# Why do we take reciprocals of multivalued functions?

Take a branch of the function $f(t) = \sqrt{1-t^2}$ on the closed upper-half plane ($\bar{H}$) so that $f(i) = \sqrt{2} > 0$. Then, we can define

$\phi(z) = \int_0^{z} \frac{dt}{f(t)}$, where the path is taken in $\bar{H}$.

This function maps $\bar{H}$ to a half-infinite rectangular strip $[-\pi/2,\pi/2] \times [0,\infty)$.

I don't completely understand why this is the case.

Wouldn't

$\psi(z) = \int_0^{z} f(t) dt$ map $\bar{H}$ onto $[-\pi/4,\pi/4] \times (-\infty,0]$?

The argument for this is the same for that of $\phi$: The function $\psi$ maps $[-1,1]$ to $[-\pi/4,\pi/4]$. Every time $z$ passes a branch point along the real axis, the argument changes by $\pm\pi$, so this results in a $\pm\frac{\pi}{2}$ turn.

So it appears $\psi$ essentially does what $\phi$ does. Then is there any reason to look at $\phi$ rather than $\psi$? (More generally, why does the Scwarz Christoffel mapping need all those things appear in the denominator rather than in the numerator?)

p.s. I don't really understand why the the turns happen in the directions that they do rather than making like a stair-case like shape in the case of $\phi$ or $\psi$.

EDIT: Here are some references that helped me clarify this:

why does the Schwarz-Christoffel mapping need all those things appear in the denominator rather than in the numerator?

It does not; this is just a traditional way of writing the formula. You can put all those factors in the numerator, with opposite sign of the exponent: $$f(\zeta) = \int^\zeta K (w-a)^{(\alpha/\pi)-1}(w-b)^{ (\beta/\pi)-1}(w-c)^{ (\gamma/\pi)-1} \cdots \,{d}w$$ Indeed, this makes the formula more transparent: we recognize $(w-a)^{ (\alpha/\pi)-1}$ as the derivative of $(w-a)^{ (\alpha/\pi)}$ (up to a constant factor), and the latter map creates an interior angle of $\alpha$ from a halfplane.

So it appears $\psi$ essentially does what $\phi$ does. Then is there any reason to look at $\phi$ rather than $\psi$?

There is a difference, masked in your text by $\pm$ signs. Namely, $\phi$ creates interior angles of size $\pi/2$, while $\psi$ creates interior angles of size $3\pi/2$. The boundary of image may stay essentially the same, but the image is on a different side of the boundary.

how to pick the right sign

Follow the real line from left to right. When you encounter a point where the integrand is $0$ or $\infty$ (such as $1$ or $-1$ here), make a small detour along a half-circle in the upper half-plane (which is the domain of our function). How much does the argument of the integrand change during this detour? Say, we go around $1$ in this fashion: then the argument of $z-1$ changes by $-\pi$ because we move in half-circle clockwise. So:

• if the integrand has $(z-1)^{1/2}$, its argument changes by $-\pi/2$. Thus, the image makes a turn by 90 degrees to the right.
• if the integrand has $(z-1)^{-1/2}$, its argument changes by $\pi/2$. The image makes a turn by 90 degrees to the left.
• generally, if the integrand has $(z-1)^{\beta}$, its argument changes by $-\pi \beta$
• The deep problem I was having with the understanding was how the different branches of the logarithm was being chosen, and how the $\frac{1}{\sqrt{1-t^2}}$ is implicitly thought as a product of $i(t-1)^{-1/2}(t+1)^{-1/2}$. You said "There is a difference, masked in your text by ± signs," but I wasn't getting how to pick the right sign. – Braindead Jan 4 '14 at 5:49
• @Braindead I edited in an explanation. – Post No Bulls Jan 4 '14 at 6:17
• Thank you! I'm still a bit fuzzy about choosing the right branches, but I think I more or less understand this now. – Braindead Jan 4 '14 at 6:48
• @Braindead It's okay to be fuzzy about branches; all the choice of a branch does is contribute a unimodular constant to the map. The change of argument of $(z-a)^\beta$ along a curve does not depend on any branch choices. – Post No Bulls Jan 4 '14 at 6:52