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 complex domain, …

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 ...

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 ...

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 lead to 2 new …

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Jan 25, 2013 · This answer is inspired by the answer of coffeemath. We know the following (from e.g. Wikipedia): \begin {align} &\sin (x) = x -\frac {x^3} {6} +\frac {x^5} {120}+ O ...

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Oct 10, 2021 · I'm trying to solve equation $\\tan(z) = 2i$ in set of complex numbers. My work so far Let's rewrite $\\sin(z)$ and $\\cos(z)$ using imaginary unit: $$\\sin(z ...

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.

Nov 16, 2014 · I need help calculating the laurent series of $\tan z$ around the points $z=0$, $z=\pi/2$, and $z=\pi$. How would one go about doing this? I solved an almost ...