Why is $i$ a removable singularity of $f(z)=\frac{\sin(z-i)}{z^2+1}$? We can find the Taylor expansions of the $\sin(z-i)$ and $\frac{1}{z+i}$ to get the Laurent series (actually Taylor series since $i$ is removable) of $f(z)$ for $0<|z-i|<2$, but is there another way to show that $f$ is removable?
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2$$f(z) = \frac{\sin(z-i)}{z^{2}+1} = \frac{\sin(z-i)}{z-i} \cdot \frac{1}{z+i}$$ and use that $$\lim_{x \to 0} \frac{\sin(x)}{x} = 1$$ – Matthew Cassell Nov 15 '18 at 07:17
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@Mattos I use Complex Variables and Applications by Brown and Churchill, and it seems that they don't introduce this method. What is this? – user398843 Nov 15 '18 at 07:19
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What method? I just used some algebraic manipulation and applied a very well known limit. – Matthew Cassell Nov 15 '18 at 07:21
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@Mattos "A discontinuity of a function is removable if the limit of the function exists at that point" stated by Carmeister in the answer. Brown and Churchill might state it in a different way, but I don't know where do the place this statement in the book... – user398843 Nov 15 '18 at 07:27
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In general, a discontinuity of a function is removable if the limit of the function exists at that point. So any method of showing that $$\lim_{z\to i}\frac{\sin(z-i)}{z^2+1}$$ exists will do the trick.
As Mattos notes in a comment, one way to do so is write $z^2+1=(z+i)(z-i)$ and then use the fact that $\lim_{z\to 0}\frac{\sin z}{z}=1$.
Carmeister
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Use the fact that$$\lim_{z\to i}\frac{\sin(z-i)}{z^2+1}=\lim_{z\to i}\frac{\sin(z-i)}{z-i}\times\frac1{z+i}=\frac1{2i}$$and Riemann's theorem on removable singularities.
José Carlos Santos
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