= Solution
A constant <real part> $\operatorname{Re}\ln z=L$ means $\ln|z|=L$, so its <level sets> are the concentric <circles>
$$
\boxed{x^2+y^2=e^{2L},\qquad L\in\mathbb R.}
$$
Their radii are strictly positive. The <real part> is independent of the logarithm's branch, so there is no branch ambiguity in these <circles>.
For $\operatorname{Re}(1/z)=K$, the equation is $x=K(x^2+y^2)$. If $K\ne0$, complete the square to obtain
$$
\boxed{\left(x-\frac1{2K}\right)^2+y^2=\frac1{4K^2},\qquad (x,y)\ne(0,0).}
$$
These <circles> have centres on the real axis and are tangent to the imaginary axis at the excluded origin. Positive $K$ puts the <circle> to the right, negative $K$ to the left. The separate case $K=0$ gives the punctured imaginary axis $\boxed{x=0,\ y\ne0}$. The sketch uses an open marker at the origin to emphasize that $1/z$ is never defined there.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-1-complex-contours.png]
{title=Constant real-part contours of the complex logarithm and reciprocal}
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