With the metric of the torus given by $ds^2=a^2d\theta^2+(b+asin\theta)^2d\phi^2$, I'm asked to find the conformal transformation which proves that a torus is conformally equivalent to a plane.
I must then find a transformation $\bar{g}_{\mu \nu}=exp(2\Phi)g_{\mu \nu}$ (as far as I know, the exponential is there just to make sure the thing is invertible, but that's the notation my professor is using) such that $\bar{R}_{\lambda \mu \nu \rho}=0$, and the new coordinates satisfy $d\bar s ^2=d\bar\theta^2 + d\bar\phi^2$ (hence rendering the metric equivalent to that of a plane).
I don't know what approach to use. I tried taking the curvature, which I calculated, and finding a variable change that would make $R_{\theta \phi \theta \phi}=\frac{b}{a}sin\theta +sin^2\theta=0$, the only non-null element of the curvature tensor. However, I saw myself wasting a lot of time there and getting me nowhere.
Then I tried using the line element $ds^2=a^2d\theta^2+(b+asin\theta)^2d\phi^2$ and try to fit a transformation to $\bar\theta$ and $\bar\phi$ such that $ds^2$ remained multiplied by some function $\Omega(\theta,\phi)$, so that $d\bar s^2=\Omega(\theta, \phi)ds^2$, which would be consistent with the "locally dilation of the metric"... But I wouldn't know how to find $\Phi$ from there.
Any help on how to work this out will be much appreciated!
HINT: Note that $$d\theta^2 + f(\theta)^2\,d\phi^2 = f(\theta)^2\left(\left(\frac{d\theta}{f(\theta)}\right)^2 + d\phi^2\right).$$