Let $\gamma_R : [0, \pi/ 4] \ni \theta \mapsto R e^{i \theta}$. I want to show that $$ \lim_{R \rightarrow \infty} \int_{\gamma_R} e^{i z^2} dz = 0 $$ for $R > 1$. In order to use Jordan's lemma I want to show that $$ \lim_{R \rightarrow \infty} \max_{z \in \gamma_R} | e^{iz^2-iz} | = 0 $$ How can I do that ?
Further thoughts: On $\gamma_R$ we have \begin{align} e^{iz(z-1)} = \exp(iRe^{i\theta}(Re^{i\theta}-1)) = \exp(iR^2e^{2i\theta}-iRe^{i\theta}) \\ = \exp(iR^2(\cos 2\theta +i \sin 2\theta)-iRe^{i\theta}) \\ \leq \exp(-R^2 \sin 2\theta) \end{align} where we take the absolute value in the second line.
Can I just ignore the case $\theta = 0 ?$
Use the relation $\sin{2 \theta} \ge \frac{4 \theta}{\pi}$ when $\theta \in [0,\pi/4]$. Then
$$\int_{C_R} dz\, e^{i z^2} = i R \int_0^{\pi/4} d\theta \, e^{i \theta} e^{-R^2 \sin{2 \theta}} e^{i R^2 \cos{2 \theta}}$$
and
$$\left | \int_{C_R} dz\, e^{i z^2}\right | \le R \int_0^{\pi/4} d\theta \, e^{-R^2 \sin{2 \theta}} \le R \int_0^{\pi/4} d\theta \, e^{-4 R^2 \theta/\pi} \le \frac{\pi}{4 R}$$
as $R \to \infty$.