# Finding the order of the pole of the complex function $f(z)=\frac{1}{\cos(z)-\sin(z)}$

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

• $\cos(z)-\sin(z) = -\sqrt{2}\sin(z-\pi/4)$. – Somos Apr 5 at 16:22
• I understood how $z=\pi/4$ is a singularity , I am not understanding how it became a pole of order one – Vinay Varahabhotla Apr 5 at 16:50
• What about $1/\sin(z)$ at $z=0$? – Somos Apr 5 at 17:21
• 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 $y=(x-π/4)$. – Alexandros Apr 5 at 20:01

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 $$g'(\pi/4)\ne 0$$.