= Solution
Use a radial <geometric Kähler potential>. Write $s=|z_1|^2+|z_2|^2$ and $y(s)=\varphi'(s)$. The coefficient <Hermitian matrix> of $i\partial\bar\partial\varphi(s)$ is
$$
G_{j\bar k}=y\delta_{jk}+y'\bar z_jz_k.
$$
Its <eigenvalues> are $y$ in the complex tangential direction and $y+sy'$ in the complex radial direction. A <Kähler metric> is positive precisely when both are positive. The standard form has coefficient <matrix> $\frac12 I$, so equality of the <Riemannian volume forms> is the condition
$$
y(y+sy')=\frac14.
$$
Indeed, the volume form is $\omega^2/2$ and its coefficient is proportional to $\det G$. Multiplying the determinant equation by $2s$ gives $(s^2y^2)'=s/2$, so we can take
$$
y(s)=\frac{\sqrt{s^2+a^2}}{2s}\qquad(a>0).
$$
An explicit antiderivative is
$$
\boxed{\varphi_a(s)=\frac12\left[\sqrt{s^2+a^2}+a\log\frac{s}{\sqrt{s^2+a^2}+a}\right],\qquad
\omega'=i\partial\bar\partial\varphi_a(s).}
$$
It is smooth for $s>0$. Its two <eigenvalues> are
$$
\lambda_{\mathrm{tan}}=\frac{\sqrt{s^2+a^2}}{2s}>0,
\qquad
\lambda_{\mathrm{rad}}=\frac{s}{2\sqrt{s^2+a^2}}>0,
\qquad \lambda_{\mathrm{tan}}\lambda_{\mathrm{rad}}=\frac14.
$$
The form is real and closed because it is $i\partial\bar\partial$ of a real function, so it is a <Kähler metric> on the punctured space. Since $a>0$, neither <eigenvalue> equals $1/2$, and the metric differs from the Euclidean one. \b[The radial and tangential stretches compensate exactly, preserving the volume form:]
$$
\boxed{\frac{(\omega')^2}{2}=\frac{\omega^2}{2},\qquad \omega'\ne\omega.}
$$
This is a <radial Kähler metric with Euclidean volume in complex dimension two>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/iii/paper-118-radial-metric.png]
{title=Radial and tangential metric eigenvalues relative to the Euclidean metric, with unchanged volume}
Back to article page
