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Resolver cosh(iz) = 0 https://forumdematematica.org/viewtopic.php?f=20&t=5476 |
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Autor: | Sobolev [ 19 mar 2014, 22:06 ] |
Título da Pergunta: | Re: Resolver cosh(iz) = 0 [resolvida] |
\(\cosh (iz) = 0 \Leftrightarrow \frac{1}{2}(e^{iz} + e^{-iz}) = 0 \Leftrightarrow e^{ix -y} + e^{-ix + y} = 0 \Leftrightarrow e^{-y}(\cos x + i \sin x) + e^{y}(\cos x - i \sin x)=0\Leftrightarrow (e^{-y}+e^y) \cos x - i(e^y-e^{-y}) \sin x = 0\Leftrightarrow \cos x = 0 \wedge (e^y-e^{-y})\sin x = 0 \Leftrightarrow x = \frac{\pi}{2} + k \pi \wedge e^{y} = e^{-y}\Leftrightarrow x = \frac{\pi}{2} + k \pi \wedge y = 0\Leftrightarrow z = \frac{\pi}{2} + k \pi, k \in \mathbb{Z}\) |
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