# Laurent Series of $~\tan(z)~$ expanded in $\frac{\pi}{2} < |z| < \frac{3\pi}{2}~$?

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!

• you cannot use the same because you pass through the singularities at $\pm {\frac{\pi}{2}}$ – Conrad Jul 11 '19 at 17:09
• The radius of convergence of your $\tan$ series is $\pi/2$, so it diverges in the region $\frac{\pi}{2} < |z| < \frac{3\pi}{2}$ – GEdgar Jul 12 '19 at 13:34

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 $$z$$.
• 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! – Kuei Kao Jul 12 '19 at 3:55
• It's not just two new terms, it's infinitely many: you will get new contributions for every odd $n$, positive or negative. – Robert Israel Jul 12 '19 at 12:20
• The Taylor series of $\sin(z)$ and $\cos(z)$ are valid everywhere, but the series of $1/\cos(z)$ is not. – Robert Israel Jul 12 '19 at 12:21