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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.

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    $\begingroup$ Generally, you don't have $\lvert \sin z\rvert \leqslant 1$ for $z \notin\mathbb{R}$. If you use Euler's formula for $\sin z$, you can get a bound for the integrand. $\endgroup$ Jun 17, 2014 at 21:52
  • $\begingroup$ $\sin (x+iy) = \frac{1}{2i}(e^{x+iy}-e^{-x-iy})$ $\endgroup$ Jun 17, 2014 at 22:07
  • $\begingroup$ @DanielFischer Euler's formula for $\sin(z)$ is $\sin(z)=z\prod_{n=1}^{\infty} (1-\dfrac{z^2}{n{\pi}^2})$. I am not so sure how to get a bound for $\prod_{n=1}^{\infty} (1-\dfrac{z^2}{n{\pi}^2})$, I was going to affirm $\prod_{n=1}^{\infty} (1-\dfrac{z^2}{n{\pi}^2})\leq \prod_{n=1}^{\infty} 1$, but this isn't always true for $z \in \mathbb C$ (it would be true for real numbers). Can I get some help with that expression? $\endgroup$
    – user156441
    Jun 17, 2014 at 23:36
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    $\begingroup$ @user156441 Euler had many formulae. I meant $$\sin z = \frac{1}{2i}\left(e^{iz} - e^{-iz}\right).$$ $\endgroup$ Jun 18, 2014 at 9:30
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    $\begingroup$ You need $\lvert e^w\rvert = e^{\operatorname{Re} w}$. With that, a simple application of the triangle inequality yields exactly the given bound. The inequality $\lvert \sin z\rvert \leqslant \sinh \lvert z\rvert$ yields a sharper bound. One can, however, also evaluate the integral exactly, as per Michael's comment, or by expanding $\sin z$ into its Taylor series. $\endgroup$ Jun 19, 2014 at 11:45

1 Answer 1

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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.

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