If you haven't gone very far in algebraic geometry, then this example is probably off your radar. Suppose you have a "plane curve" in $\mathbb{P}^2$ given by $f(x,y)=0$. We can study it "analytically locally" (which is usually interpreted as locally in the analytic topology as opposed to Zariski topology).
Suppose we want to study a neighborhood of a closed point defined by $\frak{m}$. To get a Zariski neighborhood we just localize the ring at $\frak{m}$. But now we have a local ring and we could take the $\frak{m}$-adic completion. This "zooms in" to an analytic neighborhood of the point.
If the point is regular/smooth, then it will look "flat" since the ring will just be isomorphic to $k[[s,t]]$ (by Cohen structure theorem or probably something weaker). So analytic locally all smooth points of plane curves look the same (which we'd expect of course).
The interesting thing is with singularities. It would be odd to try to classify singularities Zariski locally. To see this draw (over $\mathbb{R}$) the curve $y^2=x^2$ which is just a pair of lines and draw $y^2=x^2-x^3$ the nodal curve. The singularity at $(0,0)$ "ought" to be the same because locally they are two lines intersecting even with the same slopes!!
By passing to the completion we remember this information. Try completing the rings at $(x,y)$ and doing a change of coordinates to show they are isomorphic. There is a vast literature out there on this topic so I won't go any further. The idea of classifying singularities analytic locally has had great success and the theory of just plane curve singularities is already interesting and should be approachable for you if you have a handle on completions.