Euler Formula for Sine Function/Complex Numbers/Proof 2

Proof
For $z \in \C$ and $n \in \N_{> 0}$, let:


 * $\map {f_n} z = \dfrac 1 2 \paren {\paren {1 + \dfrac z n}^n - \paren {1 - \dfrac z n}^n}$

Then $\map {f_n} z = 0$ :

Let $n = 2 m + 1$.

Then the roots of $\map {f_{2 m + 1} } z$ are:


 * $\paren {2 m + 1} i \map \tan {\dfrac {k \pi} {2 m + 1} }$

for $- m \le k \le m$.

Observe that $\map {f_{2 m + 1} } z$ is a polynomial of degree $2 m + 1$.

Then for some constant $C$, we have:

It can be seen from the Binomial Theorem that the coefficient of $z$ in $\map {f_{2 m + 1} } z$ is $1$.

Hence $C = 1$, and we obtain:


 * $\displaystyle \map {f_{2 m + 1} } z = z \prod_{k \mathop = 1}^m \paren {1 + \frac {z^2} {\paren {2 m + 1}^2 \map {\tan^2} {k \pi / \paren {2 m + 1} } } }$

First we consider $z = x$ where $x$ is a non-negative real number.

Let $l < m$.

Then:


 * $\displaystyle x \prod_{k \mathop = 1}^l \paren {1 + \frac {x^2} {\paren {2 m + 1}^2 \map {\tan^2} {k \pi / \paren {2 m + 1} } } } \le \map {f_{2 m + 1} } x$

Taking the limit as $m \to \infty$ we have:

By Tangent Inequality, we have:


 * $\map \tan {\dfrac {k \pi} {2 m + 1} } \ge \dfrac {k \pi} {2 m + 1}$

hence:


 * $\displaystyle \map {f_{2 l + 1} } x \le x \prod_{k \mathop = 1}^l \paren {1 + \frac {x^2} {k^2 \pi^2} } \le \sinh x$

Taking the limit as $l \to \infty$ we have by Squeeze Theorem:


 * $(1): \quad \displaystyle x \prod_{k \mathop = 1}^\infty \paren {1 + \frac {x^2} {k^2 \pi^2} } = \sinh x$

Now let $1 < l < m$.

By Complex Modulus of Product of Complex Numbers and the Triangle Inequality, we can deduce:

Taking the limit as $m \to \infty$ we have:


 * $\displaystyle \cmod {\sinh z - z \prod_{k \mathop = 1}^l \paren {1 + \frac {z^2} {k^2 \pi^2} } } \le \sinh \cmod z - \cmod z \prod_{k \mathop = 1}^l \paren {1 + \frac {\cmod z^2} {k^2 \pi^2} }$

Now take the limit as $l \to \infty$.

By $(1)$ and Squeeze Theorem, we have:


 * $\displaystyle \sinh z = z \prod_{k \mathop = 1}^\infty \paren {1 + \frac {z^2} {k^2 \pi^2} }$

Finally, substituting $z \mapsto i z$, we obtain:


 * $\displaystyle \sin z = z \prod_{k \mathop = 1}^\infty \paren {1 - \frac {z^2} {k^2 \pi^2} }$