Follow that dot

By the arc length formula for a parametrically-defined curve, we have $\begin{aligned} L_a &= \int_0^a \sqrt{(x'(t))^2+(y'(t))^2}dt \\ &=\int_0^a \sqrt{\left( -5e^{-t}\cos t-5e^{-t} \sin t \right)^2 + \left( -5e^{-t} \sin t +5e^{-t}\cos t \right)^2}dt \\ &=\int_0^a 5e^{-t} \sqrt{\left(\cos t- \sin t \right)^2 + \left(\sin t +\cos t \right)^2}dt \\ &=\int_0^a 5e^{-t} \sqrt{\cos^2 t-2\sin t \cos t+\sin^2 t + \sin^2 t + 2\sin t \cos t + \cos^2 t}dt \\ &=\int_0^a 5e^{-t} \sqrt{2\sin^2 t + 2\cos^2 t}dt \\ &=5\sqrt{2} \int_0^a e^{-t}dt = -5\sqrt{2} \left[ e^{-t} \right]_0^a =-5\sqrt{2} \, e^{-a} +5\sqrt{2}. \end{aligned}$

Now we take the limit: $z=\lim_{a \to \infty} L_a = \lim_{a \to \infty} \left( -5 \, \sqrt{2}e^{-a} +5\sqrt{2} \right)=5\sqrt{2}.$ So, $z^2=\boxed{50}$ .