= Solution
Use metric signature $(+,-,-,-)$ and the model's rescaled units $c=\hbar=1$. Its exponential implicitly uses dimensionless rescaled field and coupling combinations; the numerical coefficients of the potential are those supplied. Treat $\phi$ and $\phi^*$ as independent variables in taking canonical derivatives:
$$
\pi_\phi=\dot\phi^*,\qquad\pi_{\phi^*}=\dot\phi.
$$
The <Hamiltonian density> is
$$
\boxed{\mathcal H=|\dot\phi|^2+|\nabla\phi|^2+V(\phi)
=\pi_\phi\pi_{\phi^*}+|\nabla\phi|^2-|\phi|^2+e^{\lambda|\phi|^2}}.
$$
The kinetic and gradient terms are nonnegative, so a static homogeneous minimum is obtained by minimizing the potential. Put $s=|\phi|^2\geq0$. For $\lambda=0$, $V=1-s$ is unbounded below. For $\lambda<0$, the exponential decays while $-s$ tends to minus infinity. Thus \b[a stable vacuum requires $\boxed{\lambda>0}$].
For positive $\lambda$,
$$
\frac{dV}{ds}=-1+\lambda e^{\lambda s},\qquad
\frac{d^2V}{ds^2}=\lambda^2e^{\lambda s}>0.
$$
The <exponential complex-scalar potential> consequently has
$$
\boxed{\begin{cases}
|\phi|^2=-\ln\lambda/\lambda,&0<\lambda<1,\\
\phi=0,&\lambda\geq1.
\end{cases}}
$$
For $0<\lambda<1$, the minima form a circle $\phi=ve^{i\theta}$, with $v^2=-\ln\lambda/\lambda$ and $V_{\min}=(1+\ln\lambda)/\lambda$. For $\lambda\geq1$ the minimum has $V_{\min}=1$. The <symmetry> remains global <phase> rotation in either <phase>, but a chosen nonzero vacuum breaks it spontaneously. At $\lambda=1$ the origin is a unique minimum with vanishing quadratic curvature and a positive leading quartic term; this critical masslessness should not be confused with a Goldstone mode.
\Image[/past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2017-q3-scalar-vacua.png]
{title=Original radial potentials showing the continuous and explicitly twofold vacuum structures and their thresholds}
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