0
$\begingroup$

Hi guys I was wondering how I can understand if the sin and the cos has essential singularities. for instance if I want to understand if 0 which singularity is i, can write the Laurent series only of the sin (centred in 0) and see how it works , or MUST write the Laurent series of all the function (centered in zero) ? Same for cos , help I want to understand this topic very well. Thk.

$$\int_{+\partial D}\dfrac{\sin\left(\dfrac{1}{z}\right)\cos\left(\dfrac{1}{z-2}\right)}{z-5}\,\mathrm{dz}$$

$\endgroup$
  • $\begingroup$ ok sorry , i was tryng to write the laurent series but it's difficoult so i want to understand if i have to write the resultin series of all the functions o i can just write the series of the sin( for instance) to understan which singularity is $\endgroup$ – xmaionx Jul 20 '18 at 9:04
  • $\begingroup$ you only need to write down the Laurent series for $\sin (1/z)$ centered at $0$. The other part $\cos (1/(z-2))/(z-5)=c_0+ c_1z+c_2z^2+\text{...}$ is analytic inside the unit disk. Then multiply the two series $\endgroup$ – Lozenges Jul 20 '18 at 10:38
  • $\begingroup$ @Lozenges so for instance if i want to see if z=2 is an essential singularity i just wtrite the cos (1/(z-2)) series centered in 2 ?? $\endgroup$ – xmaionx Jul 24 '18 at 14:56
0
$\begingroup$

Suppose that $f$ is an entire function, let $a \in \mathbb C$ and define $g(z):=f(\frac{1}{z-a})$ for $z \ne a$.

We have $f(z)= \sum_{n=0}^{\infty}a_nz^n$ for all $z$ (Taylor).

Then we get

$g(z)=\sum_{n=0}^{\infty}\frac{a_n}{(z-a)^n}$ for $z \ne a$ (Laurent).

Now you see: $a$ is an essential singularity of $g \iff f$ is not a polynomial.

$\endgroup$
  • $\begingroup$ so to see if they are essential singularities i can use the laurent serie only of the sin or the cos ? o o i have to write the serie of all the functions and see the resulting series? $\endgroup$ – xmaionx Jul 20 '18 at 9:03
  • $\begingroup$ and why i can study only the sin if it's possible ? $\endgroup$ – xmaionx Jul 20 '18 at 10:15

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.