= Solution
For fixed terminal time and fixed endpoint, use the maximizing form of the <Pontryagin maximum principle>. For dynamics $\dot q=f(q,u,t)$ and running cost $L$, there are multipliers $(p,p_0)$, not all zero, with $p_0\ge0$, and <Hamiltonian of an optimal-control problem> $\mathscr H=p\cdot f-p_0L$. Along an optimum, $\dot q=\mathscr H_p$, $\dot p=-\mathscr H_q$, and the chosen control maximizes $\mathscr H$ over admissible controls almost everywhere. Fixed terminal state imposes no prescribed terminal value of $p$; fixed time imposes no free-time condition $\mathscr H(T)=0$. The normal case scales $p_0=1$; abnormal extremals have $p_0=0$.
For this problem the normal <Hamiltonian of an optimal-control problem> is
$$
\mathscr H=p_xu+p_yv+k(uy-vx)-\frac12(u^2+v^2),\qquad \dot k=0.
$$
Maximization gives $u=p_x+ky$, $v=p_y-kx$. The <costate> equations give $\dot p_x=kv$, $\dot p_y=-ku$, hence $\dot u=2kv$, $\dot v=-2ku$. Nonzero velocities rotate at constant angular rate, and rotational symmetry permits $u(0)=A$, $v(0)=0$. For a closed path, $2k=2\pi n$, with integer $n\ne0$. Then $z(1)=A^2/(2\pi n)$, so the required positive endpoint has $n>0$, $A^2=2\pi n$ and cost $\pi n$.
To prove a global minimum rather than only classify stationary extremals, write $r=(x,y)$ and subtract its time average; this changes neither its derivative nor $\int(y\dot x-x\dot y)dt$. The resulting closed mean-zero path satisfies the periodic <Wirtinger inequality> $\int|r|^2\le(4\pi^2)^{-1}\int|\dot r|^2$. This follows immediately by its Fourier expansion: each nonzero mode of the derivative has squared multiplier at least $4\pi^2$. The <Cauchy-Schwarz inequality> now gives
$$
1=|z(1)|\le\left(\int|r|^2\right)^{1/2}\left(\int|\dot r|^2\right)^{1/2}
\le\frac1{2\pi}\int_0^1(u^2+v^2)dt.
$$
The argument extends to square-integrable controls by the periodic Sobolev/Fourier interpretation. Thus the cost is at least $\pi$ for every feasible control. Equality is attained by
$$
\boxed{u=\sqrt{2\pi}\cos(2\pi t),\quad v=-\sqrt{2\pi}\sin(2\pi t),\quad
x=\frac{\sin(2\pi t)}{\sqrt{2\pi}},\quad y=\frac{\cos(2\pi t)-1}{\sqrt{2\pi}},\quad
z=t-\frac{\sin(2\pi t)}{2\pi}.}
$$
These satisfy every endpoint and have cost $\pi$. Therefore \b[the minimum is $\pi$], an <area-constrained quadratic control cost>. The global inequality also prevents an overlooked abnormal candidate from lowering it; in fact an abnormal finite maximizing <Hamiltonian of an optimal-control problem> would force both linear control coefficients to vanish and then force zero velocity, incompatible with $z(1)=1$.
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