Inverse cosine of a complex number, take $\cos z=\sqrt{2}$ for $z$

Question

If I am given $\cos z=\sqrt{2}$ for $z$ and asked to solve it using the following: $$ \cos^{-1} z =-i \log\sqrt{z+i(1-z^2)} $$

I've only gotten as far as taking $\cos z=\sqrt{2}$ and changing it to $z=\cos^{-1} \sqrt{2}$. from here I'm not sure how to plug into the formula. Am I using $z=x+iy$ or is $z=\sqrt{2}$??

Answer 1

As log(x) is the natural logarithm, we get:

Real solution:

$$z=1$$

Complex solution:

$$z=(\frac{4}{51}-\frac{i}{51})((-2-5i)+(\frac{33+8i}{(((295-88i)+6((1380-2154i)^{\frac{1}{2}}))^{\frac{1}{3}}))})+(((295-88i)+6((1380-2154i)^{\frac{1}{2}}))^{\frac{1}{3}}))$$