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I have the following function:

$f(z) = \frac{\cos z}{\left(z-\frac{\pi}{2}\right)^3}$

Where I have to classify the singularity $z_o=\frac{\pi}{2}$

I've taken the limit to see if it's a removable singularity since it is indeterminate when plugging in $z_o$ but the limit does not exist.

I'm conflicted here -- I know it's not removable because the limit diverges, but then what type of singularity is it? Is it just a pole of order 3?

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3 Answers 3

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It is a pole of order $2$$$\lim_{z\to\dfrac{\pi}{2}}(z-\dfrac{\pi}{2})^2f(z)=\lim_{z\to\dfrac{\pi}{2}}(z-\dfrac{\pi}{2})^2\frac{\cos z}{\left(z-\frac{\pi}{2}\right)^3}=\lim_{z\to\dfrac{\pi}{2}}\frac{\cos z}{z-\frac{\pi}{2}}=\lim_{z\to\dfrac{\pi}{2}}\frac{-\sin z}{1}=-1$$

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  • $\begingroup$ Thanks! But what about it tells you it's a pole of order 2? $\endgroup$
    – mathwiz97
    Mar 20, 2018 at 0:19
  • $\begingroup$ In fact a pole $x_0$ of order $m$ of function $f(x)$ is such that $$\lim_{x\to x_0}(x-x_0)^mf(x)\in\Bbb R-\{0\}$$ $\endgroup$ Mar 20, 2018 at 7:34
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Hint: $\cos(\pi/2) = 0$ and the order of this zero is...

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Note that\begin{align}\cos(z)&=\cos\left(\left(z-\frac\pi2\right)+\frac\pi2\right)\\&=-\sin\left(z-\frac\pi2\right)\\&=-\left(z-\frac\pi2\right)+\frac1{3!}\left(z-\frac\pi2\right)^3-\frac1{5!}\left(z-\frac\pi2\right)^5+\cdots\end{align}Can you take it from here?

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