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Prove that if $f$ has an essential singularity at $z_0$ and $g$ has a pole at $z_0$, then $f+g$ has an essential singularity at $z_0$

I know that if $z_0$ is an essential singularity then |f(z)| is neither bounded near $z_0$ nor goes to infinity as $z \to z_0$ and vice versa. But evaluating $|f+g|$ as $z \to z_0$ we see $|f+g| \leq |f|+|g|$ which clearly is unbounded (since both $f$ and $g$ are unbounded) but it is not clear that $|f+g|$ does not approach infinity. Any ideas?

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    $\begingroup$ Look at the Laurent series? $\endgroup$ – Jacky Chong Oct 17 '16 at 17:05
  • $\begingroup$ @JackyChong I get $$\sum_{j=k}^{\infty} a_j(z-z_0)^j + \sum_{j=-\infty}^{\infty} b_j(z-z_0)^j = \sum_{j=-\infty}^{\infty} b_j(z-z_0)^j+u_{k}a_j(z-z_0)^j$$ for some negative integer $k$ and where $u$ is the Heaviside-function. Would this make sense? $\endgroup$ – Lozansky Oct 17 '16 at 17:19
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By definition, a function $f$ has an essential singularity in $z_0$ if its Laurent's expansion in $z_0$ has infinite non-null negative terms, while the function $g$ has a pole, so its Laurent expansion has a finite number of non-null negative terms in $z_0$. It's now clear that the Laurent expansion of the sum has infinite non-null negative terms. (When I say "negative terms" I mean terms with negative exponent).

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