Why arccos(2) have value in complex numbers even though its $D_f [-1,1]$

$\begingroup$

I've seen that arccos has $D_f [-1,1]$ however when putting arccos(2) into wolframalpha, I get the complex number result if I understand it correctly. Why is it so?

If true, should not $D_f$ be differently precised?

$\endgroup$ 3

1 Answer

$\begingroup$

If you are working within $\mathbb R$ then $arcos (2)$ does not exist. But in the complex plane there are infinitely many solutions of the equation $\cos(z)=2$.

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy