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I have studied the following version of the Great Picard theorem:

Suppose $f(z)$ is meromorphic on a punctured neighborhood $\left\{0<\left|z-z_{0}\right|<\delta\right\}$ of $z_{0}$. If $f(z)$ omits three values at $z_{0}$, then $f(z)$ extends to be meromorphic at $z_{0}$.

So suppose meromorphic function has essential singularity at some point then image of any neighbourhood of that singularity can possibly take all complex plane except two points.

From Wikipedia: Great Picard's Theorem: If an analytic function $f$ has an essential singularity at a point $w$, then on any punctured neighborhood of $w, f(z)$ takes on all possible complex values, with at most a single exception, infinitely often.

I know that the first version is for meromorphic function, and another one is for analytic function. Using the definition of meromorphic function as it is written as ratio of two analytic function, is there any way to conclude result of Picard theorem for analytic function from that of meromorphic one.

Thank you so much for investing your valuable time.

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Suppose that $f$ is analytic in a punctured neighborhood $\left\{0<\left|z-z_{0}\right|<\delta\right\}$ of $z_0$ and omits two distinct values $a, b \in \Bbb C$.

As a meromorphic function, $f$ omits the three values $a, b, \infty$, and it follows from the Great Picard theorem for meromorphic function that $f$ extends meromorphically at $z_0$.

This means that $f$ has a removable singularity or a pole at $z_0$, but not an essential singularity.

Remark: The versions of the Great Picard theorem for analytic and meromorphic functions are in fact equivalent. If an analytic function omits two finite values then $\infty$ is the third omitted value. Conversely, if a meromorphic function $f$ omits three distinct values $a, b, c$ then $1/(f-a)$ is analytic and omits the two values $1/(b-a)$ and $1/(c-a)$.

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  • $\begingroup$ Thank you so much Sir $\endgroup$ Commented Nov 9, 2021 at 10:13

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