Parabolic Branch Cut The following is problem 19 from page 87 of Saff and Snider's "Fundamentals of Complex Analysis for Mathematics, Science, and Engineering,"

How would you construct a branch of $\log z$ that is analytic in the
  domain D consisting of all points in the plane except those lying on
  the half-parabola $\lbrace x+iy: x \ge 0, y = \sqrt{x}\rbrace$?

Saff and Snider have defined all logarithms to be taken to the base of $e$ unless otherwise mentioned.  I understand the idea of a branch cut and its purpose in constructing a single-valued function from a multi-valued one.  Saff and Snider have also defined the principal logarithm of $z$ as,
$$ \text{Log}\;z = \text{Log}\;|z| + i\;Arg\; z $$
Where $\text{Arg} \; z$ lies in the half-open interval $(-\pi,\pi]$.  This function has a branch cut on the nonpositive real axis.  So I thought of considering $\text{Log}\;z^2.$  I figured that substituting $z^2$ for $z$ might result in a branch cut that resembled a quadratic.  Unfortunately, $\text{Log}\;z^2$ just has two branch cuts.  They lie on the nonnegative and nonpositive imaginary axes, respectively.  I also considered using a branch cut of $\log z$ whereby the arguement is taken to be on the half-open interval $(\pi/4,9\pi/4]$.  This is close to the answer, but the branch cut is still in the shape of a ray rather than a half-parabola.  I'm not sure what else to try...
 A: To construct that branch of log $z$, you just define the half-parabola to be the branch cut. This would mean log $z$ = Log|$z$| + $i\theta$ where $\theta$ equals the value of arg $z$ between $\frac{\pi}{2}$ and $2\pi$ for $z$ in the second, third, or fourth quadrant, the value of arg $z$ between $0$ and $\frac{\pi}{2}$ for $z$ in the first quadrant above the half parabola, and the value of arg $z$ between $2\pi$ and $\frac{5\pi}{2}$ for z in the first quadrant below the half parabola.
Explicitly, you can solve for $\theta$ in terms of $r$ where $z=re^{i\theta}$. By the equation of the half parabola, $\frac{y}{x}=\frac{1}{y}$. Then, $\theta=\arctan(\frac{y}{x})=\arctan(\frac{1}{y})$. 
$r=\sqrt{x^2+y^2}=\sqrt{x^2+x}$ , then  $x^2+x-r^2=0$. By the quadratic formula,
$$x=\frac{-1\pm\sqrt{1+4r^2}}{2}$$
You choose the positive square root because you want the upper half parabola so,
$$y=\sqrt{\frac{\sqrt{1+4r^2}-1}{2}}$$
Therefore, the entire branch would be defined as
$$\log z = \mbox{Log }|z| + i\theta , \mbox{where } \theta = \arctan \bigg(\sqrt{\frac{2}{\sqrt{1+4r^2}-1}}\bigg)$$
A: Just define the half-parabola to be the branch cut and you're done. A branch cut isn't intrinsic to a function, you choose it in whatever way you like that prevents you from circling a branch point. For example, $\log z$ has branch points at $0$ and $\infty$, so any unbounded curve that hits zero (and doesn't let you circle the origin) would separate the branches of this function.
