Define $f(z) = \frac{e^{iz}}{z} $ and given $R > 0$, let $γ_R$ is the upper semicircle of radius $R$ and center $0$.
Show that $\int_{γ_R} f(z)dz → 0$ as $R → ∞$.
(Hint: after a suitable parametrization $γ_R : [0, π] → \mathbb C$, divide the interval of integration in the interval [$\frac{1}{\sqrt{R}}$, $π$ − $\frac{1}{\sqrt{R}}$] and the rest).
I didn't use the hint because I didn't understand it. Though, I found a parametrization of $γ_R$: $γ_R(\theta) = cos(\theta) + i\sqrt{1-cos^2(\theta)}$ for $\theta \in [0, π]$
Here's my attempt without using the hint:
$\int_{γ_R} f(z)dz = \int_{γ_R} \frac{e^{iz}}{z} dz$ and thus $|\int_{γ_R} f(z)dz| = |\int_{γ_R} \frac{e^{iz}}{z} dz|$
Let $z=x+iy$, then $|\int_{γ_R} f(z)dz| = |\int_{γ_R} \frac{e^{iz}}{z} dz| ≤ \int_{γ_R} |\frac{e^{i(x+iy) }}{x+iy}|dx + idy = \int_{γ_R} \frac{|e^{i(x+iy)}|}{|x+iy|}dx+idy$.
But $|e^{i(x+iy)}| = e^{-y} ≤ 1$ for $y \geq 0$ (i.e in the upper half plane, specifically in the upper semicircle).
And $|x+iy| = \sqrt{x^2+y^2} ≤ R$ for $R$ big enough.
Hence, $|\int_{γ_R} f(z)dz| ≤ \int_{γ_R} \frac{|e^{i(x+iy)}|}{|x+iy|}dx+idy ≤ \int_{γ_R} \frac{1}{\sqrt{R}} dx+idy = \int_{γ_R} \frac{1}{\sqrt{R}} dx + i\int_{γ_R} \frac{1}{\sqrt{R}} dy$
$-(\int_{γ_R} \frac{1}{\sqrt{R}} dx + i\int_{γ_R} \frac{1}{\sqrt{R}} dy) ≤ \int_{γ_R} f(z)dz ≤ \int_{γ_R} \frac{1}{\sqrt{R}} dx + i\int_{γ_R} \frac{1}{\sqrt{R}} dy$
Now taking lim as $R → ∞$, we deduce by the sandwich theorem that $\int_{γ_R} f(z)dz = 0$
Is my attempt correct?
Any other answers? And what about the hint?
Your answer has a few mistakes in it.
$1)$ When you say $$\int_{\gamma_R}\frac{e^{iz}}{z}dz\leq\int_{\gamma_R}\left| \frac{e^{i(x+iy)}}{x+iy}\right|dx+idy$$
You seem to be trying to use the fact that, in general: $$\left|\int_C f(z)\mathrm{d}z\right|\leq \int_C |f(z)||\mathrm{d}z|$$
Notice the $|\mathrm{d}z|$ term above - it is an important part of the inequality. It cannot simply be replaced with $\mathrm{d}z=\mathrm{d}x+i\mathrm{d}y$.
In your question, this lack of care leads to: $$|\int_{γ_R} f(z)dz| ≤ \int_{γ_R} \frac{|e^{i(x+iy)}|}{|x+iy|}dx+idy ≤ \int_{γ_R} \frac{1}{\sqrt{R}} dx+idy = \int_{γ_R} \frac{1}{\sqrt{R}} dx + i\int_{γ_R} \frac{1}{\sqrt{R}} dy$$
which is an inequality between a real number and a complex one.
In more extreme cases, the misuse of the inequality can be truly misleading.
For instance, consider $\int_{S_1} z^*\mathrm{d}z$ ($S_1$ denotes the complex unit circle at the origin). This integral equals $$\int_{S_1} (x-iy)(\mathrm{d}x+i\mathrm{d}y)=\int_{S_1} (x\mathrm{d}x+y\mathrm{d}y)+i\int_{S_1} (x\mathrm{d}y-y\mathrm{d}x)$$ and by Green's theorem, this evaluates to $2i$, so $\left|\int_{S_1} z^*\mathrm{d}z\right|=|2i|=2$.
But $\int_{S_1}\left|z^*\right|\mathrm{d}z=\int_{S_1} 1\mathrm{d}z=0$. So, if we applied $\left|\int_C f(z)\mathrm{d}z\right|\leq \int_C |f(z)|\mathrm{d}z$ in this case (where $f(z)=z^*$), we'd get $2\leq 0$ (this is just to emphasize the point).
