## Loser66 one year ago Find z : cos z = 3. I need the logic for the last part. Please, help. I will post my work

1. Loser66

$$cos z =\dfrac{e^{iz}+e^{-iz}}{2}=3\implies e^{iz}+e^{-iz}=6$$

2. Loser66

multiple both sides by e^(iz) , I got $$e^{2iz} -6e^{iz}+1=0$$ Let $$w = e^{iz}$$ and solve for w

3. Loser66

I have $$w = 3\pm 2\sqrt 2$$ Do I have to solve one by one or I combine them on 1 argument? I meant: $$e^{iz}= 3 + 2\sqrt 2$$ and then $$e^{iz}= 3-2\sqrt 2$$separately, right?

4. Loser66

From $$e^{iz} = 3 + 2\sqrt 2$$, $$iz = log (3+2\sqrt2 ) + i 2\pi n, n\in \mathbb Z$$, hence $$z = 2\pi n - ilog (3+2\sqrt 2) , n\in \mathbb Z$$

5. Loser66

But if $$e^{iz} = 3-2\sqrt 2$$, with the same argument, I have $$z = 2\pi n -i log(3-2\sqrt2)$$ I do not know how to do. Should I combine them (how?) or let them as they are?