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If $f_1,f_2$ are meromorphic functions in $D$ and there exists a sequence of pairwise distinct points $z_n \in D$ such that $z_n \to z_o \in D$ and $f_1(z_n)=f_2(z_n),$ then $f_{1} \equiv f_2$ on $D.$

Appreciate if someone could advise me on how to prove this. Thank you.

Alexy Vincenzo
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  • Consider $g(z) = f_1(z) - f_2(z)$. Can $z_0$ be a pole of $g$? – Daniel Fischer Aug 21 '14 at 18:40
  • @DanielFischer: Suppose $g$ has pole of order $m$ at $z_o.$ Then $g(z)=\dfrac{k(z)}{(z-z_o)^m},$ for some $k$ analytic at $z_0$ and $k(z_o) \neq 0.$ Since $g(z_k)=0, \ k(z_k)=0, \forall k.$ Since $k$ is continuous at $z_o, \ k(z_o)=0 \ $ (Contradiction) Since $f_1, \ f_2$ have poles of some order, $z_o$ is not essential singularity of $g. \ $ Hence, $z_o$ is removable singularity of $g$ and $g$ is now holomorphic over $D.$ Hence, $f_1 \equiv f_2$ over $D \ ?$ – Alexy Vincenzo Aug 21 '14 at 20:17
  • Not quite, we need a step in between. So far, we only know that $g$ is holomorphic in $D\setminus P$, where $P$ is a closed discrete subset of $D$ that doesn't contain $z_0$. ($P$ is the union of the pole sets of $f_1$ and $f_2$, minus ${z_0}$). And from $g(z_k) = 0$ for all $k$ we conclude that $g\equiv 0$ on $D\setminus P$ by the identity theorem for holomorphic functions. Now we remove the removable singularities of $g$ in the points of $P$, and then we have $g\equiv 0$ on $D$. And hence $f_1 = f_2 + g = f_2 + 0 = f_2$. – Daniel Fischer Aug 21 '14 at 20:38
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    What happens if $f_{1}$ has an essential singularity? – Diego Fonseca May 09 '16 at 18:05

1 Answers1

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Since $f_1$ and $f_2$ are meromorphic in $D$ then there exists a holomorphic functions $a,b,c,d$ in $D$ such that $f_1=a/b$ and $f_2=c/d$, now let $h=ad-bc$ it is a holomorphic function on $D$, and for all $z_k$ we have $h(z_k)=0$, by continuity of $h$ and the fact that $z_k\to z_0$, we get $h(z_0)=0$, hence $h=0$, so that $a/b=c/d$ i.e $f_1=f_2$.

Hamou
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  • there exists $k$ such that $f_1(z) (z-z_0)^k$ and $f_2(z) (z-z_0)^k$ are holomorphic, and as you wrote $f_1(z) (z-z_0)^k - f_2(z) (z-z_0)^k = 0$ hence $f_1 = f_2$ – reuns May 11 '16 at 04:10