Real and Imaginary Parts of tan (z) - Mathematics Stack Exchange
网页2016年4月7日 · Applying the above, with a little manipulation, gives me: tan(z) = i(e − iz − eiz) eiz + e − iz. My thoughts are that I could use ez = ex + iy = ex(cos(y) + isin(y)) to express both the numerator and denominator in trig form. Then I could times both by the denominator's complex conjugate as to get a real denominator, which would then, in ...
$\\tan(z)=i$, no solution - Mathematics Stack Exchange
网页Here's a simpler way to proceed: tan(z) = i implies that sin(z) = icos(z). Multiplying by i on both sides yields isin(z) = − cos(z). Now consider Euler's identity: eiz = cos(z) + isin(z), but substituting our earlier equation in gives us that eiz = cos(z) + (− cos(z)) = 0.
$\\tan(z)$ with residue theorem - Mathematics Stack Exchange
网页These are the poles of the function tanz Among these the two poles z = ± π / 2 lie inside the circle | z | = 2. Residue of f(z) at π / 2 = − 1 and Residue of f(z) at − π / 2 = − 1. By the Residue Theorem, we have. the answer to be = 2πi(sum of the residues ) …
Laurent Series of - Mathematics Stack Exchange
网页2019年7月11日 · $\begingroup$ 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.
Type of singularity for $\\tan(z)$ at $z = \\frac{\\pi}{2}$
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How do i find domain of - Mathematics Stack Exchange
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Solving equation $\\tan(z) = 2i$ - Mathematics Stack Exchange
网页2021年10月10日 · 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) = eix − e − ix 2i. cos(z) = eix + e − ix 2. Out of these facts we have that tan(z) = sin(z) cos(z) = eix − e − ix eix + e − ix ⋅ 1 2. eix − e − ix eix + e − ix ⋅ 1 2 = 2i. eix − ...
Proving $\\tan(z)$ is analytic. - Mathematics Stack Exchange
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Are all singularities of - Mathematics Stack Exchange
网页2023年5月3日 · 2. They are all simple poles because the poles are due to the cosine function (the sine is entire and is nonzero at all of the poles of the tangent), which only has simple zeroes. This last fact can be verified by, say, L'Hopital's rule. limz→π 2(2n+1) cos(z) z − π 2(2n + 1) = (−1)n+1 ∈C ∖ {0} lim z → π 2 (2 n + 1) cos (z) z − ...
complex analysis - How to determine the Laurent expansion of …
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