Describe the Riemann surface: $$W = \sqrt{1-z^2}$$ 
I would like hints only.
Using @Dr.MV's hint, I get two factors: the first is
$$\sqrt{(x-1)+y^2}^{\frac{1}{2}}e^{i\frac{\theta}{2}}$$, which, when we let theta range from 0 to 2$\pi$, describes a semi-circular arc in quadrant I, centered at (1,0).
Similarly, the second factor $$\sqrt{(x-(-1))+y^2}^{\frac{1}{2}}e^{i\frac{\phi}{2}}$$ describes a semi-circular arc in quadrant II, centered at (-1,0), when we let $\phi$ range from 0 to 2$\pi$.
However, these two factors are to be multiplied:
If I multiply the factors, then the resulting argument is $e^{i(\frac{\theta + \phi}{2} )}$, which, when we let both theta and phi range from 0 to 2pi, describes a semi-circular arc that runs through quadrant I and quadrant II - starting at (1,0), and ending at (-1,0)
Where can I go from here? 
Thanks,
 A: HINT:
$$W=(1-z)^{1/2}(1+z)^{1/2}$$
has branch points at both $z=1$ and $z=-1$.  Now, you can write $z-1=\rho e^{i\phi}$ and $z+1=re^{i\theta}$.  You will find that one suitable branch cut is from $(-1,0)$ to $(1,0)$.
SPOILER ALERT
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First, for both the branch point at $(-1,0)$ and at $(1,0)$, cut the plane along the negative real axis.  Then, let $z+1=re^{i\theta}$ ($z-1=\rho e^{i\phi}$) for $-\pi<\phi<\pi$ and $-\pi<\theta<\pi$. Now, let's see if $W$ is single valued for points just above and just below the branch cut for $z+1$.  For points just  above (below) that branch cut, $\theta =\pi$ ($\theta =-\pi$) and $\phi =\pi$ ($\phi=-\pi$).  Thus, $\arg(W)$ is $\pi$ ($-\pi$) and $W=-(r\rho)^{1/2}$ is single valued.  Now, let's see if $W$ is single valued for points just above and just below the line segment $(-1,0)$ to $(1,0)$.  For points just  above (below) that segment, $\theta =0$ ($\theta =0$) and $\phi =\pi$ ($\phi=-\pi$).  Thus, $\arg(W)$ is $\pi/2$ ($-\pi/2$) and $W=\pm i (r\rho)^{1/2}$ is multi-valued.  Thus, the effective branch cut is the segment $(-1,0)$ to $(1,0)$ and $W$ is analytic in the plane less this branch cut.

