The Weierstrass product for sine

For $z=x+iy$, [F4] gives $\sin (\pi z)=(e^{i\pi z}-e^{-i\pi z})\mathrm{/}(2i)$, so $\mathrm{\mid }\sin (\pi z)\mathrm{\mid }\le \frac{1}{2}(e^{-\pi y}+e^{\pi y})\le e^{\pi \mathrm{\mid }z\mathrm{\mid }}$. Along the imaginary axis one has $\mathrm{\mid }\sin (\pi ir)\mathrm{\mid }=(e^{\pi r}-e^{-\pi r})\mathrm{/}2$ for $r>0$. Therefore [F3] makes $f(z)=\sin (\pi z)$ an entire function of order $1$.

By [F1], the zero at $0$ has order $1$ and the nonzero zeros are exactly $\pm 1,\pm 2,\dots$. Applying [F2] with $\rho =1$ yields a polynomial $Q$ of degree at most $1$ such that $\sin (\pi z)=z{e}^{Q(z)}{\prod }_{n\ge 1}{E}_1(z\mathrm{/}n)E_1(-z\mathrm{/}n)$. Since $E_1(w)E_1(-w)=(1-w)e^w(1+w)e^{-w}=1-w^2$, this becomes $\sin (\pi z)=z{e}^{Q(z)}{\prod }_{n\ge 1}(1-z^2\mathrm{/}{n}^2)$.

By [F4], the function $\sin (\pi z)$ is odd, while ${\prod }_{n\ge 1}(1-z^2\mathrm{/}{n}^2)$ is even. Therefore the quotient $e^{Q(z)}=\frac{\sin (\pi z)}{z{\prod }_{n\ge 1}(1-z^2\mathrm{/}{n}^2)}$ is even. Writing $Q(z)=az+b$, this means $e^{az+b}=e^{-az+b}$ for every $z$, so $e^{2az}=1$ on $\mathbb{C}$. Therefore $a=0$, and $Q$ is constant.

Dividing the power series in [F4] by $\pi z$ gives $\sin (\pi z)\mathrm{/}(\pi z)=1-{\pi }^2{z}^2\mathrm{/}6+O(z^4)$. Step 2.1 with step 3.1 gives $\sin (\pi z)\mathrm{/}z=e^b{\prod }_{n\ge 1}(1-z^2\mathrm{/}{n}^2)$, and substituting $z=0$ shows $e^b=\pi$. Therefore $\sin (\pi z)=\pi z{\prod }_{n\ge 1}(1-z^2\mathrm{/}{n}^2)$. The convergence is locally uniform because step 2.1 is a normally convergent canonical-product factorization.