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


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

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


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

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?

  • $\begingroup$ 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. $\endgroup$
    – saulspatz
    May 31 '21 at 19:39
  • $\begingroup$ @saulspatz I didn't quite get why it has pole only at $z=i$ $\endgroup$
    – Snowball
    May 31 '21 at 19:46
  • $\begingroup$ 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 $-1$. $\endgroup$ May 31 '21 at 19:57
  • $\begingroup$ $z^4-1=(z^2+1)(z^2-1)=(z-i)(z+i)(z-1)(z+1)$ $\endgroup$ May 31 '21 at 19:58
  • $\begingroup$ 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) $\endgroup$
    – Snowball
    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:

$$a_{-1}=\lim_{z\to i}(z-i)f(z) =\lim_{z\to i}\frac{z^2}{(z+i)(z^2-1)}=\frac{-1}{2i(-1-1)}=\frac1{4i}$$

and thus yes: your work is correct.

  • $\begingroup$ Aren't $z = -i,1,-1$ also poles in given domain $0 < |z-i| < \sqrt{2}$? $\endgroup$
    – Snowball
    May 31 '21 at 19:45
  • 1
    $\begingroup$ @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 $\endgroup$
    – DonAntonio
    May 31 '21 at 19:55
  • $\begingroup$ Sorry, I just noticed it now. Thanks for the answer. $\endgroup$
    – Snowball
    May 31 '21 at 19:57

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