On an upper semicircle, $e^{iaz}$ supplies exponential decay, bounding the arc integral by $\pi M_R/a$.Suggest a slogan
Let \(a>0\) and \(R>0\), let \(C_R\) be the semicircle \(t\mapsto Re^{it}\), \(t\in[0,\pi]\), and let \(g\) be continuous on \(C_R\). For \(f(z)=e^{iaz}g(z)\), \[ \abs{\int_{C_R}f(z)\dz}\le\frac{\pi M_R}{a},\qquad M_R\coloneqq\max_{t\in[0,\pi]}\abs{g(Re^{it})}. \]
For \(z=Re^{it}\), \(\abs{e^{iaz}}=e^{-aR\sin t}\) and \(\abs{\dz}=R\dt\), so \[ \abs{\int_{C_R}f(z)\dz}\le M_R\int_0^\pi e^{-aR\sin t}R\dt=2M_R\int_0^{\pi/2}e^{-aR\sin t}R\dt. \] On \([0,\pi/2]\) concavity of \(\sin\) gives \(\sin t\ge\frac{2t}{\pi}\), hence \[ 2M_R\int_0^{\pi/2}e^{-aR\sin t}R\dt\le2M_R\int_0^{\pi/2}e^{-2aRt/\pi}R\dt=\frac{\pi M_R}{a}\big(1-e^{-aR}\big)\le\frac{\pi M_R}{a}. \]