Jul 9, 2018 · I would like to expand a bit on this answer, because the lack of a solution for $\tan z=i$ dovetails with the characteristics of the singularity of the function at infinity. In the …

Jul 28, 2018 · This is where I'm at: I know $$ \\cos(z) = \\frac{e^{iz} + e^{-iz}}{2} , \\hspace{2mm} \\sin(z) = \\frac{e^{iz} - e^{-iz}}{2i}, $$ where $$ \\tan(z) = \\frac{\\sin(z ...

Jan 3, 2015 · Using residue theorem, evaluate the following; $C:\left| {z - 1} \right| = 2$ $$\int_c { { {\tan z} \over z}dz}$$ I want you guys to check my answer.Is it correct ...

Oct 20, 2014 · You can split up the limit into: $$2\pi i\lim_ {z\rightarrow \frac {\pi} {2}} \frac {\sin z \cdot (z-\pi/2)} {\cos z} \: = 2\pi i (\lim_ {z\rightarrow \frac {\pi} {2 ...

Nov 22, 2020 · What are the poles of tanz/z and what is the best way to find them? I know what the answer is and the way I found it was to rewrite tanz/z as sinz/cosz and got 0 and (2n+1)pi/2.

Jul 11, 2019 · 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 …

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Aug 13, 2015 · You could explore $\tan (x_1+x_2+\dots x_r)$ - in terms of the $\tan x_i$. The result is a fraction with "odd" combinations of the $\tan x_i$ in the numerator and "even" …

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My professor told me that there's a non-isolated singularity of $\tan (z)$ as $z$ tends to infinity. However, I'm wondering why is this the case, since for $\tan (z ...