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 $\pi cot(\pi \alpha)$ 
 A: There is an error with the problem itself. A constant factor of $\sin\pi\alpha$ is missing.
Noticing that $\sin \pi(z+\alpha)$ has simple zeros at $a_n = -n-\alpha$, as well as $\sum_{n=1}^\infty 1/|a_n|^{h+1} < \infty$ with $h=1$, we know from Weierstrass' theorem the following
$$\sin \pi(z+\alpha) = e^{g(z)} \prod_{n\in Z}\left(1+\frac{z}{n+\alpha}\right)e^{-z/(n+\alpha)},$$
where $g(z)$ is an entire function.
We take the logarithmic derivatives of both sides, and obtain
$$\pi \cot \pi(z+\alpha) = g'(z) + \sum_{n\in Z} \left(\frac{1}{z+n+\alpha} - \frac{1}{n+\alpha} \right).$$
where the procedure is easy to justify by uniform convergence on any compact set which does not contain the points $z = a_n$.
The LHS is periodic with the period 1. Taking $z = x + iy$, we have (Ahlfors, Chap. 2, Sec. 3.2, Ex. 4)
$$ |\sin \pi z|^2 = \cosh^2 \pi y - \cos^2 \pi x$$
and hence also
$$ |\cos \pi z|^2 = \cosh^2 \pi y - \sin^2 \pi x.$$
Therefore the LHS tends uniformly to $\pi$ as $|y|\rightarrow \infty$.
For the infinite series in the RHS, notice that it is absolutely convergent and therefore we can take the sum in whatever order we like. We can define $S_N(z) = \sum_{n=-N}^{N}\left(\frac{1}{z+n+\alpha} - \frac{1}{n+\alpha} \right)$. For any fixed $z=x+iy$, it is easy to show that $\lim_{N\rightarrow\infty}S_N(z+1)-S_N(z) = 0$. Similarly, we can show that the series uniformly tends to a finite value depending on $\alpha$ as $|y| \rightarrow \infty$.
Therefore $g'(z)$ has the following properties.

*

*It is entire, since it is the derivative of an entire function.

*It is periodic with the period 1, since it is the difference between two periodic functions with the same period.

*It tends uniformly to a finite value, which depends on $\alpha$, as $|y| \rightarrow \infty$, since it is the difference between two functions of such properties.

These properties combined guarentees that $g'(z)$ is bounded. By Liouville's theorem $g'(z)$ must reduce to a constant.
Setting $z=0$, we have
$$g'(z) = g'(0) = \pi \cot \pi\alpha,$$
and therefore
$$g(z) = \pi z \cot \pi\alpha + C$$, where $C$ is a constant.
Substituting it into the first equation with $z=0$, we have $C = \log\sin\pi\alpha$. So we have
$$\sin \pi(z+\alpha) = \sin\pi\alpha e^{\pi z \cot \pi\alpha } \prod_{n\in Z}\left(1+\frac{z}{n+\alpha}\right)e^{-z/(n+\alpha)}.$$
A: Here is the solution I came up with (which received full marks, though I am still shaky on a bit of it.)
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 $g(z)=\pi z \cot(\pi\alpha)$
  $\blacksquare$
