What Is the Residue of f(z) at z=0?

[Shackleford]

Homework Statement

Find Res(0;f) for

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

Homework Equations

Residue Theorem

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.

 

[Dick]

Homework Statement

Find Res(0;f) for

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

Homework Equations

Residue Theorem

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.

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

 

[Shackleford]

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

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.

 

[Dick]

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.

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.

 

[Shackleford]

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.

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.

 

[Dick]

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.

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.

 

[Shackleford]

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.

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.

 

[Dick]

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.

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 $a_0=\frac{f(0)}{g(0)}$.

 

[Shackleford]

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 $a_0=\frac{f(0)}{g(0)}$.

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.