Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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...

share|cite|improve this question
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
up vote 2 down vote accepted

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.

share|cite|improve this answer
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
@RobertIsrael Ah, I didn't see what it was asking. Should I update this or did you want to put that down as a separate answer? – Robert Mastragostino Sep 11 '12 at 22:37

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 $$\log z = \mbox{Log }|z| + i\theta , \mbox{where } \theta = \arctan \bigg(\sqrt{\frac{2}{\sqrt{1+4r^2}-1}}\bigg)$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.