ACOS logarithmic form

[Claudio L.]

Sorry if I turned this into a monologue, but I'll leave it written for the curious reader.
So how's the 50g and everybody else doing it?

It seems everybody has a slightly different formula for ACOS, but everybody agrees 100% identical for ASIN.
ASIN doesn't have a problem, whether you use the calculator, or do it "by hand" following the formula you get the same result.

Here, all the way to the bottom, we have a formula from Mathworks.
And here is the implementation from Wolfram.

Basically, Wolfram just does pi/2-asin(Z), so the branch chosen is consistent with the ASIN results.
Mathworks uses a formula slightly different from Wikipedia:

Wikipedia uses sqrt(Z^2-1), while the other formula has i*sqrt(1-Z^2).

Before somebody jumps and says "it's the same!", let's try a couple of cases:

Z=2:
sqrt(2^2-1)=1.73...
i*sqrt(1-2^2)=i*sqrt(-3) = i*(i*1.73...) = -1.73...

Z=2+3*i:
sqrt(Z^2-1)=sqrt(-6+12*i)=(1.92...,3.11...)
i*sqrt(1-Z^2)=sqrt(6-12*i)=i*(3.11...,-1.92...)=(1.92..., 3.11...)

Z=2-3*i:
sqrt(Z^2-1)=sqrt(-6-12*i)=(1.92...,-3.11...)
i*sqrt(1-Z^2)=sqrt(6+12*i)=i*(3.11...,1.92...)=(-1.92..., 3.11...)

Very subtle... the second form gives results consistent with the 50g for all values.
It would never occur to me that such a trivial expression manipulation would push you through a different solution. Very sneaky.