Define square root function over the complex numbers. Define a holomorphic function $f\colon\mathbb{C}\setminus[-1,1] \longrightarrow \mathbb{C}$ such that $\forall z \in \mathbb{C}\setminus[-1,1] \ \left( (f(z))^{2} = z^{2} - 1\right)$ and $f(2)=\sqrt{3}$.
 A: For any path between $2$ and $z$ that doesn't intersect the real line between $+1$ and $-1$, define
$$
f(z)=\frac{\log(3)}{2}+\int_2^z\frac{\zeta\,\mathrm{d}\zeta}{\zeta^2-1}
$$
Since $f'(z)=\frac{z}{z^2-1}=\frac{1/2}{z-1}+\frac{1/2}{z+1}$, after accounting for the constant of integration at $z=2$, we get that $f(z)=\frac12\log(z^2-1)$ .
Since the sum of the residues of $\frac{\zeta}{\zeta^2-1}=\frac{1/2}{\zeta-1}+\frac{1/2}{\zeta+1}$ at $\zeta=+1$ and $\zeta=-1$ is $1$, the difference of the integral between two different paths that don't intersect the real line between $+1$ and $-1$ must be an integral multiple of $2\pi i$. Thus, $e^f$ is the same over both paths.
Therefore, $e^{f(z)}=\sqrt{z^2-1}$ is well-defined independent of the path taken.
A: Let $\operatorname{Log} z$ denote the principal branch of the complex logarithm, i.e. the holomorphic choice of logarithm that is defined everywhere except on the negative real axis.
Note that $z^2-1 = z^2(1-\frac{1}{z^2})$ and the second factor is negative real if and only if $z \in [-1,1]$. This observation indicates that our "square root" should be
$$g(z) = z\sqrt{1-\frac{1}{z^2}} = z e^{\frac12 \operatorname{Log}(1-\frac{1}{z^2})}.$$
Make sure to check the details for yourself.
A: I would try the following. Define:
$$f(z)=f(x+iy)=u(x,y)+iv(x,y)$$
where $x,y,u(x,y)$ and $v(x,y)$ are real. Now impose Cauchy-Riemman equations 
$$u_{x}(x,y)=v_{y}(x,y)\\ u_{y}(x,y)=-v_{x}(x,y)$$
As there are differential equations you might expect to have one undetermined constant. It will be fixed by $f(2)=\sqrt{3}$
A: Consider the square root function defined on $\mathbb C \setminus \mathbb R_-$ such that $\sqrt{1}=1$ (so it's an extension of the usual square root defined on $\mathbb R_+^*$). Define $f_1(z) = \sqrt{z-1} \sqrt{z+1}$ on $\mathbb C \setminus (- \infty ; 1]$ and $f_2(z) = f_1(-z)$ on $\mathbb C \setminus [- 1 ; \infty)$.
For $z \in \mathbb C \setminus \mathbb R$, we have $f_1(z)^2 = z^2-1 = (-z)^2-1 = f_2(z)^2$ so we must have $f_1/f_2 = \pm 1$ on each half-plane. Since $f_2(z)/f_1(z) = f_1(-z)/f_2(-z)$, we must have that $f_1/f_2$ is constant on $\mathbb C \setminus \mathbb R$. It turns out that $f_1 = - f_2$, so you can glue $f_1$ and $-f_2$ together to obtain a holomorphic function $f$ on $\mathbb C \setminus [-1 ; 1]$ satisfying $f(z)^2 = z^2-1$
Then you check that $f(2) = f_1(2) = \sqrt 1 \sqrt 3 = \sqrt 3$
