Power Series for Complex Number z: f(z) and Radius of Convergence Calculation

[fauboca]

z is a complex number.

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

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

$$\frac{1}{(z + 2)^2} = -\frac{d}{dz} \ \frac{1}{1 - (-z - 1)} = -\frac{d}{dz}\sum_{n = 0}^{\infty}(-z - 1)^n = \sum_{n = 0}^{\infty}-n(-z-1)^{n-1}$$

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

Radius of convergence by the ratio test:

$$\lim_{n\to\infty}\left|\frac{-(4+3z)(n+1)(-z)^{n+1}(-z-1)^n}{-(4+3z)n(-z)^n(-z-1)^{n-1}}\right|=\lim_{n\to\infty}\left|\frac{(n+1)(z)(z+1)}{n}\right|$$

$$|z(z+1)|<1$$

Is the above correct?

If so, how do I find the Radius of Convergence from that?

 

[Char. Limit]

No, that doesn't look right. For one thing, you can't just combine summands like that. The product of (z+1)(z+2)^2 is going to be the product of the two sums, and you're probably going to end up with a double sum or something.

 

[fauboca]

No, that doesn't look right. For one thing, you can't just combine summands like that. The product of (z+1)(z+2)^2 is going to be the product of the two sums, and you're probably going to end up with a double sum or something.

Then how can I find the power series of that expression?

 

[Char. Limit]

I'd split the function up using Partial Fractions. Then you'll get three functions that are easy to devise a power series for.

 

[fauboca]

I'd split the function up using Partial Fractions. Then you'll get three functions that are easy to devise a power series for.

Could I instead use the rule $R\geq\min{R,S} = 1$?

Isn't there a way to do this without the use of partial fractions?

 

[fauboca]

By partial fractions, we can re-write $f(z)$ as
$$\frac{A}{z + 1} + \frac{B}{z + 2} + \frac{C}{(z + 2)^2} = \frac{4 + 3z}{(z + 1)(z + 2)^2}.$$
By direct computations, $A(z + 2)^2 + B(z + 2)(z + 1) + C(z + 1)$ $= Az^2 + 4Az + 4A + Bz^2 + 3Bz + 2B + Cz + C.$
To find the coefficients, we solve the system of linear equations.
$$\begin{pmatrix}1 & 1 & 0 & 0\\ 4 & 3 & 1 & 3\\ 4 & 2 & 1 & 4\end{pmatrix}\Rightarrow \begin{pmatrix}1 & 0 & 0 & 1\\ 0 & 1 & 0 & -1\\ 0 & 0 & 1 & 2\end{pmatrix}$$
So the coefficients are $A = 1$, $B = -1$, and $C = 2$.
Therefore, $\displaystyle\frac{4 + 3z}{(z + 1)(z + 2)^2} = \frac{1}{z + 1} - \frac{1}{z + 2} + \frac{2}{(z + 2)^2}$ $\displaystyle\Rightarrow\frac{1}{1 - (-z)} - \frac{1}{2}\cdot\frac{1}{1 - \left(-\frac{z}{2}\right)} - \frac{d}{dz}\frac{1}{1 - \left(-\frac{z}{2}\right)}$.
The rational expressions can be represented as the following power series
$$\sum_{n = 0}^{\infty}(-z)^n - \frac{1}{2}\sum_{n = 0}^{\infty}\left(-\frac{z}{2}\right)^n - \sum_{n = 0}^{\infty}n\left(-\frac{z}{2}\right)^{n - 1}.$$

Now how do I find the radius of convergence

 

[Char. Limit]

First express it as a single sum. Due to associativeness, you can easily rewrite it like this:

$$\sum_{n=0}^{\infty} (-1)^n z^n - (-1)^n z^n 2^{-n-1} - (-1)^n n z^n 2^{1-n}$$

Which, in a simpler form, is:

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

Now how would you find the radius of convergence of a single sum?

 

[fauboca]

First express it as a single sum. Due to associativeness, you can easily rewrite it like this:

$$\sum_{n=0}^{\infty} (-1)^n z^n - (-1)^n z^n 2^{-n-1} - (-1)^n n z^n 2^{1-n}$$

Which, in a simpler form, is:

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

Now how would you find the radius of convergence of a single sum?

Ratio test

So we end up getting

$$|z| < 1$$

and R = 1

Correct?