# Find the residue for

1. Apr 14, 2015

### Shackleford

1. The problem statement, all variables and given/known data

Find Res(0;f) for

$f(z) = \frac{e^{4z} - 1}{sin^2(z)}.$

2. Relevant equations

Residue Theorem

3. The attempt at a solution

If there's a nice (z-a)n singularity in the denominator, then I can simply use the Residue Theorem. However, I'm skeptical on what I'm doing:

The singularity occurs at z = 0 and its order is two. Using the Residue Theorem, I can take $\ d/dz(sin^2(z)f(z) = d/dz(e^{4z} - 1)$ evaluated at z = 0 which is 4.

Expressing the terms in a power series gets messy.

2. Apr 14, 2015

### Dick

I guess I'm not sure why you are skeptical of that. It looks fine to me.

3. Apr 14, 2015

### Shackleford

Okay. I wasn't sure if I could multiply the sin2z term because it's not of the form (z-a)n as given in the theorem.

4. Apr 14, 2015

### Dick

Actually, that is a correct concern. I was being sloppy. Since you know the pole is of order 2 you should be multiplying by $z^2$ and then taking the derivative and the limit. You'll get the same result.

5. Apr 14, 2015

### Shackleford

I thought about that but I'd still have a sin2z term in the denominator if I take the derivative of $f(z) = \frac{z^2(e^{4z} - 1)}{sin^2(z)}$, using the quotient rule.

6. Apr 14, 2015

### Dick

You are probably better off actually working on the series expansion. Remember you only need the coefficient of the 1/z term. $e^{4z}-1$ has the form $4z f(z)$ where $f(z)$ is analytic and $f(0)$ is not zero. $sin(z)^2$ has the form $z^2 g(z)$ where $g(z)$ is analytic and $g(0)$ is not zero. There may actually not be much series expansion to do.

7. Apr 14, 2015

### Shackleford

Hm. Expanding the numerator of $f(z) = \frac{e^{4z} - 1}{sin^2(z)}$ is easy, i.e. $f(z) = \frac{1 + 4z + \frac{4^2z^2}{2} + ... \frac{4^nz^n}{n!} - 1}{sin^2(z)}.$ I'm not sure how expanding sin2(z) in the denominator would allow me to find the z-1 coefficient.

8. Apr 14, 2015

### Dick

Ok, I'll spell it out a little more. I claim that your function can be written as $\frac{4z f(x)}{z^2 g(z)}$ where $\frac{f(z)}{g(z)}$ is analytic at $z=0$. So $\frac{f(z)}{g(z)}=a_0+a_1 z+...$. So the coefficient you are looking for is just $4 a_0$ and $a_0=\frac{f(0)}{g(0)}$.

9. Apr 14, 2015

### Shackleford

Okay. I think I follow you now. Of course, multiplying the analytic quotient $\frac{f(z)}{g(z)}$ by $\frac{4z}{z^2}$ gives me the n = -1 or z-1 coefficient in the Laurent expansion, which should just be $4\frac{f(0)}{g(0)} = 4\frac{1}{1} = 4.$