Analytic Continuity via Reflections: Exercises, derivations

Vasco,

I had not given any thought to the status of the sign of the square root of a complex number. It seems as mysterious as the idea of sqrt(-1). But (you know all this; I'm thinking aloud) we visualize the negative of a complex number as an antipode of z on the circle $|z| = r$, and the square root of z is just a complex number that multiplied by itself gives z, so both $\sqrt{z}^2$ and $(-\sqrt{z})^2$ equal z. But I suppose that representing z as $re^{i\theta}$ raises a problem. We need $\sqrt{r}$ to be positive unless we think of r as a vector on the whole real line, and $\sqrt{z}$ is usually just $+\sqrt{r}e^{i\theta/2}$, not $\pm\sqrt{r} e^{i\theta/2}$, but $(re^{i(\theta + Pi)})^2$ and $(-re^{i\theta})^2$ are still $r^{2}e^{i2\theta} $ (in some branch). Why would we omit the plus sign when taking the square root of a complex number in the quadratic equation?

As a bit of concrete evidence for retaining both signs in the square root, I found that replotting Figure [31] requires both roots in the quadratic equation.

Gary

Yes. It is now fixed.


Thank you. I'm working on finding a way to display the tex file in OS X.



Are you referring to the root in the calculation of y, $S_E(z)$, or $\mathfrak{R}_{E}(z)$? If I omit one of the branches of y(x), I get only half an ellipse. When I plot $S_E(z)$ on the points of the ellipse, the entire image with both positive and negative roots plots outside and below the ellipse in a shape that resembles the top half of a bow-tie. When I plot $\mathfrak{R}_{E}(z)$ on the points of the ellipse, the image of the positive branch applied to $Re(z) \gt  0$ lies outside and below the plot range. The image of the negative branch applied to $Re(z) \leq 0$ also lies outside and below the plot range.

Gary