Find the Laurent Series representations in powers of $z$ for

i) $\frac{ \cos{z}}{z}$

ii) $z^4 \cosh{\frac{1}{z^2}}$

Where do they converge?

I found the Laurent Series for each of the functions, yielding the result

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

ii) $\sum_{n=0}^\infty \frac{z^4}{(2n)!(z^2)^{2n}}$

Both of these functions have a pole at $z=0$, so I want to say that the answer to where they converge is $z \neq 0$. However, I am under the impression that a Laurent series should have a convergence along the lines of $r< |z-z_0| < R$.


1 Answer 1


Just take $z_0:=0$ giving $$r< |z-0| < R$$ or $$r< |z| < R$$ for the desired set of convergence.

  • $\begingroup$ Yes, but what is $r$ and $R$? That is what I am trying to figure out. $\endgroup$ Mar 11, 2015 at 22:23
  • $\begingroup$ @Snagglewhen $r$ and $R$ are just any positive real numbers: $0<r<R$. $\endgroup$ Mar 11, 2015 at 22:24
  • $\begingroup$ Yes, but in this particular case? Is what I assumed above true and the series converges for $z \neq 0$? In that case, how would I represent that in the above form? I could say that it converges for $0 < |z| < \infty$. $\endgroup$ Mar 11, 2015 at 22:27
  • $\begingroup$ @Snagglewhen Yes, you can say it like this:the Laurent series is convergent on the set $0 < |z| < \infty$. Thanks. $\endgroup$ Mar 11, 2015 at 22:32
  • $\begingroup$ Thanks. I was simply trying to make sure that I was not mistaken. $\endgroup$ Mar 11, 2015 at 22:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .