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As we know, we can get Laurent series of $~\tan(z)~$ expanded in $~0 \le |z| \lt \frac{\pi}{2}~$ by dividing Taylor series expansion of $~\sin(z)~$ by Taylor series expansion of $~\cos(z)~$, and we'll get $$\tan(z) = \frac{\sin(z)}{\cos(z)} = \frac{z-\frac{z^3}{3!}+\frac{z^5}{5!}+\dots}{1-\frac{z^2}{2!}+\frac{z^4}{4!}+\dots} = z+\frac{z^3}{3}+\frac{2z^5}{15}+\dots$$

Now I am curious about the Laurent series of $~\tan(z)~$ expanded in $~\frac{\pi}{2}~< |z| < \frac{3\pi}{2}$.

Since Taylor series of $~\sin(z)~$ and $~\cos(z)~$ is valid everywhere on the complex plane, I originally think the answer will be the same as above.

But one exercise on Brown&Churchill's Complex variables and applications states that value of the integral $$\oint\limits_C \tan(z)\,dz = -4\pi i$$ where path $~C~$ is the positively oriented circle $~|z| = 2~$, which lies in the domain $~\frac{\pi}{2}~ < |z| < \frac{3\pi}{2}~$.

If Laurent series of $~\tan(z)~$ expanded in $~\frac{\pi}{2}~ < |z| < \frac{3\pi}{2}~$ is really $$z+\frac{z^3}{3}+\frac{2z^5}{15}+\dots$$, then the residue (coefficient of order $~-1~$) is zero, which means the integral should be $~0~$, contradicting the result of the exercise.

So it is clear that the above series form is not the correct Laurent series of $~\tan(z)~$ in domain $~\frac{\pi}{2}~ < |z| < \frac{3\pi}{2}~$.

Then what is the correct Laurent Series of $~\tan(z)~$ in domain $~\frac{\pi}{2}~ < |z| < \frac{3\pi}{2}~$?

And can somebody help me explain why I can't use Taylor Series of $~\sin(z)~$ and $~\cos(z)~$ to derive the Laurent Series?

What's wrong with my concept?

Thank you very much!

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  • $\begingroup$ you cannot use the same because you pass through the singularities at $\pm {\frac{\pi}{2}}$ $\endgroup$ – Conrad Jul 11 at 17:09
  • $\begingroup$ The radius of convergence of your $\tan$ series is $\pi/2$, so it diverges in the region $\frac{\pi}{2} < |z| < \frac{3\pi}{2}$ $\endgroup$ – GEdgar Jul 12 at 13:34
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The Laurent series of $f(z)$ in an annulus $a < |z| < b$ (assuming the function is analytic there) is $\sum_{n=-\infty}^\infty c_n z^n$ where $$ c_n = \dfrac{1}{2\pi i} \oint_C f(z) z^{-n-1}\; dz $$ for a simple closed positively-oriented contour $C$ that goes around $0$ in this annulus. In your case, you can compute this using residues: the poles to consider are $0$ and $\pm \pi/2$. The residues at $0$ give you the same coefficients as you would get for the Maclaurin series. The residues at $\pm \pi/2$ give you something new.

EDIT: Another way of looking at it: $g(z) = \tan(z) + \frac{1}{z-\pi/2} + \frac{1}{z+\pi/2}$ is analytic in $|z|<3\pi/2$ (check that the singularities at $\pm \pi/2$ are removable). Its Maclaurin series is the sum of the Maclaurin series of $\tan(z)$, $1/(z-\pi/2)$ and $1/(z+\pi/2)$. To get a series for $f(z)$, you need to subtract the $1/(z-\pi/2)$ and $1/(z+\pi/2)$, and to do this in a way that is convergent in your annulus, you must use their Laurent series for $|z| > \pi/2$, which involve negative powers of $z$.

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  • $\begingroup$ Thank you for your reply. Now I know what you're saying. We can derive Laurent series of tan(z) through direct integration, and the integration around two poles $\pm \pi/2$ will lead to 2 new terms. But could you help me explain why I can't use Taylor series of sin(z) and cos(z) to get Laurent series of tan(z) in that domain? Since the Taylor series for the two function is valid everywhere, I don't understand why I can't do such way. Thank you very much! $\endgroup$ – Kuei Kao Jul 12 at 3:55
  • $\begingroup$ It's not just two new terms, it's infinitely many: you will get new contributions for every odd $n$, positive or negative. $\endgroup$ – Robert Israel Jul 12 at 12:20
  • $\begingroup$ The Taylor series of $\sin(z)$ and $\cos(z)$ are valid everywhere, but the series of $1/\cos(z)$ is not. $\endgroup$ – Robert Israel Jul 12 at 12:21

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