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Locate, name the singularities and determine if they are isolated or not $$f(z)= \frac{(z^2+9)^2}{(z^2+iz+6)^2} $$ Singularity when: $$z^2+iz+6=0 \\ -(z+3i)(-z+2i)=0 \\ \implies z_1=2i, z_2=-3i $$

for $z_1=2i$

$$\lim_{z \to +2i}(z-2i)^n \cdot \frac{(z^2+9)^2}{(z+3i)^2(-z+2i)^2} \neq 0 \mathrm{ \ \ \ \ for \ \ }n=2 $$

A Pole of order two.

Is this isolated or not? How to determine it?

(Isolated $\because$ defined in a Nhood)

for $z_2=-3i$

$$\lim_{z \to -3i}(z+3i)^n \cdot \frac{(z^2+9)^2}{(z+3i)^2(-z+2i)^2} $$

$$\lim_{z \to -3i}(z+3i)^n \cdot \frac{(z+3i)^2(z-3i)^2}{(z+3i)^2(-z+2i)^2} $$

$$\lim_{z \to -3i}(z+3i)^n \cdot \frac{(z-3i)^2}{(-z+2i)^2} $$

What type of singularity am I dealing with here and is it isolated or not?

(Removable singularity)

Thanks.

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1 Answer 1

up vote 1 down vote accepted

A singularity is isolated if the function is defined and nonsingular in a neighborhood of the singularity. So, how does your function look, near $2i$?

For the second question, factor the numerator.

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1  
Function looks defined and non-singular in a Nhood near $2i$. 2nd one we are dealing with a removable singularity. Thanks. –  user42538 Feb 7 '13 at 22:44

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