Since ez e z is never zero for any z ∈C z ∈ C, no solution exists for e−z = 0 e − z = 0:

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

Sep 7, 2015 · The hyperbolic functions are quite different from the circular ones. For one thing, they are not periodic. For your equation, the double-"angle" formula can be used: sinh x cosh x = 0 sinh x cosh x = 0 12sinh 2x = 0 1 2 sinh 2 x = 0 sinh 2x = 0 sinh 2 x = 0 The only solution to that is 2x = 0 x = 0 2 x = 0 x = 0. Alternatively, you can simply observe that cosh x cosh x is always non-zero, and ...

So if cosh x = 0 cosh x = 0, then ex +e−x = 0 e x + e − x = 0. But this can't happen because both ex e x and e−x e − x are positive, no matter what real value x x has. And the sum of two positive numbers can never be zero. Therefore cosh x cosh x can never be zero for real values of x x. Also, when doing the second equation, you get cos(x) sinh(y) = 0 cos (x) sinh (y) = 0, but this ...

But if we restrict the domain of cosh cosh suitably, then there is an inverse. The usual definition of cosh−1 x cosh − 1 x is that it is the non-negative number whose cosh cosh is x x.

Feb 4, 2017 · For your first point, this depends on what else you know about cosh cosh. It is true that f(z)2 ≥ 1 f (z) 2 ≥ 1 does not imply f(z) ∈ [1, ∞) f (z) ∈ [1, ∞). But you may know that cosh cosh already is positive. For the second point let a <b a <b, then b − a b − a is positive and cosh(b) = cosh(a + (b − a))> cosh(a) cosh (b) = cosh (a + (b − a))> cosh (a) follows from the ...

Oct 1, 2015 · I know that for large positive numbers cosh (x) and sinh (x) would almost be equal to ex/2 e x / 2 as e−x/2 e − x / 2 would become negligible given the magnitude of x in both cases. And so for a number like 31427.7920639882, sinh (x) and cosh (x) are equal. Apart from numbers being large, are there any other conditions at which sinh and cosh would be equal?

Oct 7, 2016 · If f(x) = 0 f (x) = 0 then x x satisfies cos x = − 1 cosh x cos x = − 1 cosh x. For large x x, cosh x cosh x is large, so x x must be such that cos x cos x is small.

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

This gives solutions x = 0, x = ln(25 7 ± 24 7) x = 0, x = ln (25 7 ± 24 7) However, when solving for cosh instead initially (and working in terms of sinh), the solutions are x = 0, x = ln(25 7 + 24 7) x = 0, x = ln (25 7 + 24 7) only. 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 ...