Complex Numbers (Laurent Series)

[jojosg]

Homework Statement

Am attempting a Laurent Series question but I'm not confident in my answers. would appreciate any corrections.

Relevant Equations

f(z) = sigma (cnz^n)

see attached.

 

[FactChecker]

Are you sure this problem is valid? The domains of the two expansions would make more sense to me if the function was
$\frac {z+1}{z^3(z^2-1)}$
or if the domains were $0\lt z \lt \sqrt 2$ and $\sqrt 2 \lt z \lt \infty$

PS. I can not read your work.

 

[jojosg]

Are you sure this problem is valid? The domains of the two expansions would make more sense to me if the function was
$\frac {z+1}{z^3(z^2-1)}$
or if the domains were $0\lt z \lt \sqrt 2$ and $\sqrt 2 \lt z \lt \infty$

PS. I can not read your work.

It was given by my teacher so I assume the problem is valid. Hope these are clearer.

 

[fresh_42]

Hope these are clearer.

Not at all. You should try writing them in LaTeX code here, see
https://www.physicsforums.com/help/latexhelp/

If you only want to check the result, then WA does a good job:
https://www.wolframalpha.com/input?i=laurent+series+of+(z+1)/(z^3(z^2-2))

 

[FactChecker]

It was given by my teacher so I assume the problem is valid. Hope these are clearer.

I don't believe there is a Laurent series for that function that converges for $1 \lt |z| \lt \infty$

 

[jojosg]

$\boxed{ \textbf{Problem 3: Find the Laurent series of } f(z) = \frac{z+1}{z^3(z^2-2)} \text{ for:} \quad \text{(1) } 0 < |z| < 1 \quad \text{(2) } 1 < |z| < +\infty }$

**Partial Fraction Decomposition**

We begin with the partial fraction decomposition:

$$f(z) = \frac{z+1}{z^3(z^2-2)} = \frac{A}{z} + \frac{B}{z^2} + \frac{C}{z^3} + \frac{Dz+E}{z^2-2}$$

Multiply both sides by $z^3(z^2-2)$:

$$z+1 = Az^2(z^2-2) + Bz(z^2-2) + C(z^2-2) + (Dz+E)z^3$$

$$z+1 = A(z^4-2z^2) + B(z^3-2z) + C(z^2-2) + Dz^4 + Ez^3$$

$$z+1 = (A+D)z^4 + (B+E)z^3 + (-2A+C)z^2 - 2Bz - 2C$$

Comparing coefficients:

$$\begin{cases}
z^4: & A + D = 0 \\
z^3: & B + E = 0 \\
z^2: & -2A + C = 0 \\
z^1: & -2B = 1 \\
\text{Constant}: & -2C = 1
\end{cases}$$

Solving:

$$B = -\frac{1}{2}, \quad C = -\frac{1}{2}, \quad -2A - \frac{1}{2} = 0 \Rightarrow A = -\frac{1}{4}$$
$$D = -A = \frac{1}{4}, \quad E = -B = \frac{1}{2}$$

Thus:

$$f(z) = -\frac{1}{4z} - \frac{1}{2z^2} - \frac{1}{2z^3} + \frac{\frac{1}{4}z + \frac{1}{2}}{z^2-2}$$

**1. Laurent Series for $0 < |z| < 1$**

For $|z| < 1$, we expand $\frac{1}{z^2-2}$:

$$\frac{1}{z^2-2} = -\frac{1}{2} \cdot \frac{1}{1 - \frac{z^2}{2}} = -\frac{1}{2} \sum_{n=0}^{\infty} \left(\frac{z^2}{2}\right)^n = -\sum_{n=0}^{\infty} \frac{z^{2n}}{2^{n+1}}$$

Then:

$$\frac{\frac{1}{4}z + \frac{1}{2}}{z^2-2} = \left(\frac{1}{4}z + \frac{1}{2}\right) \left(-\sum_{n=0}^{\infty} \frac{z^{2n}}{2^{n+1}}\right) = -\frac{1}{4} \sum_{n=0}^{\infty} \frac{z^{2n+1}}{2^{n+1}} - \frac{1}{2} \sum_{n=0}^{\infty} \frac{z^{2n}}{2^{n+1}}$$

So the Laurent series for $0 < |z| < 1$ is:

$$f(z) = -\frac{1}{4z} - \frac{1}{2z^2} - \frac{1}{2z^3} - \frac{1}{4} \sum_{n=0}^{\infty} \frac{z^{2n+1}}{2^{n+1}} - \frac{1}{2} \sum_{n=0}^{\infty} \frac{z^{2n}}{2^{n+1}}$$

**2. Laurent Series for $1 < |z| < +\infty$**

For $|z| > 1$, we expand $\frac{1}{z^2-2}$ differently:

$$\frac{1}{z^2-2} = \frac{1}{z^2} \cdot \frac{1}{1 - \frac{2}{z^2}} = \frac{1}{z^2} \sum_{n=0}^{\infty} \left(\frac{2}{z^2}\right)^n = \sum_{n=0}^{\infty} \frac{2^n}{z^{2n+2}}$$

Then:

$$\frac{\frac{1}{4}z + \frac{1}{2}}{z^2-2} = \left(\frac{1}{4}z + \frac{1}{2}\right) \sum_{n=0}^{\infty} \frac{2^n}{z^{2n+2}} = \frac{1}{4} \sum_{n=0}^{\infty} \frac{2^n}{z^{2n+1}} + \frac{1}{2} \sum_{n=0}^{\infty} \frac{2^n}{z^{2n+2}}$$

