Mathematician Finds Easier Way to Solve Quadratic Equations

[Albert Chan]

(05-01-2024 04:57 PM)Gerson W. Barbosa Wrote:  Use the identity

\(\cosh^{-1}{(z)}=\ln\left({z+\sqrt{z^2-1}}\right)\)

or

\(e^{\cosh^{-1}{(z)}}=z+\sqrt{z^2-1}\)

Above identity does not hold, if sign(Re(z)) = -1

acosh(z) = ln(z + √(z+1) * √(z-1))      // 2 square roots to ensure acosh non-negative real part

Example: z = -2

ln(-2 + √(-1) * √(-3)) = ln(-2 − √3) ≈ +1.31696 + pi*i
ln(-2 + √((-2)^2 - 1)) = ln(-2 + √3) ≈ −1.31696 + pi*i

For acosh √(z²-1) form, we need to match sign, to ensure non-negative real part.
Again, assume sign(x) = ±1, never 0

\(e^{\cosh^{-1}{(z)}}=z + sign(\Re(z)) \;\sqrt{z^2-1}\)

It is now trivial to show e^acosh(z) is an odd function:

e^acosh(z) = - ((-z) + sign(Re(-z)) * √((-z)²-1)) = - e^acosh(-z)

Update: Perhaps better proof

e^z = (e^z+1/e^z)/2 + (e^z-1/e^z)/2 = cosh(z) + sinh(z)

e^acosh(z) = cosh(acosh(z)) + sinh(acosh(z)) = z + sign(Re(z)) * √(z²-1)