# 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...

• I'm not sure I understand the question. What prevents you from just choosing the half-parabola to be the branch cut of $\log z$? – Robert Mastragostino Sep 11 '12 at 21:13
• @RobertMastragostino: I think you're right. Actually, I think I had trouble understanding the question, and what you wrote is the correct interpretation. If you post your comment as an answer, I'll be happy to mark it as the solution. – Andrew Sep 11 '12 at 21:18

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.
• Actually you should say a bit more, e.g. specify the value of $\log(z)$ at some point not on the curve. – Robert Israel Sep 11 '12 at 22:33
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 $$\log z = \mbox{Log }|z| + i\theta , \mbox{where } \theta = \arctan \bigg(\sqrt{\frac{2}{\sqrt{1+4r^2}-1}}\bigg)$$