Zeros of a complex function

Let $z = a + bi$ be any complex number such that:

$sin(z) = z \Rightarrow \frac{e^{iz} - e^{-iz}}{2i} = a + bi \Rightarrow \frac{e^{i(a+bi)} - e^{-i(a+bi)}}{2i} = a + bi \Rightarrow e^{-b} \cdot e^{ia} - e^{b} \cdot e^{-ia} = 2i \cdot (a + bi)$

or finally, $e^{-b} \cdot (cos(a) + i sin(a)) - e^{b} \cdot (cos(a) - i sin(a)) = -2b + 2ai.$ Setting the real and the imaginary parts equal to each other now yields:

REAL: $(e^{-b} - e^{b}) \cdot cos(a) = -2b \Rightarrow -2\cdot sinh(b) cos(a) = -2b \Rightarrow sinh(b)cos(a) = b;$

IMAGINARY: $(e^{-b} + e^{b}) \cdot sin(a) = 2a \Rightarrow 2\cdot cosh(b) sin(a) = 2a \Rightarrow cosh(b)sin(a) = a.$

Knowing that $cos(a), sin(a) \in [-1, 1]$ for all $a$ , we now express:

$cos(a) = \frac{b}{sinh(b)} \Rightarrow -1 \le \frac{b}{sinh(b)} \le 1 \Rightarrow -sinh(b) \le b \le sinh(b);$

$sin(a) = \frac{a}{cosh(b)} \Rightarrow -1 \le \frac{a}{cosh(b)} \le 1 \Rightarrow -cosh(b) \le a \le cosh(b).$

The first inequality pair is satisfied iff $b \ge 0$ . For every non-negative value of $b$ that is substituted into the second inequality pair, there exists at least one value of $a \in [-cosh(b), cosh(b)]$ that satisfies $sin(a + bi) = a + bi$ other than the trivial solution $z = 0.$