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I'm looking at the wolframalpha residue calculator for the following equation: $$\frac{\tan(z)+2}{4z^2+\pi z}$$ which gives me the residue: $2/\pi $ for Root 1 which I found, but it also says residue: $-1/\pi$ for Root 2.

Root 1: $z=0$

Root 2: $z=-\pi/4$

If you insert Root 2 into the equation. $\frac{\tan(z)+2}z$ you get the following: $\frac{\tan(-\pi/4)+2}{-\pi/4}$

The curious thing is that I found the residue to be $-4/\pi$ and not $-1/\pi$ for Root 2.

Is there a step that I am missing or is wolframalpha wrong in this case?

Link to wolframalpha residue calculator: https://www.wolframalpha.com/widgets/gallery/view.jsp?id=186fb23a33995d91ce3c2212189178c8

Gary
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Claire
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  • We have $$ (z - ( - \pi /4))\frac{{\tan z + 2}}{{4z^2 + \pi z}} = \frac{{\tan z + 2}}{{4z}}. $$ Now let $z\to -\pi/4$. – Gary Mar 17 '23 at 00:47

1 Answers1

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Here's how to find the the residue at $-\frac{\pi}{4}$, Write $z=z-(-\frac{\pi}{4})+\frac{\pi}{4} = z+\frac{\pi}{4}-\frac{\pi}{4}$. Let $w=z+\frac{\pi}{4}$. We have to write your expression as a Laurent series in $w$ and find the coefficient of $w^{-1}$ in that Laurent series. A little bit of basic algebra shows that the denominator of your expression is $-\pi(1-\frac{4}{\pi}w)w$. The numerator of your expression is \begin{align*} \tan(w-\frac{\pi}{4})+2 & =\frac{\tan w-1}{1+\tan w}+2 \\ & =\frac{-1+w(1+w^2/3+\ldots)}{1+w(1+w^2/3+\ldots)}+2\\ &=(-1+w+\ldots)(1-w+\ldots)+2\\ & =1+2w+\ldots \end{align*} So your whole expression is \begin{align*}& =\frac{1+2w+\ldots}{\pi(1-\frac{4}{\pi}w)w}\\ & =\frac{(1+2w+\ldots)(1+\frac{4}{\pi}w+\ldots)}{-\pi w} \end{align*} and the cefficient of $w^{-1}$, i.e. the coefficient of $(z+\frac{\pi}{4})^{-1}$ is $-\frac{1}{\pi}$, which is the residue at the pole $-\frac{\pi}{4}$.

Gary
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P. Lawrence
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  • This works but it is much simpler just to multiply by $z+\pi/4$ and substitute $-\pi/4$ as in my comment above. – Gary Mar 17 '23 at 05:31