Hi I have attached 3 questions which I am very stuck with. I have looked up the definitions and such for Q1 and dont know where else to start with any of them. Thanks for looking!

Edit: Have successfully Solved Q1 with the kind help of shawsend.

Q2 here I come

2. For 1:

You need to look at these in terms of limits and Laurent series and focus on just the term that has the singular component. For example, at z=0, the exp term is a constant so when looking at the function at z=0, you could just as well consider $\frac{z^2}{\sin^2(z)}k$.

If the pole is removable, then $\lim_{z\to z_0} f(z)$ exists. So what is $\lim_{z\to 0}\frac{z^2}{\sin^2(z)}$?

If it has a pole of order one, then $\lim_{z\to z_0} (z-z_0) f(z)$ exists and is not zero. And if it has a pole of order k, then $\lim_{z\to z_0} (z-z_0)^k f(z)$ exists and is not zero.

Now, what about all the pi's? They are all the same in terms of singular order right so just consider the one at $\pi$. What is:

$\lim_{z\to\pi} (z-\pi)^k \frac{z^2}{\sin^2(z)}$ for k=1, k=2, . . . until it exists and is not zero?

Well consider:

$\lim_{z\to\pi} (z-\pi) \frac{z^2}{\sin^2(z)}$

The $z^2$ term just goes to $\pi^2$ so can consider:

$\lim_{z\to\pi} (z-\pi) \frac{1}{\sin^2(z)}$

That's infinity right?

$\lim_{z\to\pi} (z-\pi)^2 \frac{1}{\sin^2(z)}$

Do a double L'Hospital on it and get one. So it's a pole of order 2.

As far as the $\pm i$, well, $e^{f(z)}$ has an essential singularity anywhere $f(z)$ is singular. You can prove this by writing it's Laurent series.

3. Thank you very much! I had a couple of steps correct so not as bad as I thought. I am going to work more on this now. Thanks again!!

4. Hello Again. I have worked out all of question 1! Woo thanks again for your help and not just giving me the answer as I have learned a lot working out the rest!

5. Thanks solved the rest. Used residue Thm on Q 3.