Complex

 

Complicated.

Why'd you have to go and make things so complicated?

I see the way you're acting like you're somebody else

Gets me frustrated.

Life's like this you,

And you fall, and you crawl, and you break

And you take what you get, and you turn it into

Honesty and promise me I'm never gonna find you faking

No, no, no

What if things are complex, complicated, and you have to consider imaginary numbers.

The formula for the hypotenuse, c, of a right triangle with sides a and b,  on a hyperbolic surface is

cosh(c)=cosh(a) * cosh(b)

Using the formula for the inverse of the hyperbolic cosine this becomes

c=ln(cosh(a) * cosh(b) ± √((cosh(a) * cosh(b))²-1))

Using the identity for  hyperbola, cosh²-sinh²=1 which also means that cosh²-1=sinh², this becomes

c=ln(cosh(a) * cosh(b) ± sinh(a) * sinh(b)).

Using Euler’s Formula which involves imaginary numbers, e^iz=cos(z)+sin(z)*i; the definitions of cosh(x) = ½(e^x+e^-x) and sinh(x) = ½(e^x‑e^-x); and letting z = ix, this becomes

cosh(-x) = sin(ix) and sinh(-x) = cos(ix).

Using the fact that cosh is symmetrical, reflects, with respect to 0, while sinh is not symmetrical with respect to 0, this becomes

cosh(x) = sin(ix) and sinh(x) = -cos(ix)

As a result,  the hypotenuse of a right triangle on a hyperbolic surface can be stated as the natural logarithm of regular circular trigonometric functions as

c=ln(sin(ia)*sin(ib) ± cos(ia)*cos(ib)).

For the case when a, or b, = π,  since cos(iπ) = -1 and sin(iπ)=0, this then means that

c=ln(0 ± cos(ai)) if b= π, or ln(0 ± cos(bi)) if a=π.

Equation 1

For the case when a, or b, =0,  since cos(i0)=1 and sin(i0)=0, this then means that

c=ln(sin(ai) ± 0) if b= 0, or ln(sin(bi) ±0)) if a=0.

Equation 2

This shows that for a hypotenuse as a complex number, if its real portion has a coefficient of zero, then the imaginary coefficient can take on any value, and there is zero standard deviation, i.e. the second eqaution.  If instead the real portion has a non-zero coefficient, then the imaginary coefficient must be zero, and there must be a non-zero standard deviation, i.e. the first equaion.