Laurent Series (non-homework) question

[Void123]

Homework Statement

I am trying to understand Laurent series from the few and limited examples given by my texts.

I understand the basic idea, which is to expand the series about a function's pole.

So how would one go about finding the Laurent expansion for $$\frac{z + 1}{z^{2} - z - 6}$$?

Or $$\frac{cosh z}{z^{2}}$$?

Homework Equations

$$f(z) = \sum a_{n} (z - z_{0})^{n}$$

The Attempt at a Solution

Pretty straightforward equation. Except, could someone explain the procedure of calculating $$a_{n}$$

Thanks.

 

[lippyka]

For the first equation, we can write it as \frac{z + 1}{(z - 2)(z + 3)}. Then we can use partial fraction decomposition to write it as \frac{A}{z - 2} + \frac{B}{z + 3}. Then we can solve for A and B by setting z = 2 and z = -3. This gives us A = -1/5 and B = 3/5. Now we can rewrite the equation as \frac{-1/5}{z - 2} + \frac{3/5}{z + 3}. The Laurent expansion of this equation is\frac{-1/5}{z - 2} + \frac{3/5}{z + 3} = -\frac{1}{5}\sum_{n=0}^{\infty} (z - 2)^{n} + \frac{3}{5}\sum_{n=0}^{\infty} (-1)^{n} (z + 3)^{n}.For the second equation, we can use the formula f(z) = \sum a_{n} (z - z_{0})^{n}, where z_{0} is the pole of the function. In this case, the pole is at z = 0, so we can write the equation as f(z) = \sum a_{n} z^{n}. We can find a_{n} by taking the derivative of both sides of the equation with respect to z. This gives us f'(z) = \sum na_{n} z^{n-1}. We can then equate this to the derivative of our original equation, which is f'(z) = \frac{d}{dz}\left(\frac{cosh z}{z^{2}}\right) = -\frac{2sinh z}{z^{3}}. So now we have two equations that are equal to each other, one on the left and one on the right. We can then solve for the coefficients a_{n}. For example, if we set n = 0, then we get a_{0} = \frac{sinh z}{z^{3}}|_{z=0} = \frac{1