product formula for the sum of cosine and sine 
Above is an exercise from Conway. I have proved the product formula for sine and cosine functions and they are like below. However, I am stuck at how to calculate the exercise problem. Could anyone help me how to compute product formula in the exercise?


 A: I think there is typo in Conway's book. It should be 
$$\cos\left(\frac{\pi z}{4}\right)-\sin\left(\frac{\pi z}{4}\right)=
\prod_{n=1}^{\infty}\left(1+\frac{(-1)^n z}{2n-1}\right).$$
Note that
$$\cos\left(\frac{\pi z}{4}\right)-\sin\left(\frac{\pi z}{4}\right)=\frac{\sin\left(\frac{\pi (z-1)}{4}\right)}{\sin\left(-\frac{\pi}{4}\right)}$$
Then by using
$$\sin(\pi z) = \pi z \prod_{n\neq 0} \left(1-\frac{z}{n}\right)e^{z/n},$$
we obtain
\begin{align*}
\cos\left(\frac{\pi z}{4}\right)-\sin\left(\frac{\pi z}{4}\right)
&=\frac{\frac{\pi (z-1)}{4} \prod_{n\neq 0} \left(1-\frac{(z-1)}{4n}\right)e^{(z-1)/(4n)}}{\frac{\pi (-1)}{4} \prod_{n\neq 0} \left(1-\frac{(-1)}{4n}\right)e^{(-1)/(4n)}}\\
&=(1-z) \prod_{n\neq 0} \left(\frac{4n-(z-1)}{4n+1}\right)e^{z/(4n)}\\
&=(1-z) \prod_{n\neq 0} \left(1-\frac{z}{4n+1}\right)e^{z/(4n)}\\
&=(1-z)\prod_{n\geq 1} \left(1-\frac{z}{4n+1}\right)e^{z/(4n)}\left(1-\frac{z}{-4n+1}\right)e^{-z/(4n)}\\
&=(1-z)\prod_{n\geq 1} \left(1-\frac{z}{4n+1}\right)\left(1+\frac{z}{4n-1}\right)
\end{align*}
which is equal to
\begin{align*}
\prod_{n=1}^{\infty}\left(1+\frac{(-1)^n z}{2n-1}\right)
&=\prod_{k\geq 1}\left(1+\frac{(-1)^{2k-1} z}{2(2k-1)-1}\right)\prod_{k\geq 1}\left(1+\frac{(-1)^{2k} z}{2(2k)-1}\right)
\\
&=\prod_{k\geq 1}\left(1-\frac{z}{4k-3}\right)\prod_{k\geq 1}\left(1+\frac{z}{4k-1}\right)\\
&=(1-z)\prod_{k\geq 2}\left(1-\frac{z}{4(k-1)+1}\right)\prod_{k\geq 1}\left(1+\frac{z}{4k-1}\right)\\
&=(1-z)\prod_{k\geq 1} \left(1-\frac{z}{4k+1}\right)\left(1+\frac{z}{4k-1}\right).
\end{align*}
