0
$\begingroup$

I am new to complex analysis , this was a example problem and the author just says that as $z=\pi/4$ an isolated singularity , it is clear that the order of the pole is one.

But I am not able to see why ? From what I understood , a pole of order one , means in the Laurent's expansion , the negative term's order is maximum of one . How can one deduce the expansion from just looking at an isolated singularity

$\endgroup$
  • 2
    $\begingroup$ $\cos(z)-\sin(z) = -\sqrt{2}\sin(z-\pi/4)$. $\endgroup$ – Somos Apr 5 at 16:22
  • $\begingroup$ I understood how $z=\pi/4$ is a singularity , I am not understanding how it became a pole of order one $\endgroup$ – Vinay Varahabhotla Apr 5 at 16:50
  • $\begingroup$ What about $1/\sin(z)$ at $z=0$? $\endgroup$ – Somos Apr 5 at 17:21
  • $\begingroup$ Just find its laurent expansion, if i'm not mistaken it should be $\frac{1}{\sqrt{2}y}-\frac{y}{6\sqrt{2}} + \mathcal{O}(y^3) $, where $y=(x-π/4)$. $\endgroup$ – Alexandros Apr 5 at 20:01
0
$\begingroup$

The comment of Somos gives an elegant solution, but alternatively, let be $$g(z) = \cos(z) - \sin(z).$$ $\pi/4$ is a zero of order 1 of $g$ because (check yourself) $g(\pi/4) = 0$ and $g'(\pi/4)\ne 0$.

$\endgroup$

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.