Proving the product representation of $\sin[\pi z+ \alpha]$ for non-integral $\alpha$

The objective is to show that the entire function $\sin π(z+α)$ has the following product representation (This is an exercise in Alfhors chapter 5)

$\sin \pi(z+\alpha)=e^{\pi z \cot(\pi \alpha)}\prod_{n=-\infty}^\infty \left(1+\frac{z}{n+\alpha}\right)e^{-z/(n+\alpha)}$

The problem suggests that we consider the logarithmic derivatives of the term in front of the canonical product, if we let $g(z)=e^{\pi zcot(\pi\alpha)}$. then $$\frac{g'(z)}{g(z)}=\frac{\pi \cot(\pi\alpha)e^{\pi \cot(\pi\alpha)}}{e^{\pi z\cot(\pi\alpha)}}=\pi \cot(\pi\alpha),$$ as long as $\alpha$is not an integer, $\cot[\pi(\alpha)]$ is finite.

If in the interest of comparing this to the right side, we compute the logarithmic derivative of the RHS, we get $$f(z)=\sin\pi(z+\alpha)\rightarrow\frac{f'(z)}{f(z)}=\frac{\pi \cos[\pi(z+\alpha)]}{\sin[\pi(z+\alpha)]}=\pi \cot[\pi(z+\alpha)]$$

If we can show that the logarithms of each side differ by at most a constant, than the result follows by exponentiation.

I'm not seeing how the hint implies the equality (since we don't obtain the equivalence of logarithmic derivatives), nor why the canonical product doesn't seem to have the right zeroes $(n-\alpha)$, but instead has zeroes $n+\alpha$. I've tried converting the canonical product into two sums of $$\sum_{n=1}^{\infty}\log[1-\frac{z}{n+\alpha}]$$ and looking at the power series of log to try and get the equality, but I'm not seeing why this canonical product is the same as the one you would naively construct.

Thanks

EDIT: Zero problem is not a problem, the function actually has the same zeroes, since it has zeroes whenever $\frac{z}{n+ \alpha}=-1$ and therefore when $z=-n-\alpha$. These zeroes are precisely those of our function, so all that you need to show is that the value of g(z) in

$\sin \pi(z+\alpha)=e^{g(z)}\prod_{n=-\infty}^\infty \left(1+\frac{z}{n+\alpha}\right)e^{-z/(n+\alpha)}$

is in fact $\pi cot(\pi \alpha)$

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If anyone else has been following this - I turned in a somewhat hand-wavy solution in which took the logarithmic derivative and argued that since the limits had to agree (at 0, precisely), g(z) had to equal that factor. I will post a real answer for everyone as soon as the solutions are posted – Quantumpencil Oct 19 '12 at 18:58

We see by inspection that the Canonical Product $\prod_{n=-\infty}^{\infty}(1+\frac{z}{n+\alpha}$) has zeroes when $\frac{z}{n+\alpha}=-1=-n-\alpha$ and therefore since n ranges through all the positive and negative integers, when $z=n-\alpha$ , precisely the zeroes of $\sin[\pi(z+\alpha)]$ . We know from Weierstrass' product theorem that every entire function with prescribed zeroes has such a product expansion of the form $f(z)=z^m e^g(z) \prod_1^\infty(1+\frac{z}{a_n}e^{p_n(z)})$ We therefore need to find g(z) s.t $\sin(\pi(z+\alpha)=e^{g(z)}\prod_{n=-\infty}^{\infty}(1+\frac{z}{n+\alpha})e^{-(z/n+\alpha)}$ . We take the logarithmic derivatives of both sides, and obtain $\pi \cot(\pi(z+\alpha))=g'(z)+\frac{d}{dz}\sum_{n=-\infty}^{\infty}(1+\frac{z}{n+\alpha})-\frac{z}{n+\alpha}$ . We take the logarithmic derivative of $h(z)=e^{\pi z\cot(\pi\alpha)}$ . then $\frac{h'(z)}{h(z)}=\frac{\pi \cot(\pi\alpha)e^{\pi zcot(\pi\alpha)}}{e^{\pi z\cot(\pi\alpha)}}=\pi \cot(\pi\alpha)$, as long as $\alpha$ is not an integer, $\cot[\pi(\alpha)]$ is finite. Then by considering the limit as $z\rightarrow 0$ , we have $\frac{h'(z)}{h(z)}=\pi \cot(\pi\alpha)=lim_{z\rightarrow0}\frac{f'(z)}{f(z)}=\pi \cot(\pi\alpha)$ Note that the canonical product will go to 0, since when we differentiate the sum we will remove the constant dependency and thanks to the convergence polynomial, it has no linear term. we see that $g'(z)=\pi \cot(\pi\alpha)$. That implies $g(z)=\pi z \cot(\pi\alpha)$ $\blacksquare$