MAT334--2020F > Quiz 5

LEC0101 Quiz#5 oneC

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Xun Zheng:
Locate each of the isolated singularities of the given function and tell whether it is a removable singularity, a pole, or an essential singularity (case (3)). If the singularity is removable, give the value of the function at the point; if the singularity is a pole, give the order of the pole.
$$\frac{e^z-1}{z}$$
Here is my answer:

* First, we get that z=0 is a singularity.
* Next, we have  $$\lim_{z\to 0}\frac{e^z-1}{z}=\lim_{z\to 0}\frac{e^z}{1} = \lim_{z\to 0}{e^z} = 1$$
* Hence, by definition, it is a removable singularity.

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