Exercise 15.4.10

Proof. The identity (15.29) implies that

Setting $w=z+(1+i)\frac{\varpi }{2}$, we obtain (15.48)

Note that $\phi (-z)=-\phi (z)$, and $\phi (\mathit{iz})=\mathit{i\phi }(z)$, so that

Since $\beta \equiv i^{𝜀}(\mathit{mod}2(1+i))$, we can write $\beta =i^{𝜀}+2(1+i)\alpha$, where $\alpha \in \mathbb{Z}[i]$. Then

Since $1+i$ is even, ${(1+i)}^2\alpha$ is even, thus ${(1+i)}^2\mathit{\alpha \varpi }$ is a period of $\phi$. Therefore

where $Z=i^{-𝜀}\mathit{\beta z}$.

By (15.29), where we substitute $Z=i^{-𝜀}\mathit{\beta z}$ to $z$,

therefore

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