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Let $f$ be a complex function analytic in a domain $D$. Let $γ_1$ and $γ_2$ be two rectifiable curves that are in $D$ except for their final point $P$ which is in $∂D$. The final point is common to both curves. Also, both curves have are tangent to each other at $P$: $$\left.{\frac{∂γ_1}{∂ z}}\right\vert_P = \left.{\frac{∂γ_2}{∂ z}}\right\vert_P .$$ If the values of $lim_{z→P} f(z)$ are different (at least one of them is finite) when the approach is done through each curve, then the singularity must be essential, right? Are there any other singularities that might fit those premises?

SolutionExists
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    To clarify "different": do you mean both limits exist (in particular, are not $\infty$)? – Greg Martin Jun 29 '21 at 02:24
  • @GregMartin See edit. – SolutionExists Jun 29 '21 at 02:30
  • The definition of "essential singularity" usually assumes that the function is analytic in a neighborhood of the point in question except at the point itself. In this case, the hypotheses quickly imply that the singularity can't be removable (since they're different) and can't be a pole (since one of them is finite). – Greg Martin Jun 29 '21 at 02:33
  • @GregMartin Are logarithmic singularities essential? – SolutionExists Jun 29 '21 at 04:20
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    Perhaps strangely, no! See this question and its answer, for example. One can also see that it's not an essential singularity because its values near the singularity are confined to the left half-plane (say), which violates the "Great Picard's Theorem". – Greg Martin Jun 29 '21 at 07:16
  • @GregMartin 1) I didn't realize that isolated singularities excluded multi-valued functions even though I knew the definition. 2) Are you sure that $\log$ does not take on every complex value? I might be wrong but $\log\exp z = z$ works for every $z \in \mathbb{C} \setminus {0}$. – SolutionExists Jun 29 '21 at 18:24
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    The theorem says that a function takes on every value (with at most one exception) in every neighborhood of an essential singularity, which is much stronger than just being globally surjective. – Greg Martin Jun 29 '21 at 18:52
  • @GregMartin My bad. I meant to imply that by going around the different branches you can get any argument you want in the neighborhood of zero. What I now realize is that there is no direction to approach zero and get the log to map to a finite value. Thank you. You basically answered the original question too. – SolutionExists Jun 29 '21 at 19:26

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