$|\int_{0}^{a+bi}\cos(z^2)\,dz|\leq(a^2+b^2)^{1/2}\sinh(2ab)/(2ab)$ Assuming that $a>0$ and $b>0$, derive the estimate
$$\left|\int_{0}^{a+bi}\cos(z^2)\,dz\right|\leq \frac{(a^2+b^2)^{1/2}\sinh(2ab)}{2ab}$$
I know $|\int_{\gamma}f(z)\,dz|\leq \int_{\gamma}|f(z)|\,dz$, but I do not know how to apply it here, I have tried the following:
$$\left|\int_{0}^{a+bi}\cos\left(z^2\right)\,dz\right|\leq \int_{0}^{a+bi}\left|\frac{e^{iz^2}+e^{-iz^2}}{2}\right|\,dz,$$
but I do not know what else to do.  Could anyone help me with this please? Thank you very much.

First note that \begin{align}
\int_0^{a+bi} \cos (z^2) dz&=\int_0^1\cos((a+bi)^2t^2)\cdot (a+bi)dt\quad (z=(a+bi)t,\, 0\le t\le 1)\\
&=(a+bi)\int_0^1\cos((a^2-b^2)t^2+2abt^2i)dt.
\end{align}
Using the inequality \begin{align}
|\cos (\alpha +\beta i)|&=\left|\frac{\exp(i(\alpha +\beta i))+\exp(-i(\alpha +\beta i))}{2}\right|\le \frac{|\exp (i\alpha -\beta )|+|\exp(-i\alpha +\beta )|}{2}\\
&=\frac{\exp(-\beta )+\exp(\beta )}{2},
\end{align}
we have \begin{align}
\left|\int_0^{a+bi} \cos (z^2) dz\right|&\le \sqrt{a^2+b^2}\int_0^1\left|\cos((a^2-b^2)t^2+2abt^2i)\right|dt\\
&\le \frac{\sqrt{a^2+b^2}}{2}\int_0^1 \left(\exp(2abt^2)+\exp(-2abt^2)\right)dt. 
\end{align}
Now we use the inequality $$\exp(2abt^2)+\exp(-2abt^2)\le \exp(2abt)+\exp(-2abt)\quad (0\le t\le 1).$$
This follows the fact that $f(x)=e^x+e^{-x}$ is increasing for $x>0$.
Thus we have \begin{align}
\left|\int_0^{a+bi} \cos (z^2) dz\right|&\le \frac{\sqrt{a^2+b^2}}{2}\int_0^1 \left(\exp(2abt)+\exp(-2abt)\right)dt\\
&=\frac{\sqrt{a^2+b^2}}{2}\cdot \frac{\exp(2ab)-\exp(-2ab)}{2ab}.
\end{align}
