Statement Let $\gamma$ be the curve that goes through the upper unit circle counterclockwise (positive orientation). Prove that $$\left|\int_{\gamma} \dfrac{\sin(z)}{z^2}dz\right|\leq \pi\dfrac{1+e}{2}$$
The attempt at a solution
What I did is:$$\left|\int_{\gamma} \dfrac{\sin(z)}{z^2}dz\right|=\left|\int_0^{\pi} \dfrac{\sin(e^{i\theta})ie^{i\theta}}{e^{i2\theta}}d\theta\right|$$
The right member equals to $$\left|\int_0^{\pi} \dfrac{\sin(e^{i\theta})ie^{i\theta}}{e^{i2\theta}}d\theta\right|=\left|\int_0^{\pi} \sin(e^{i\theta})e^{-i\theta}d\theta\right|$$
Now, $$\left|\int_0^{\pi} \sin(e^{i\theta})e^{-i\theta}d\theta\right|\leq \int_0^{\pi}\left|\sin(e^{i\theta})\right|d\theta$$
For real numbers, I know that $|\sin(x)|\leq 1$, but I don't know if this inequality holds for complex numbers as well. If that was the case, then I could say that the last member is less than or equal to $$\int_0^{\pi} d\theta=\pi < \pi\dfrac{1+e}{2}$$ and then the original statement would be proved. If $|\sin(z)| \not \leq 1$ for all $z \in \mathbb C$ (or at least in the unit circle), then I don't know what to do next. I would appreciate some help.
From Euler's formula
$$\sin z = \frac{1}{2i}\left(e^{iz} - e^{-iz}\right)$$
and $\lvert e^{\pm iz}\rvert = e^{\pm \operatorname{Re} (iz)} = e^{\mp \operatorname{Im} z}$, we obtain the estimate
$$\lvert \sin (x+iy)\rvert \leqslant \frac{1}{2}\left(e^{-y} + e^y\right).$$
On the unit circle, we have $\lvert y\rvert \leqslant 1$, so
$$\lvert \sin z\rvert \leqslant \frac{1+e}{2}$$
there, which gives exactly the suggested bound.