Write the function $\frac{1}{(z+1)(3-z)}$ as a Laurent series.

Question

$$f(z)=\frac{1}{(z+1)(3-z)}=\frac{1}{4z+4} + \frac{1}{12-4z}$$

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

$$\frac{1}{12-4z}=\frac{1}{12}\frac{1}{1-\frac{z}{3}}=\frac{1}{12}\sum_{k=0}^{\infty} \left(\frac{z}{3}\right)^k$$

$$f(z)=\frac{1}{4z}\sum_{k=0}^{\infty} \left(\frac{-1}{z}\right)^k+\frac{1}{12}\sum_{k=0}^{\infty} \left(\frac{z}{3}\right)^k$$

I can rewrite that as $$f(z)=\frac{1}{4z}\sum_{k=-\infty}^{0} (-1)^k z^k+\frac{1}{12}\sum_{k=0}^{\infty} \left(\frac{1}{3}\right)^k z^k$$.

I need to move the $\frac{1}{4z}$ and $\frac{1}{12}$ into the sums but finding a series that will converge to each, but I have no idea what to use for either. Any suggestions? Am I taking a wrong approach or is there an obvious series to use for this?

Edit: The center is $0$ and the region in $1 \le |z| \le 3$.

I think I can use a geometric sequence to say $\frac{1}{4z}=\sum_{k=0}^{\infty}\frac{1}{2}(\frac{1}{2})^{k-1}\frac{z}{k}$ and $\frac{1}{12}=\sum_{k=0}^{\infty}\frac{1}{36}(\frac{2}{9})^{k-1}\frac{z}{k}$. I'm pretty sure that's true, but it seems like it makes the whole thing a complicated mess.

Answer 1

I can write the expression as: $$\frac{1}{(z+1)(-z+3)} = \frac{-1}{(z+1)(z-3)}$$

Use partial fractions: $$\frac{-1}{(z+1)(z-3)} = \frac{A}{z+1}+\frac{B}{z-3}$$ $$-1 = A(z-3)+B(z+1)$$ If z=3, then $$-1 = B(3+1)$$ $$-1 = B(4)$$ $$B=\frac{-1}{4}$$

If z=1, then $$-1 = A(1-3)$$ $$-1 = A(-2)$$ $$1 = A(2)$$ $$A = 1/2$$

So, we have: $$\frac{-1}{(z+1)(z-3)} = (\frac{1}{2})\frac{1}{z+1}+(\frac{-1}{4})\frac{1}{z-3}$$

The previous expression is a Laurent Series. Because the expression has no expressions of the form $$\frac{1}{z^{n}}$$ The Laurent Series behaves as a Taylor Series.

If we suppose that in z=0 the expression is not analytic, then we can write it as

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

Answer 2

The function

\begin{align*} f(z)&=\frac{1}{(z+1)(3-z)}\\ &=\frac{1}{4(z+1)}-\frac{1}{4(z-3)} \end{align*} has two simple poles at $-1$ and $3$.

Since we want to find a Laurent expansion with center $0$, we look at the poles $-1$ and $3$ and see they determine three regions.

\begin{align*} |z|<1,\qquad\quad 1<|z|<3,\qquad\quad 3<|z| \end{align*}

• The first region $ |z|<1$ is a disc with center $0$, radius $1$ and the pole $-1$ at the boundary of the disc. In the interior of this disc all two fractions with poles $-1$ and $3$ admit a representation as power series at $z=0$.

• The second region $1<|z|<3$ is the annulus with center $0$, inner radius $1$ and outer radius $3$. Here we have a representation of the fraction with poles $-1$ as principal part of a Laurent series at $z=0$, while the fraction with pole at $3$ admits a representation as power series.

• The third region $|z|>3$ containing all points outside the disc with center $0$ and radius $3$ admits for all fractions a representation as principal part of a Laurent series at $z=0$.

A power series expansion of $\frac{1}{z+a}$ at $z=0$ is \begin{align*} \frac{1}{z+a}&=\frac{1}{a}\cdot\frac{1}{1+\frac{z}{a}}\\ &=\sum_{n=0}^{\infty}\frac{1}{a^{n+1}}(-z)^n \end{align*} The principal part of $\frac{1}{z+a}$ at $z=0$ is \begin{align*} \frac{1}{z+a}&=\frac{1}{z}\cdot\frac{1}{1+\frac{a}{z}}=\frac{1}{z}\sum_{n=0}^{\infty}\frac{a^n}{(-z)^n} =-\sum_{n=0}^{\infty}\frac{a^n}{(-z)^{n+1}}\\ &=-\sum_{n=1}^{\infty}\frac{a^{n-1}}{(-z)^n} \end{align*}

We can now obtain the Laurent expansion of $f(z)$ at $z=0$ for all three regions.

• Region 2: $1<|z|<3$

\begin{align*} f(z)&=\frac{1}{4(z+1)}-\frac{1}{4(z-3)}\\ &=-\frac{1}{4}\sum_{n=1}^\infty\frac{1}{(-z)^n}-\frac{1}{4}\sum_{n=0}^\infty \frac{1}{(-3)^{n+1}}(-z)^n\\ &=\frac{1}{4}\sum_{n=1}^\infty(-1)^{n+1}\frac{1}{z^n}+\frac{1}{4}\sum_{n=0}^\infty \frac{1}{3^{n+1}}z^n\\ \end{align*}

The Laurent expansion for the other regions can be calculated similarly.