Mathematician Finds Easier Way to Solve Quadratic Equations

[Albert Chan]

Trivia: with signed zero, e^acosh(z) is an odd function!

acosh(z) = acos(z) * ±i       // sign to ensure acosh non-negative real part.
acos(z) + acos(-z) = pi       // acos has non-negative real part too: 0 .. pi

acosh(z) * ∓i + acosh(-z) * ±i = pi      // im(acos(z)) = - im(acos(-z))
acosh(z) - acosh(-z) = ±pi*i                // multiply both side by ±i
e^acosh(z) / e^acosh(-z) = -1            // Euler Identity, e^(pi*i) = -1

e^acosh(z) = - e^acosh(-z)

Identity might still hold without signed zero, but I have not checked edge cases.

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Another proof for acosh(z) - acosh(-z) = ±pi*i

versin(x) = 1 - cos(x) = 2*sin(x/2)^2      → acos(1-x) = 2*asin(sqrt(x/2))

acos(z) = 2*asin(sqrt((1-z)/2)) = 2*atan(sqrt((1-z)/(1+z)))
acosh(z) = 2*asinh(sqrt((z-1)/2)) = 2*atanh(sqrt((z-1)/(z+1)))

acosh(z) - acosh(-z)
= 2*atanh(u) - 2*atanh(1/u))      , where u = sqrt((z-1)/(z+1))
= ln(v) - ln(-v)                            , where v = (1+u)/(1-u)
= ±pi*i