Finding the order of a zero

Question

Studying for my complex analysis final, and I came across this question:

Find the order of the zero of

$$ f(z) = \frac{\cos(z)}{z-\pi/2} $$

at $z=\pi/2$.

The order of the zero would just be one, right? This is because the polynomial in the denominator is of degree one.

Answer 1

Hint: Use the Taylor series of $\cos(x)$ at the point $z=\pi/2$

$$ \cos(x) = -(z+\pi/2) +O \left( (z-\pi/2 \right) ^{3} ). $$

Check other techniques.

Answer 2

$x=\pi/2$ is a simple zero both for $\cos x$ and $x-\pi/2$.

Hence $x=\pi/2$ is not a zero of your function, as you can see by computing: $$\lim_{x\to\pi/2}\frac{\cos x}{x-\pi/2}=\lim_{x\to\pi/2}\frac{-\sin x}{1}=-1.$$

Answer 3

You can think this way:

$f(z)$ has removable singularity in $z=\frac{\pi}{2}$, because:

$$\lim_{z \to \frac{\pi}{2}} \frac{\cos z}{z-\frac{\pi}{2}}=\lim_{z \to 0}\frac{-\sin z}{z}=-1$$

So the order of $z=\frac{\pi}{2}$ is zero.

Answer 4

$f(\pi/2)\neq 0$ then it's a zero of order $0$.