How to determine the type of singularities

Solution 1:

a) $\displaystyle{f(z)=\dfrac{1}{e^{1/z}-1}}$.

It says $f:\mathbb C\setminus\{0\}\to\mathbb C$, but this is incorrect, because $f$ has a simple pole at $z=\dfrac{1}{2\pi ki}$ for each nonzero integer $k$, and $z=0$ is not even an isolated singularity. If you change the codomain to $\mathbb C\cup\{\infty\}$ and think of $f$ as a meromorphic function, then it has an essential singularity at $0$.

In fact, you can show that $f(D(0,r)\setminus\{0\})=(\mathbb C\cup\{\infty\})\setminus\{0,-1\}$ for all $r>0$, using elementary properties of the exponential function.

b) $\displaystyle f:\mathbb{C}\backslash\{0,2\}\rightarrow\mathbb{C},\ f(z)=\frac{\sin z ^2}{z^2(z-2)}$

Evaluate $\lim\limits_{z\to 0}f(z)$ and $\lim\limits_{z\to 2}f(z)$. One is finite, the other is $\infty$, so you have a removable singularity and a pole.

c) $\displaystyle f:\mathbb{C}\backslash\{0\}\rightarrow\mathbb{C},\ f(z)=\cos\left(\frac{1}{z}\right)$

In this case, you should be able to show, even just using real variables, that $\lim\limits_{z\to 0}f(z)$ does not exist in either a finite or infinite sense. Sketch a graph of $y=\cos(1/t)$ close to $0$. Another useful tool is the Laurent series, which in this case is obtained from the power series expansion of $\cos$ by substitution of $1/z$. And similarly to a), you could use elementary properties of the exponential function along with the identity $\cos(z)=\frac{1}{2}(e^{iz}+e^{-iz})$ to find the image of a small punctured disk at $0$.

d) $\displaystyle f:\mathbb{C}\backslash\{0\}\rightarrow\mathbb{C},\ f(z)=\frac{1}{1-\cos\left(\frac{1}{z}\right)}$

Similarly to a), this is incorrect. Either the domain or the codomain should be changed. If you don't change the codomain, then $f$ is undefined where $\cos(1/z)=1$, and there is not an isolated singularity at $0$. If you allow meromorphic functions, then it is an essential singularity at $0$. (And again you could even explicitly find the range, or you could more simply show that no limit exists by choosing special values.)

e) $\displaystyle f:\mathbb{C}\backslash\{0\}\rightarrow\mathbb{C},\ f(z)=\frac{1}{\sin\left(\frac{1}{z}\right)}$

See a) and d).

Solution 2:

Thank you for all your feedback. I've decided to simplify things and use the method from Schaum's Outline on Complex Analysis. Otherwise, I am getting nowhere.

They do it like this:

(i) If $\lim_{z\rightarrow a} f(z)$ exists then we have a removal singularity.

(ii) If $\lim_{z\rightarrow a} (z-a)^n f(z) = A \neq 0$, then $z=a$ is a pole of order $n$.

If we don't have (i) or (ii), then the singularity is essential.

Question: Why are these 3 options, the only ones for isolated singularities? A short explanation in words would be nice!

So, using this we have:

a) $\displaystyle f:\mathbb{C}\backslash\{0\}\rightarrow\mathbb{C},\ f(z)=\frac{1}{e^{\frac{1}{z}}-1}$

We must check $\lim_{z\rightarrow 0} z^n \frac{1}{e^{\frac{1}{z}}-1}$

We have $\lim_{z\rightarrow 0} z^n \frac{1}{e^{\frac{1}{z}}-1}=0$ for any natural number $n$.

So, this means that 0 is an essential singularity here.

b) $\displaystyle f:\mathbb{C}\backslash\{0,2\}\rightarrow\mathbb{C},\ f(z)=\frac{\sin z ^2}{z^2(z-2)}$

$\lim_{z\rightarrow 0} z^n \frac{\sin z ^2}{z^2(z-2)}=0$

$\lim_{z\rightarrow 2} z^n \frac{\sin z ^2}{z^2(z-2)}=-\infty$

So, we have again essential singularities, I believe...

c) $\displaystyle f:\mathbb{C}\backslash\{0\}\rightarrow\mathbb{C},\ f(z)=\cos\left(\frac{1}{z}\right)$

$\lim_{z\rightarrow 0} z^n \cos\left(\frac{1}{z}\right)=0$

Uhem... essential again...

d) $\displaystyle f:\mathbb{C}\backslash\{0,\frac{1}{2k\pi}\}\rightarrow\mathbb{C},\ f(z)=\frac{1}{1-\cos\left(\frac{1}{z}\right)}$

$\lim_{z\rightarrow 0} z^n \frac{1}{1-\cos\left(\frac{1}{z}\right)}$

0 is odd here... might it be that 0 is no singularity?

For $2k\pi,\ k\neq 0$, the limit can be evaluated to something. I believe these values are the poles then. I think we have $n$ of them.

e) $\displaystyle f:\mathbb{C}\backslash\{0,\frac{1}{k\pi}\}\rightarrow\mathbb{C},\ f(z)=\frac{1}{\sin\left(\frac{1}{z}\right)}$

$\lim_{z\rightarrow 0} z^n\frac{1}{\sin\left(\frac{1}{z}\right)}$

Same as d, 0 is no singularity?

For $n = 1$, the limit is $1$. So, we got a pole of order $1$ at $z=0$.

Do you agree with the above?