$2)$ That $\sqrt{x^2+y^2}\leq R$ when $R$ is big enough is true. But you use this to derive (in combination with $\left|e^{i(x+iy)}\right|$) that: $$\int_{γ_R}\frac{|e^{i(x+iy)}|}{|x+iy|}dx+idy ≤ \int_{γ_R} \frac{1}{\sqrt{R}} dx+idy$$
This is faulty, most immediately, because it is an inequality involving a complex number (as mentioned before), but also because $\sqrt{x^2+y^2}\leq R\implies \frac{1}{\sqrt{x^2+y^2}}\geq \frac{1}{R}$ not $\frac{1}{\sqrt{x^2+y^2}}\leq \frac{1}{R}$.
$3)$ Finally, you give the parametrization of $\gamma_R$ as: $$\gamma_R(\theta)=cos(\theta)+i\sqrt{1-cos^2(\theta)},\quad\theta\in [0,\pi]$$ but it should actually be $$\gamma_R(\theta)=Rcos(\theta)+iR\sqrt{1-cos^2(\theta)},\quad\theta\in [0,\pi]$$ where the extra factor of $R$ accounts for the radius of the semicircle (otherwise the parametrization would always just trace out a unit circle).
I bring this up, even though you did not use the parametrization in your answer, because I will in mine (following what the "Hint" says).
A more convenient way of writing the transformation is $$\gamma_R(\theta)=Re^{i\theta},\quad \theta\in[0,\pi]$$
It is easy to see that $\mathrm{d}z=Rie^{i\theta}\mathrm{d}\theta$
So the integral becomes: $$\int_0^{\pi}\frac{e^{i\left(Re^{i\theta}\right)}}{Re^{i\theta}}\times\left(Rie^{i\theta}\mathrm{d}\theta\right)=i\int_0^{\pi}e^{i\left(Re^{i\theta}\right)}\mathrm{d}\theta=i\int_0^{\pi}e^{Ri\mathrm{cos}(\theta)-R\mathrm{sin}(\theta)}\mathrm{d}\theta$$
So (using the estimation lemma): $$\left|\int_{\gamma_R}\frac{e^{iz}}{z}\mathrm{d}z\right|=\left|i\int_0^{\pi}e^{Ri\mathrm{cos}(\theta)-R\mathrm{sin}(\theta)}\mathrm{d}\theta\right|\leq \int_0^{\pi}e^{-R\mathrm{sin}(\theta)}\mathrm{d}\theta$$
Now, using the hint, we break up the above integral: $$=\int_0^{\frac{1}{\sqrt R}}e^{-R\mathrm{sin}(\theta)}\mathrm{d}\theta+\int_{\frac{1}{\sqrt R}}^{\pi-\frac{1}{\sqrt R}}e^{-R\mathrm{sin}(\theta)}\mathrm{d}\theta+\int_{\pi-\frac{1}{\sqrt R}}^{\pi}e^{-R\mathrm{sin}(\theta)}\mathrm{d}\theta$$
It is clear that $\left|e^{-R\mathrm{sin}(\theta)}\right|\leq 1$ for all $\theta\in [0,\pi]$, so the left and right integrals are bounded above by $\frac{1}{\sqrt R}$, giving us $$\leq \frac{2}{\sqrt R}+\int_{\frac{1}{\sqrt R}}^{\pi-\frac{1}{\sqrt R}}e^{-R\mathrm{sin}(\theta)}\mathrm{d}\theta$$
Lastly, $\mathrm{sin}(\theta)\geq \frac{\theta}{2}$ whenever $\theta\geq 0$ and so long as $\theta$ is small enough. So, as long as $R$ is big enough, $\mathrm{sin}(\theta)\geq \mathrm{sin}(\frac{1}{\sqrt R})\geq \frac{1}{2\sqrt R}$ for all $\theta\in [\frac{1}{\sqrt R},\pi-\frac{1}{\sqrt R}]$, giving us a final inequality: $$\leq \frac{2}{\sqrt R}+\int_{\frac{1}{\sqrt R}}^{\pi-\frac{1}{\sqrt R}}e^{-R\left(\frac{1}{2\sqrt R}\right)}\mathrm{d}\theta\leq\frac{2}{\sqrt R}+\pi e^{-\frac{\sqrt R}{2}}$$
Remember that all of this is an upper bound on the original integral's magnitude, $\left|\int_{\gamma_R}\frac{e^{iz}}{z}\mathrm{d}z\right|$.
Then, since both $\frac{2}{\sqrt R}$ and $\pi e^{-\frac{\sqrt R}{2}}$ tend to $0$ as $R\rightarrow \infty$, the original integral too must approach $0$ as $R\rightarrow \infty$.