# Laurent series (Complex)

#### scorpion007

Using term-by-term differentiation, expand $$\displaystyle f(z)=\frac{1}{(z+3)^2}$$ in a Laurent series valid for $$\displaystyle |z| > 3$$.

Ok, so, suppose I get a few derivatives:

$$\displaystyle f'(z)=-2(z+3)^{-3}$$

$$\displaystyle f''(z)=6(z+3)^{-4}$$

$$\displaystyle f'''(z)=-24(z+3)^{-5}$$
...

There's infinitely many. What next?

#### Drexel28

MHF Hall of Honor
Using term-by-term differentiation, expand $$\displaystyle f(z)=\frac{1}{(z+3)^2}$$ in a Laurent series valid for $$\displaystyle |z| > 3$$.

Ok, so, suppose I get a few derivatives:

$$\displaystyle f'(z)=-2(z+3)^{-3}$$

$$\displaystyle f''(z)=6(z+3)^{-4}$$

$$\displaystyle f'''(z)=-24(z+3)^{-5}$$
...

There's infinitely many. What next?
Pattern.

Do you have to do it this way? Note that $$\displaystyle f(z)=\frac{1}{z^2}\frac{1}{(1+\frac{3}{z})^2}$$

#### scorpion007

I guess I have to, since that's what the question asks.

I know the pattern is:

$$\displaystyle f^{(n)}(z)=(n+1)!(-1)^n(z+3)^{-(n+2)}$$

but I don't know what to do with it.

#### scorpion007

Doing it the other way, I suppose what happens is:

$$\displaystyle f(z)=\frac{1}{z^2}\frac{1}{(1+\frac{3}{z})^2} = \frac{1}{z^2}\left(\sum_{n=0}^\infty \left(-\frac{3}{z}\right)^n\right)^2$$ for $$\displaystyle |z|>3$$.

But the whole series is squared. How am I suppose to write out the first few terms?

#### Drexel28

MHF Hall of Honor
Doing it the other way, I suppose what happens is:

$$\displaystyle f(z)=\frac{1}{z^2}\frac{1}{(1+\frac{3}{z})^2} = \frac{1}{z^2}\left(\sum_{n=0}^\infty \left(-\frac{3}{z}\right)^n\right)^2$$ for $$\displaystyle |z|>3$$.

But the whole series is squared. How am I suppose to write out the first few terms?
Hint: $$\displaystyle \frac{1}{(1+z)^2}=-\left(\frac{1}{1+z}\right)'$$...

scorpion007

#### scorpion007

Ok, so do I subs that in one of the factors and write it in terms of d/dz ?

I'm not sure what do with that...

#### Drexel28

MHF Hall of Honor
Ok, so do I subs that in one of the factors and write it in terms of d/dz ?

I'm not sure what do with that...
$$\displaystyle \left(\frac{1}{1-x}\right)'=\left(\sum_{n\in\mathbb{N}}x^n\right)'=\sum_{n\in\mathbb{N}}nx^{n-1}$$

scorpion007

#### scorpion007

I guess you want me to take the deriv of both sides?

$$\displaystyle f'(z) =\frac{2}{z^2}\sum_{n=0}^\infty \left(-\frac{3}{z}\right)^n\cdot \sum_{n=0}^\infty n \left(-\frac{3}{z}\right)^{n-1}\cdot \frac{3}{z^2}$$

is that right? I could simplify this and all, but I don't quite see where this is going...

#### Opalg

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