd}.$$

$\text{∙}$ If $z+\frac{\varpi }{2}$ is a pole of $\phi$, then

$$z+\frac{\varpi }{2}=(p+\mathit{iq})\frac{\varpi }{2},p\equiv q\equiv 1(\mathit{mod}2),$$

thus

$$\begin{array}{llll}\hfill z& =(p-1+\mathit{iq})\frac{\varpi }{2}& \hfill & \\ \hfill & =(m+\mathit{in})\frac{\varpi }{2},\text{where }m=p-1,n=q,\text{and }m+n=p+q-1\equiv 1(\mathit{mod}2).& \hfill & \end{array}$$

(This case gives the same solutions that $\phi (z)=i$, so that these two cases are equivalent.)

In all cases, $z=(m+\mathit{ni})\frac{\varpi }{2}$, where $m+n$ is odd, thus all zeros of ${\phi }^{\prime }$ are our known zeros. □