When $\cosh (z)=0$? Ask Question Asked 9 years, 1 month ago. Modified 9 years, 1 month ago. Viewed 32k ...

2015年10月8日 · Free Online Scientific Notation Calculator. Solve advanced problems in Physics, Mathematics and Engineering. Math Expression Renderer, Plots, Unit Converter, Equation Solver, Complex Numbers, Calculation Histo

2015年9月7日 · $\sinh x \cosh x = 0$ $\frac 12 \sinh 2x = 0$ $\sinh 2x = 0$ The only solution to that is $2x = 0 \implies x = 0$. Alternatively, you can simply observe that $\cosh x$ is always non-zero, and the only solution comes from $\sinh x = 0$. Updated: in the complex numbers, $2x = k\pi i \implies x = \frac 12 k\pi i, k \in \mathbb{Z}$. See my comment ...

Now here I thought this implied $\sin(x) = 0$ or $\cosh(y) = 0$. However the solutions give only $\sin(x) = 0$, why? Also, when doing the second equation, you get $\cos(x)\sinh(y) = 0$, but this time the solutions split the case into $\cos(x)=0$ and $\sinh(y)=0$ as I did above.

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$\begingroup$ The function $\cosh$ is even, and does not have an inverse. But if we restrict $\cosh$ to the interval $[0,\infty]$ (the usual choice), the inverse function has the $+$. If we restrict to the interval $(-\infty,0]$, then $-$ is right. $\endgroup$ –

2015年10月1日 · I know that for large positive numbers cosh(x) and sinh(x) would almost be equal to $e^x/2$ as $e^{-x}/2$ would become negligible given the magnitude of x in both cases.

2016年10月7日 · $\begingroup$ @Nan when $\cos(x)$ is positive, clearly the product is larger than $-1$. This happens at least at every even integer multiple of $\pi$.

Prove the following theorem: The zeros of $\sinh z$ and $\cosh z$ in the complex plane all lie on the imaginary axis.

When plugging to the original equation, the negative solution from the inverse cosh definition is the only solution that does not work. Is it safe to assume that rearranging for cosh and solving in terms of sinh will provide the correct solutions, and that …