Holomorphic branch of square root of $f$

Let $f(x) = (z-\frac{1}{z})$ , $z \in \mathbb{C}$\ {${0}$}

Let $F$ be a holomorphic branch of the square root of $f$ that is defined at $z=2$ and has the value $\sqrt{3/2}$ there.

GIve an explicit expression for $F$ as a composition of standard functions ( exponential, logarithm etc)

My attempt : $$F(z)^2 = (z-\frac{1}{z})$$ $$F(z) = (z-\frac{1}{z})^{\frac{1}{2}}$$ $$= exp(\frac{!}{2} (log (z-\frac{1}{z})))$$

I am stuck here. I know $f$ is real valued when $z$ is real valued.

Presumably, the question asks you to show that the branch of the logarithm you've defined is holomorphic around $2$. Take a ball of appropriate radius around $2$,and argue that if a function $f$ is non - zero in a simply connected domain, then it has a well-defined holomorphic branch of a logarithm there.
$\textbf{Edit With More Details}:$ First, recall that if $U$ is a simply-connected domain and $f:U\to \mathbb{C}\setminus\{0\}$ is analytic, then there exists a well-defined analytic branch of the logarithm of $f$ on $U$. Secondly, for the problem at hand, observe that the given $f$ equals $0$ at $z=\pm1$, therefore $f$ is non-zero in a ball of radius $\dfrac{1}{2}$ around $2$. Now combine these two facts.