What is the Nature of Singularity in the Function f(x)=exp(-1/z)?

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SUMMARY

The function f(x)=exp(-1/z) exhibits an essential singularity at z=0 when analyzed through series expansion, yielding the series f(z)=1-1/z+1/2!(z^2)-..., which confirms the nature of the singularity. However, when considering limits from both sides of zero, the function appears to exist finitely, suggesting it may not be a singular point. This duality in results highlights the complexity of singularities in complex analysis, particularly for functions defined over complex numbers.

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ion santra
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what is the nature of singularity of the function f(x)=exp(-1/z) where z is a complex number?
now i arrive at two different results by progressing in two different ways.
1) if we expand the series f(z)=1-1/z+1/2!(z^2)-... then i can say that z=0 is an essential singularity.
2) now again if i take the limit of the function from either sides of zero, i see it exists finitely, therefore not a singular point at all.
 
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It is an essential singularity.

ion santra said:
2) now again if i take the limit of the function from either sides of zero, i see it exists finitely, therefore not a singular point at all.

Can you expand on this? Don't forget that ##z## is a complex number, so just taking the two sided limits in ##\mathbb{R}## is not good enough.
 
ion santra said:
what is the nature of singularity of the function f(x)=exp(-1/z) where z is a complex number?
now i arrive at two different results by progressing in two different ways.
1) if we expand the series f(z)=1-1/z+1/2!(z^2)-... then i can say that z=0 is an essential singularity.
2) now again if i take the limit of the function from either sides of zero, i see it exists finitely, therefore not a singular point at all.
For Real z < 0, the limit as z ->0 is infinite.
 

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