ACOS logarithmic form

05012016, 03:58 AM
Post: #18




RE: ACOS logarithmic form
(04302016 08:57 AM)ljubo Wrote: ...it is interesting question why they have chosen this exact principal branch  or in other words, what would be broken or "ugly" if principal branch would be different, especially if they would took Im(ARCCOS(z))<0 for Re(z)>1 and Im(z)=0. Thank you, now I don't feel so dumb. The first few posts seemed to dismiss this as something trivial that I should've known since third grade. I'm glad to see now that it wasn't so trivial. Back on topic, it seems you are on the right track, I think those branches were chosen to preserve the symmetries and the relationships between asin() and acos() that we know from the real realm. I guess all you have to do is get a list of all symmetries, and test them using the Wikipedia formula for acos(). I can easily see acos(Z)=pi/2asin(Z) failing if you have the other branch, since the sign of the imaginary part is opposite, you'd get acos(Z)=pi/2conj(asin(Z)). 

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