Help to understand this comment in Conway's complex analysis book

My question is really simple. I'm trying to understand this comment on page 45 in Conway's complex analysis book:

Consider the function defined by $f(z) = z^2$. If $z = x+iy$ and $\mu + i\nu = f(z)$ then $\mu = x^2-y^2$, $\nu = 2xy$. Hence, the hyperbolas $x^2-y^2 = c$ and $2xy = d$ are mapped by $f$ into the straight line $\mu = c, \nu =d$. One interesting fact is that for $c$ and $d$ not zero, these hyperbolas intersect at right angles, just as their images do. This is not an isolated phenomenon and this property will be explored in general later in this section.

Now examine what happens to the lines $x = c$ and $y = d$. First consider $x = c$ ($y$ arbitrary); $f$ maps this line into $\mu = c^2-y^2$ and $\nu = 2cy$. Eliminationg $y$ we get that $x = c$ is mapped onto the parabola $\nu^2 = -4c^2(\mu-c^2)$. Similarly, $f$ takes the line $y = d$ onto the parabola $\nu^2 = 4d^2(\mu+d^2)$. These parabolas intersect at $(c^2-d^2,\pm 2|cd|)$. It is relevant to point out that as $c \to 0$ the parabola $\nu^2 = -4c^2(\mu-c^2)$ gets closer and closer to the negative real axis. This corresponds to the fact that the function $z^{1/2}$ maps $G = \mathbb{C}-\{z : z \leq 0\}$ onto $\{z : \mathrm{Re}\ z > 0\}$. Notice also that $x = c$ and $x = -c$ (and $y = d$, $y = -d$) are mapped onto the same parabolas.

Could someone explain why the fact the parabola $\nu^2=-4c^2(\mu-c^2)$ gets closer and closer to the negative real axis, when $c \to 0$ corresponds to the fact that the function $z^{1/2}$ maps $G$ onto $\{z : \mathrm{Re}\ z>0\}$?

See the following picture: The function $f(z)=z^2$ maps the vertical lines to the right hand side parabolas. You see that $x=\pm c$ are mapped onto the same parabola. The line $x=0$ is mapped to $\{z: z\leq 0\}$. It is a degenerate case since it is in fact folded on to the negative part of the real axis.

Now consider the inverse map, which is $z^{\frac{1}{2}}$. We ignore the degenerate line segment $\{z: z\leq 0\}$. The rest of the part is then either mapped onto the right half plane or left half plane since $x=\pm c$ are mapped to the same parabola. The author chooses the right half of the plane as the main branch by convention.

• Great answer! thank you! Is the author using the fact $z^2$ and $z^{1/2}$ are inverses of each other? – user42912 Jan 1 '16 at 18:34
• Yes. It is just as the inverse of $\sin x$ is assigned such that the range is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. The range is chosen by convention. It could be any other ranges that gives the value $[-1,1]$ for $\sin x$. – KittyL Jan 1 '16 at 18:54

The author is trying to explain that under the map $z = w^{1/2}$ the open right half-plane $Re(z)>0$ corresponds to what remains in the $w$ plane after deleting the slit $S$ on the negative real axis, defined by $S=\{ w: Re(w)\leq 0 \}$.) This correspondence can be seen by progressively exploring a family of half-planes in the $z$ plane, defined by $Re(z)>c>0$, and then letting $c\to 0^+$. The boundary of each such half-plane is one of the lines described in the textbook. It corresponds to a parabolic region in the $w$ plane that encloses the slit $S$. As $c\to 0^+$ the author is observing that the parabolic regions gobble up everything in the $w$ plane except the slit. The author failed to emphasize that $c>0$, and this crucial point made the explanation hard to follow. Computer graphics can also be of great help in visualizing such maps.

Think about what squaring does to the positive $y$-axis. It takes it to the negative $x$-axis right? And the same with the negative $y$-axis, it also goes to the negative $x$-axis. So it's like the $y$-axis is folded in half at the origin and laid over the negative $x$-axis. If you think about it that's a degenerate parabola. A degenerate parabola is just a half line whose (one) end is the (degenerate) vertex. Now move the $y$-axis ever so slightly to the right, to $x=c$ for $c$ small. You'll get a nearly degenerate parabola with a very sharp vertex at the origin. As $c$ grows the parabola widens. Does that help you picture this at all?

• "Think about what squaring does to the positive y-axis. It takes it to the negative x-axis right?" I didn't understand what you meant by this. Thank you for your answer! – user42912 Jan 1 '16 at 18:42
• I was referring to the imaginary axis in the complex plane as the $y$-axis and the real axis as the $x$-axis. Does that clarify it? If you square $xi$ where $x$ is positive real (so $xi$ is a point on the positive $y$-axis), then you get $-x^2$ which lies on the negative $x$-axis. Same with $-xi$. – Gregory Grant Jan 1 '16 at 18:44
• Thank you again, now I understood what you meant! However I did understand this part of the comment of the author. What I didn't understand was why the fact the parabola $\nu^2=-4c^2(\mu-c^2)$ gets closer and closer to the negative real axis, when $c \to 0$ corresponds to the fact that the function $z^{1/2}$ maps $G$ onto $\{z : \mathrm{Re}\ z>0\}$ – user42912 Jan 1 '16 at 18:51
• Right, I guess he just wants you to see how the square root unwinds the plane by half, taking stuff close to the negative real axis close to the positive imaginary axis. It's obviously not enough to just know it takes the plane to the positive real part of the plane, since it really depends specifically how it does that. – Gregory Grant Jan 1 '16 at 18:55
• But I may well be missing the point, I don't want to give the impression that I'm sure that's what he had in mind. – Gregory Grant Jan 1 '16 at 18:56