So the Laurent series for $1 < |z| < +\infty$ is:

$$f(z) = -\frac{1}{4z} - \frac{1}{2z^2} - \frac{1}{2z^3} + \frac{1}{4} \sum_{n=0}^{\infty} \frac{2^n}{z^{2n+1}} + \frac{1}{2} \sum_{n=0}^{\infty} \frac{2^n}{z^{2n+2}}$$

**Final Answer:**

**(1) For $0 < |z| < 1$:**
$$f(z) = -\frac{1}{4z} - \frac{1}{2z^2} - \frac{1}{2z^3} - \frac{1}{4} \sum_{n=0}^{\infty} \frac{z^{2n+1}}{2^{n+1}} - \frac{1}{2} \sum_{n=0}^{\infty} \frac{z^{2n}}{2^{n+1}}$$

**(2) For $1 < |z| < +\infty$:**
$$f(z) = -\frac{1}{4z} - \frac{1}{2z^2} - \frac{1}{2z^3} + \frac{1}{4} \sum_{n=0}^{\infty} \frac{2^n}{z^{2n+1}} + \frac{1}{2} \sum_{n=0}^{\infty} \frac{2^n}{z^{2n+2}}$$

 

[FactChecker]

Series (2) can not converge at $z=\sqrt 2$ because the function is not defined there. The region of convergence ($1 \lt |z| \lt +\infty$) of a Laurent series can not include a pole ($z=\sqrt 2$) of the original function, .
I repeat that the problem statement is wrong.

 

[pasmith]

Yes. Expanding $(z - a)^{-k}$ in binomial series can be done in two ways; one converges for $|z| < |a|$ and the other for $|z| > |a|$.

 

[wrobel]

I repeat that the problem statement is wrong.

I agree

 

[vela]

It was given by my teacher so I assume the problem is valid. Hope these are clearer.

Your teacher is not infallible, so I'd reach out for clarification about the intended function and intended domain.

Note that your first series converges when $\left\lvert \frac{z^2}2 \right\rvert < 1$ or, equivalently, $\lvert z \rvert < \sqrt 2$.

 

[FactChecker]

I haven't looked at your work, but if it is correct, then it will be valid for the modified problem with domains specified as $|z|\lt \sqrt 2$ and $\sqrt 2 \lt |z| \lt +\infty$.

 

[jojosg]

I haven't looked at your work, but if it is correct, then it will be valid for the modified problem with domains specified as $|z|\lt \sqrt 2$ and $\sqrt 2 \lt |z| \lt +\infty$.

can I use this theorem to prove that its not analytic in that range for part 2?

 

[fresh_42]

WA delivers a series for all values, just not the same. However, it does not provide the convergence radii.

 

[FactChecker]

$\boxed{ \textbf{Problem 3: Find the Laurent series of } f(z) = \frac{z+1}{z^3(z^2-2)} \text{ for:} \quad \text{(1) } 0 < |z| < 1 \quad \text{(2) } 1 < |z| < +\infty }$

**Partial Fraction Decomposition**

We begin with the partial fraction decomposition:

$$f(z) = \frac{z+1}{z^3(z^2-2)} = \frac{A}{z} + \frac{B}{z^2} + \frac{C}{z^3} + \frac{Dz+E}{z^2-2}$$

Multiply both sides by $z^3(z^2-2)$:

$$z+1 = Az^2(z^2-2) + Bz(z^2-2) + C(z^2-2) + (Dz+E)z^3$$

$$z+1 = A(z^4-2z^2) + B(z^3-2z) + C(z^2-2) + Dz^4 + Ez^3$$

$$z+1 = (A+D)z^4 + (B+E)z^3 + (-2A+C)z^2 - 2Bz - 2C$$

Comparing coefficients:

$$\begin{cases}
z^4: & A + D = 0 \\
z^3: & B + E = 0 \\
z^2: & -2A + C = 0 \\
z^1: & -2B = 1 \\
\text{Constant}: & -2C = 1
\end{cases}$$

Solving:

$$B = -\frac{1}{2}, \quad C = -\frac{1}{2}, \quad -2A - \frac{1}{2} = 0 \Rightarrow A = -\frac{1}{4}$$
$$D = -A = \frac{1}{4}, \quad E = -B = \frac{1}{2}$$

Thus:

$$f(z) = -\frac{1}{4z} - \frac{1}{2z^2} - \frac{1}{2z^3} + \frac{\frac{1}{4}z + \frac{1}{2}}{z^2-2}$$

It is simpler to add the series expansions of $\frac{z}{z^3(z^2-2)}$ and $\frac{1}{z^3(z^2-2)}$. Both series are simple to derive and valid for $|z^2/2| \lt 1$. Your restriction of $|z| \lt 1$ is wrong.

 

[fresh_42]

I don't think we should use the words wrong or false. The answer is simply not as simple as expected, since the crucial point is different from the one given by the problem statement.

 

[FactChecker]

I don't think we should use the words wrong or false. The answer is simply not as simple as expected, since the crucial point is different from the one given by the problem statement.

Maybe I should have said, unnecessary or misleading or arbitrary.
In any case, IMO, it is important to know and use the radius of convergence of the geometric series.