# Find the principal part for the Laurent series

Find the principal part for the following Laurent series: $$\frac{z^{2}}{z^{4}-1}$$ for $$0< |z-i| <\sqrt{2}$$

Attempt:

$$\frac{z^{2}}{z^{4}-1} = \frac{1}{2(z^2+1)} + \frac{1}{2(z^2-1)} = \frac{1}{2(z+i)(z-i)} + \frac{1}{2(z-1)(z+1)} \text{ (1)}$$

Also, $$\frac{1}{z+1} = \frac{1}{(1+i)(1 + \frac{z-i}{1+i})} = \frac{1}{1+i}\sum_{n=0}^{\infty}(-1)^n(\frac{z-i}{1+i})^n$$

and

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

Therefore, principal part from LHS in (1) is $$\frac{1}{4i(z-i)}$$ and from RHS, it will be $$0$$ I guess. Therefore, principal part should be $$\frac{1}{4i(z-i)}$$.

Am I correct? I am still not sure about the last part (principal part of $$\frac{1}{2(z-1)(z+1)}$$ being $$0$$). Can anyone give tips?

• You are correct, but notice that the function has a simple pole at $z=i$, so all you have to do is compute the residue. May 31 '21 at 19:39
• @saulspatz I didn't quite get why it has pole only at $z=i$ May 31 '21 at 19:46
• That's not what @saulspatz wrote. It was that it has a simple pole at $i$. It also has simple poles at $-1$, at $1$, and at $-1$. May 31 '21 at 19:57
• $z^4-1=(z^2+1)(z^2-1)=(z-i)(z+i)(z-1)(z+1)$ May 31 '21 at 19:58
• Okay I understood the solution using poles. But is there any solution following my attempt? (without residue, or checking singularities, just by using taylor series) May 31 '21 at 20:00

Observe that in the given domain your function only has one simple pole, namely $$\;z=i\;$$, and thus the principal part of its Laurent series there will contain only one single element, namely $$\;\cfrac{a_{-1}}{z-i}\;$$ . In this case, thus, the easiest way to solve your problem is evaluating the residue $$\;a_{-1}\;$$ using Cauchy's Formula:
$$a_{-1}=\lim_{z\to i}(z-i)f(z) =\lim_{z\to i}\frac{z^2}{(z+i)(z^2-1)}=\frac{-1}{2i(-1-1)}=\frac1{4i}$$
• Aren't $z = -i,1,-1$ also poles in given domain $0 < |z-i| < \sqrt{2}$? May 31 '21 at 19:45
• @Snowball Nop, they aren't as you can easily check by sustituing. For example, for $\;z=-1\;$ we get $$|-1-i|=\sqrt2\nless\sqrt 2$$ The same is true with those other two May 31 '21 at 19:55