Complex Variable Taylor Expansion at z=2i

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SUMMARY

The discussion centers on determining the order of the pole of the function \(\frac{e^{iz} - 1}{z^2 + 4}\) at \(z = 2i\). It is established that Taylor's theorem can be applied to complex variables, allowing for the analysis of the function's behavior near the pole. The conclusion is that \(z = 2i\) is a pole of order 1, as the modified function \(f(z)(z - 2i) = \frac{e^{iz} - 1}{z + 2i}\) is analytic at this point.

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  • Understanding of complex analysis concepts, specifically poles and analyticity.
  • Familiarity with Taylor's theorem and its application in complex variables.
  • Knowledge of exponential functions in the context of complex numbers.
  • Basic proficiency in manipulating complex functions and limits.
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  • Study the application of Taylor's theorem in complex analysis.
  • Learn about the classification of singularities in complex functions.
  • Explore the properties of exponential functions in the complex plane.
  • Investigate the methods for determining the order of poles in complex functions.
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Students and professionals in mathematics, particularly those focusing on complex analysis, as well as anyone interested in understanding the behavior of complex functions near singularities.

sachi
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I'm having trouble determining the order of the pole of

[exp(iz) - 1]/((z^2) + 4) at z=2i

I know I can't just expand the exponential as 1 + iz + [(iz)^2]/2 ...
because this formula only works near the origin. Can I still use Taylor's theorem to find the expansion at z=2i (i.e does Taylor's theorem still work for complex variables?)
thanks for your help.
 
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It's pretty obvious, isn't it, that f(z)(z-2i)= (eiz-1)/(z+2i) is analytic at z= 2i while f(z) itself is not. z= 2i is a pole of order 1.
 

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