trigonometry - Evaluate $cos(z)$, given that $z = i log(2+sqrt{3})$

Q. Evaluate $\cos(z)$, given that $z = i \log(2+\sqrt{3})$

I did it in 2 ways, and the results do not match.

Method 1

Multiply both sides by $i^3$, we get

$$-iz = \log(2+\sqrt{3})$$ and so,

$$2+\sqrt{3} = e^{-iz} = \cos(z) - i \sin(z)$$

from which we have $\cos(z)= 2+\sqrt{3}$ ?!

Method 2

I added $i\log(2-\sqrt{3})$ both sides, to get:

$$i\log(2-\sqrt{3})+z=i\log(4)$$

from which we have, following similar procedure, $\cos(z) = 2-\sqrt{3}$, which fortunately is $< 1$

What actually is happening? And is $\cos(z)$ even defined for complex number $z$?

Answer

Hint: For complex $z$, we simply have $$\cos z=\frac{e^{iz} + e^{-iz}}{2} \text{ and } \sin z=\frac{e^{iz} - e^{-iz}}{2i}$$