0
$\begingroup$

Im attempting to solve a problem of defining the types of singularities of a complex function. I found that if the function has a Laurent series expansion with a finite number of terms in the principal part, the singularities are "poles". If none exist they they are "removable", and if infinitely many exist they are "essential". To find these expansions I decomposed my function into parts and most of them checked out nicely. I seem to be having an issue however finding the correct Laurent series of $$\frac{3z}{2z+3} \quad at \quad z=-\frac{3}{2}$$ I've attempted to expand it in a geometric series fashion. $$\frac{3z}{2z+3} = \frac{3}{2+\frac{3}{z}}\\\rightarrow\frac{3}{2}\times\frac{1}{1+\frac{3}{2z}} = \frac{3}{2}\sum_{n=0}^{\infty}\left(\frac{-3}{2 z}\right)^{n}$$ Which would go to $$\frac{3}{2}\sum_{n=0}^{\infty}\left(\frac{-3}{2 (z+\frac{3}{2})}\right)^{n}$$I'm assuming a lapse in logic here somewhere as obviously this gives me infinitely many terms in the principle part of the Laurent series which I know to not be correct (Using Mathematica). The n = 0 and n= 1 terms seem to be the only ones that should exist. Thanks in advance for any advice offered.

$\endgroup$
0
$\begingroup$

This can also be written as $$\begin{align} \frac{3z}{2z+3} &=\frac32\left(\frac{z+\frac32-\frac32}{z+\frac32}\right)\\ &=\frac32\left(1-\frac32\left(\frac{1}{z+\frac32}\right)\right)\\ &=\frac32-\frac94\left(\frac1{z+\frac32}\right)\\ \end{align}$$

$\endgroup$
  • $\begingroup$ Thank you, that seems to be correct. $\endgroup$ – Aaron Fitzpatrick Apr 17 at 13:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.