I would like to expand a bit on this answer, because the lack of a solution for tan z = i tan z = i dovetails with the characteristics of the singularity of the function at infinity. In the complex domain, all periodic …

Apr 7, 2016 · My thoughts are that I could use ez = ex+iy = ex(cos(y) + i sin(y)) e z = e x + i y = e x (cos (y) + i sin (y)) to express both the numerator and denominator in trig form. Then I could times both by …

because tan(z) = sin(z) cos(z) tan (z) = sin (z) cos (z) the poles are when cos(z) = 0 cos (z) = 0 at z = ±π/2 ± nπ, n ∈ Z z = ± π / 2 ± n π, n ∈ Z Poles inside |z| = 2 | z | = 2 are ±π/2 ± π / 2 and are of first …

Jul 11, 2019 · 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 ...

Jan 2, 2015 · Res(f(z), 0) =limz→0(z − 0) × tan z z = limz→0 tan z = 0 Res (f (z), 0) = lim z → 0 (z 0) × tan z z = lim z → 0 tan z = 0

I try to expand tan x tan x in Taylor order to o(x6) o (x 6), but searching of all 6 derivative in zero (ex. tan′(0),tan′′(0) tan (0), tan ″ (0) and e.t.c.) is very difficult and slow method. Is there another way to …

Sep 18, 2017 · How can I prove that $\tan (z)$ is an analytic function? I tried with the expansion for $\tan (A+B)$. Couldn't complete. Please help.

Nov 26, 2019 · Q: If tan(z) = ∑∞ k=0akzk tan (z) = ∑ k = 0 ∞ a k z k based at z = 0 z = 0, show that ak = 0 a k = 0 for every even index k k. Furthermore, calculate a1,a3 a 1, a 3. So finding a1,a3 a 1, a 3 is …

Sep 26, 2017 · How do i find domain of tan(z) tan (z) that is differentiable?? I haven seen forums where people showed until this part

Jun 13, 2015 · I want to determine the Laurent expansion of tanz tan z around z = 0 z = 0 that is convergent in z = π z = π (only the first couple of terms). Now I know that if ∑∞ n=−∞cnzn ∑ n = ∞ ∞ …