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find the isolated singular points of $f(z)$, and what the type of this point ?

1) $f(z)={1\over {z^2{(z-3)}^3}}$

2) $f(z)={e^{2z}\over {(z-1)^2}}$

my try:

1) the isolated singular points are 0 and 3, but what about there types ?

2)

the isolated singular point is 1

and for the Type :

$e^{2z}=1+2z+{{(2z)}^2\over 2!}+{{(2z)}^3\over 3!}+...$

so

$f(z)={1\over {(z-1)}^2} (1+2z+{{(2z)}^2\over 2!}+{{(2z)}^3\over 3!}+...)$

but what about $(z-1)^2$ ?

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    When does $\frac1{(z-1)^2}$ have singularities? – 2017-01-16
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    1, but I don't know how can I explain the type :( – 2017-01-16
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    For what smallest value of $n$ does $\lim_{z\to1}(z-1)^nf(z)$ exist? – 2017-01-16
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    ummm, maybe 2 ? – 2017-01-16
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    Yup. :-) You can find poles that way as well – 2017-01-16
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    Thanks :), so for (1) : 0 is pole (order 2) and 3 is pole (order 3) ? – 2017-01-16
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    Yup, it's pretty much that simple. Like simpleart :-) – 2017-01-16
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    No problem :-) Just be careful though, as you need $\lim_{z\to a}(z-a)^nf(z)$ to exist from all directions with $z\in\mathbb C$, and it doesn't work on essential singularities. – 2017-01-16
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    I will be careful for that :), Thank you. – 2017-01-16

1 Answers 1

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I hope I remember everything correctly from a Complex Analysis course I took and I hope I'm not making mistakes here. I understand this answer may not be the best, but I'm trying to help anyway. :) Corrections from the community are very welcome.

First question: You can write the Taylor series of the function around the singularity like $$\sum_{n = 0}^{N}a_n(z-z_0)^n +\sum_{m=1}^{M}b_m(z-z_0)^{-m}$$ If the series has infinitly many $b_m \neq 0$, then the singularity is essisential. If $b_m = 0$ for all $m$, then the singularity is isolated. If only finitely there are only finitely many $b_m \neq 0$, then the singularity is a pole of order $m$ with $m$ the highest integer for which $b_m \neq 0$

Second question: Let's say $f(z) = \frac{p(z)}{q(z)}$ with $p(z) = e^{2z}$ and $q(z) = (z-1)^2 = z^2 - 2z + 1$ and call the singularity $z_0$.

The singularity is a pole. ($\lim_{z\to z_0} (z-z_0)^2f(z)$ is neither $0$ nor $\infty$.)

If $p(z_0) \neq 0$ then the order of the pole is the lowest integer $n$ for which $$q^{(n)}(z_0) \neq 0$$

Let's see: $$q^{(1)}(z) = 2z - 2$$ $q^{(1)}(1) = 0$, so we have to differentiate more: $$q^{(2)}(z) = 2$$

Obviously $q^{(2)}(1) = 2$ (since it equals $2$ for any $z$), so the singularity is a pole of order $2$.

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    You're welcome. :) But please check if what I said is correct, I don't want you to make any mistakes because of me. ;) – 2017-01-16
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    (And if it helped, a +1 is appreciated :$ ) – 2017-01-16
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    Is the power $m$ in first question negative? – 2017-01-16
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    @MyGlasses Indeed, it should be negative. Thanks! I fixed it. – 2017-01-16