How can I solve $\cosh(z) = 2$ for $z\in \mathbb{C}?$ My steps have only led to values of $z$ which are not imaginary, those being $\ln(2+\sqrt3)$ and $\ln(2-\sqrt3)$. How do I find solutions to the equations that are imaginary?
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https://math.stackexchange.com/questions/1625848/when-cosh-z-0 – G_D Oct 02 '17 at 05:45
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Sorry, I accidentally typed in $0$. Fixed it now – Anonymous Oct 02 '17 at 05:47
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3$\cosh$ has period $2\pi i$. – Angina Seng Oct 02 '17 at 05:49
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@LordSharktheUnknown sorry I still don't understand – Anonymous Oct 02 '17 at 05:50
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Using the fact that $e^{2 \pi i}=1$, if $e^x=2+\sqrt{3}$, then $e^{x+2\pi i}=e^{x+4\pi i}=e^{x+2n\pi i}....=2+\sqrt{3}$.
A similar argument for $e^{x}=2-\sqrt{3}$.
Therefore, the solutions are $\ln(2 \pm \sqrt{3})+2n\pi i$
Saketh Malyala
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