Exercise 15.3.6

Proof. If $z={(-1)}^{m+n}{z}_0+(m+\mathit{in})\varpi$, then the periodicity and odd parity of $\phi$ shows that

Conversely, suppose that $\phi (z)=\phi (z_0)$ (where $z,z_0$ are not poles of $\phi$). By Proposition 13.3.1, the addition law gives, for all $x,y\in ℂ$ such that both members are defined,

thus, by subtraction,

Take $x=\frac{z+z_0}{2},y=\frac{z-z_0}{2}$ in this formula. We obtain

(This formula is analogous to the trigonometric formula $\sin <!--\; FUNCTION\; APPLICATION\; -->p-\sin <!--\; FUNCTION\; APPLICATION\; -->q=2\cos <!--\; FUNCTION\; APPLICATION\; -->\frac{p+q}{2}\sin <!--\; FUNCTION\; APPLICATION\; -->\frac{p-q}{2}$.) Therefore,

By the Principle of Analytic Continuation (as in the proof of Proposition 15.3.1), this formula is true for all $z$ such that both members are defined, including the points such that $1+{\phi }^2\left(\frac{z+z_0}{2}\right)=0$. In others words, this is true for all $z\in \mathrm{\Omega }$ such that $\frac{z-z_0}{2},\frac{z+z_0}{2}$ are not poles of $\phi$.

Thus $\phi (z)=\phi (z_0)$ implies that

$\text{∙}$

Suppose that $\phi \left(\frac{z-z_0}{2}\right)=0$.

By Proposition 15.3.2, the zeros of $\phi$ are $z=(p+\mathit{iq})\varpi ,p,q\in \mathbb{Z}$, thus

$$\begin{array}{llll}\hfill \phi \left(\frac{z-z_0}{2}\right)=0& ⟺\frac{z-z_0}{2}=(p+\mathit{qi})\varpi ,p,q\in \mathbb{Z}& \hfill & \\ \hfill & ⟺z=z_0+2\mathit{p\varpi }+2\mathit{q\varpi i},p,q\in \mathbb{Z}.& \hfill & \\ \hfill & & \hfill \end{array}$$

This shows that

$$z={(-1)}^{m+n}{z}_0+(m+\mathit{in})\varpi ,\text{where }m=2p,n=2q\in \mathbb{Z}.$$

$\text{∙}$

Suppose that ${\phi }^{\prime }\left(\frac{z+z_0}{2}\right)=0$.

By Exercise 4, we know that the points $(m+\mathit{in})\frac{\varpi }{2}$, $m+n$ odd, are zeros of ${\phi }^{\prime }$, and we showed that they are the only zeros of ${\phi }^{\prime }$ (without using Theorem 15.3.3). Therefore,

$$\begin{array}{llll}\hfill {\phi }^{\prime }\left(\frac{z+z_0}{2}\right)=0⟺& \frac{z+z_0}{2}=(m+\mathit{in})\frac{\varpi }{2},m+n\equiv 1(\mathit{mod}2)& \hfill & \\ \hfill ⟺& z=-z_0+(m+\mathit{in})\varpi ,m+n\equiv 1(\mathit{mod}2)& \hfill & \\ \hfill & & \hfill \end{array}$$

This shows that

$$z={(-1)}^{m+n}{z}_0+(m+\mathit{in})\varpi ,m,n\in \mathbb{Z}.$$

$\text{∙}$

Suppose that $\frac{z-z_0}{2}\notin \mathrm{\Omega }$. Then $\frac{z-z_0}{2}$ is a pole. By Theorem 15.3.2, $$\frac{z-z_0}{2}=(m+\mathit{in})\frac{\varpi }{2},m\equiv n\equiv 1(\mathit{mod}2),$$

thus

$$z=z_0+(m+\mathit{in})\varpi ,m\equiv n\equiv 1(\mathit{mod}2),$$

so that

$$z={(-1)}^{m+n}{z}_0+(m+\mathit{in})\varpi .$$

$\text{∙}$

Suppose at last that $\frac{z+z_0}{2}\notin \mathrm{\Omega }$ (this case is more tricky). Then $$\frac{z+z_0}{2}=(m+\mathit{in})\frac{\varpi }{2},m\equiv n\equiv 1(\mathit{mod}2),$$

thus

$$z=-z_0+(m+\mathit{in})\varpi ,m\equiv n\equiv 1(\mathit{mod}2),$$

where the sign $(-1)$ before $z_0$ is not equal to ${(-1)}^{m+n}$ !?!

But fortunately, in this case, by Proposition 15.3.1,

$$\phi (z)=\phi (-z_0+(m+\mathit{in})\varpi )={(-1)}^{m+n}\phi (-z_0)=\phi (-z_0)=-\phi (z_0).$$

Since by hypothesis $\phi (z)=\phi (z_0)$, we obtain $\phi (z_0)=-\phi (z_0)$, thus $\phi (z_0)=0$, and $\phi (z)=\phi (z_0)$ is equivalent to $\phi (z)=0$.

By Proposition 15.3.2, $z_0=(p+\mathit{iq})\varpi ,z=(r+\mathit{is})\varpi$, where $p,q,r,s$ are integers. Thus

$$z=z_0+(r-p+i(s-q))\varpi =z_0+(m^{\prime }+i{n}^{\prime })\varpi ,\text{where }{m}^{\prime }=r-p,n^{\prime }=s-q\in \mathbb{Z},$$

and

$$z=-z_0+(r+p+i(s+q)\varpi =-z_0+(m^{″}+i{n}^{″}),\text{where }{m}^{″}=r+p,n^{″}=s+q\in \mathbb{Z}.$$

Note that $m^{\prime }+n^{\prime }\equiv m^{″}+n^{″}(\mathit{mod}2)$. If $m^{\prime }+n^{\prime }$ is even then $z={(-1)}^{m^{\prime }+n^{\prime }}{z}_0+(m^{\prime }+i{n}^{\prime })\varpi$, and if $m^{\prime }+n^{\prime }$ is odd, then $z={(-1)}^{m^{″}+n^{″}}{z}_0+(m^{″}+i{n}^{″})\varpi .$

In all cases $z={(-1)}^{m+n}{z}_0+(m+\mathit{in})\varpi$, for some $m,n\in \mathbb{Z}$. □