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calculate the principal part of $\tan(z)$ at $\frac{\pi}{2}$.

of course $\tan(z) = \frac{\sin(z)}{\cos(z)}.$ Because $\cos(z)$ is of order 1 in $\frac{\pi}{2}$ we know that our primal part must look like:

$c_{-1}(z-\frac{\pi}{2})^{-1}$.

The main problem here is that i don't seem to get how to write the sine and cosine. Are there any neat tricks to see this?

Kees

janmarqz
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Kees Til
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    Some English math terminology fixes: Principal part, not primal part, and "at $\pi/2$", not "in $\pi/2$". – aes May 01 '15 at 19:52
  • thanks, my textbook is dutch so these mistakes can be made fast :) – Kees Til May 01 '15 at 19:53
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    More terminology fixes: "sine," not "sinus," and "cosine," not "cosinus." – Jack Lee May 01 '15 at 19:53
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    To get the principal part of $\tan(z)$ you only need to compute $\lim_{z\to\pi/2}(z-\pi/2)\tan(z)=\lim_{z\to\pi/2}\frac{(z-\pi/2)}{\cos(z)}=\lim_{z\to\pi/2}\frac{1}{-\sin(z)}=-1$. The existence of this limit tells you that $(z-\pi/2)\tan(z)$ is analytic (removable singularity) at $z=\pi/2$ and that the principal part of $\tan(z)$ is $-1/(z-\pi/2)$. – Alamos May 01 '15 at 20:21
  • hmmm i haven't heard of this trick. If one wants to calculate principal parts of a function at a certain point c, it is enough to calculate $\lim_{z \rightarrow c}(z-c)f(z)$? – Kees Til May 01 '15 at 20:28

2 Answers2

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Just compute the taylor series (which equals the function because these are analytic):

$f(z) = f(c) + f'(c) (z-c) + \frac{f''(c)}{2} (z-c)^2 + \cdots$

Use $c = \pi/2$.

$\cos(z) = \cos(\pi/2) - \sin(\pi/2)(z-c) + \ldots = -(z-c) + \ldots$

$\sin(z) = \sin(\pi/2) + \ldots = 1 + \ldots$

Now you can get the principal part by dividing these (you don't need any more terms than I gave to know the degree $-1$ part).

aes
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  • hmmm i already thought about this, but does one not get any trouble in the denominator. We can't just ignore the restterms right? – Kees Til May 01 '15 at 20:05
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    @KeesTil Sure you can. Factor $z-c$ out of the denominator, and then it's $\frac{1}{z-c}\frac{1+o(1)}{-1 + o(1)}$ and the $\frac{1+o(1)}{-1+o(1)} = -1 + o(1)$ (also/alternatively, that part is analytic near $c$). – aes May 02 '15 at 04:36
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$$\lim_{z\to \pi/2}(z-\pi/2)\frac{\sin z}{\cos z} = \lim_{z\to \pi/2} \frac{z-\pi/2}{\cos z - \cos (\pi/2)}\sin z$$ $$ = \frac{1}{\cos '(\pi/2)}\cdot 1= -1.$$ It follows that $\tan z$ has a simple pole at $0,$ hence the prinicpal part is $-1/(z-\pi/2).$

zhw.
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  • hmmm i haven't heard of this trick. If one wants to calculate principal parts of a function at a certain point c, it is enough to calculate limz→c(z−c)f(z)?

    Or is this only usefull if you have a simple pole?

     – Kees Til May 01 '15 at 20:59
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    It works for simple poles only. With a higher order pole, the limit will be $\infty.$ – zhw. May 01 '15 at 21:06