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Find all singular points of f(z), classify them and inspect the behaviour at infinity

$f(z)=sin(1/z)+1/z^2$

Here i'm stuck to find singularities since z=0'd be a singularity however we have $+sin(1/z)$ so i guess z has to be a value to make whole function singular. Then i realized i may be rewrite this function as a series expansion but i failed. So i need some suggestions or guidance.

Aegean
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1 Answers1

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You've got the right idea.

Note that since you need to inspect the behavior at infinity, you'll want to consider the function $$g(z):=f\left(\frac1z\right)=\sin(z)+z^2.$$

I would start by finding the series expansion for $g$ about $0,$ then noting that $f(z)=g(1/z).$ That will help you show that $z=0$ is the only singularity of $f,$ as well as to classify it.

Cameron Buie
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  • i tried this. But i stuck somewhere so i tried taylor expansion for sin(1/z) and found z=0 is singularity. However, this method didn't look good to me. But i'll retry. Thanks – Aegean Jan 17 '21 at 16:39
  • If you get stuck finding the Taylor expansion, or just want to check that you've done it correctly, let me know! – Cameron Buie Jan 17 '21 at 16:40
  • i just expand as sin(1/z)=1/z-(1/z)^3(1/3!)+(1/z)^5(1/5!)+... then the function became f(z)=1/z-(1/z)^3(1/3!)+(1/z)^5(1/5!)+...+1/z^2 so clearly it seems z=0 is the only singular point. At this point i have no idea about the behaviour at infinity – Aegean Jan 17 '21 at 16:48
  • That's the idea, yes. Can you tell what kind of singularity it is from that, since there are infinitely-many singular terms? – Cameron Buie Jan 17 '21 at 16:49
  • it looked like many poles but i had no idea about its behaviour – Aegean Jan 17 '21 at 16:53
  • or maybe it acts like f(z)=e^(1/z) so it'd be an essential singularity – Aegean Jan 17 '21 at 16:55
  • In order for it to be a pole, there would have to be only finitely-many singular terms, so it's an essential singularity. – Cameron Buie Jan 17 '21 at 16:59
  • okay i understand now. Thanks a lot. What should i say for "behaviour"? I'm not expecting the answer but i didn't understand exactly what the question expects me to answer – Aegean Jan 17 '21 at 17:08
  • The "behavior at infinity" of $f(z)$ is the limit of $g(z)$ as $z\to 0,$ if it exists. See this related question for more. – Cameron Buie Jan 17 '21 at 17:10
  • ah okay it's now crystal clear. Thanks for all your helps. – Aegean Jan 17 '21 at 17:12
  • You're welcome! – Cameron Buie Jan 17 '21 at 17:14