Why is $e^{g(x)} = \pi$ where $g(x)$ is holomorphic in Weierstrass factorization of sine function? I just can't get why it's true.
2026-05-15 04:39:02.1778819942
Why is $e^{g(x)} = \pi$ where $g(x)$ is holomorphic in Weierstrass factorization of sine function?
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This is more of textbook material than Math.SE material. The entire Chapter 6 of Complex Made Simple by David C. Ullrich is devoted to the factorization of sine. He gives three different, and very detailed, proofs. Luckily, this chapter is in the free preview of the book.