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Let be given in $\mathbb{C}\setminus\{0\}$ the functions : $\displaystyle{\frac{1}{\sin(z)^{2}}; \frac{\tan(z)}{z^{2}}, \frac{1}{e^{\frac{1}{z^{2}}}}, \frac{z^{3}}{1-\cos(z)}} $ and the book says to evaluate the type of singularity at $z_{0}=0$ without the usage of Laurent series.

My solutions are as follows:

$1.$ since $\lim \limits_{z \rightarrow z_{0}} |f(z)| \rightarrow \infty$ this must be a pole. I guess it is order 2 , but how to proove that?

$2.$ since $\lim\limits_{z\rightarrow z_{0}} |\frac{\tan(z)}{z^{2}}| \rightarrow \infty$ so this must be a pole, again i dont know what the order is.. but I guess it is 2 .

$3.$ since $\lim\limits_{z\rightarrow z_{0}} \left|\frac{1}{e^{1/z^2}}\right| \rightarrow 0$ this exists in $\mathbb{R}$ so this must be a removable singularity.

$4.$ using de lhopital 4 times gives : $\lim |\frac{6}{-\sin(z)}| \rightarrow \infty$ so this is also a polish singularity.

Again I don't know how to determine the order of the polish ones.

Please tell me.

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2 Answers 2

Can you find a value of $n$ such that $z^{n-1}f(z)$ is singular at zero but $\lim_{z\to0}z^nf(z)$ exists? Does the answer to that question tell you about the nature of the singularity?

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De Lhopitals rule??? –  VVV Nov 3 '11 at 23:15
2  
Try it and see! –  Gerry Myerson Nov 3 '11 at 23:16

So :

a) because $lim _{z \rightarrow 0} \frac{1}{sin(z)^{2}}$ goes to infinity, it is a pole. of order 2.

c) is an essential singularity, because its not a pole but also not removable

b) is removable, set for f(0) = 1

d) lhoptial gives that it is a pole

I dont think this is right! Whats the right way?

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What makes you think $f(0)=1$ removes the singularity in b)? In c), have you tried evaluating the limit as $z\to0$? In d), can you show your work? –  Gerry Myerson Nov 4 '11 at 5:45
    
I have edited my question. Is it correct? –  VVV Nov 6 '11 at 18:36
    
Why do you think there is any kind of singularity in c)? By what power of $z$ do you have to multiply in b) to get a finite limit? how does using l'Hopital 4 times on d) give you $6/(-\sin x)$? –  Gerry Myerson Nov 7 '11 at 0:35

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