# Complex integral (2)

• Feb 5th 2010, 10:38 AM
Bruno J.
Complex integral (2)
Show that the function

$z \mapsto f(z)=\int_0^z \frac{dt}{\sqrt{1-t^4}}$

maps the (open) unit disc onto the interior of a square.

This problem is difficult. Good luck! (Wink)
• Feb 5th 2010, 01:37 PM
Laurent
Not a full proof (a few topological details missing), but convincing arguments...

Spoiler:

a) Since $\sqrt{1-t^4}=\sqrt{1-t}\sqrt{1+t+t^2+t^3}\sim 2\sqrt{1-t}$ as $t\to 1^-$, it makes sense to consider $f(1)$ (i.e. the integral converges). In addition, $\alpha:=f(1)\in(0,+\infty)$.

b) Symmetry of the image: Let $z=r e^{i\theta}\in\mathbb{D}$. Considering the radial path $t\mapsto itz$ from 0 to $iz$, we get $f(iz)=\int_0^1\frac{iz\ dt}{\sqrt{1-(itz)^4}}$ $=\int_0^1\frac{iz\ dt}{\sqrt{1-(tz)^4}}=i f(z)$. Thus, it suffices to prove that one quarter of disc is sent to one quarter of square.

c) Of course, $f(0)=0$. For $u\in(0,1)$, we even have $f(u)=\int_0^{u}\frac{dt}{\sqrt{1-t^4}}\in(0,\alpha)$, hence $f(iu)=if(u)\in i(0,\alpha)$.

d) Let's consider the image of the arc of circle from 1 to $i$. For $\theta\in(0,\pi/2)$, we have (considering the path from 0 to 1 and then to $e^{i\theta}$ along the unit circle) $f(e^{i\theta})=f(1)+\int_0^\theta \frac{ie^{i\gamma}\ d\gamma}{\sqrt{1-e^{4i\gamma}}}$. And $1-e^{4i\gamma}=-2i\sin(2\gamma)e^{2i\gamma}=2\sin(2\gamma)e^{2i\ga mma+\frac{-i\pi}{2}}$ hence $\sqrt{1-e^{4i\gamma}}=\sqrt{2\sin(2\gamma)}e^{i\gamma}e^{-i\frac{\pi}{4}}$. Thus $e^{i\gamma}$ simplify and $f(e^{i\theta})=f(1)+ie^{-i\frac{\pi}{4}}\int_0^\theta\frac{d\gamma}{\sqrt{2 \sin(2\gamma)}}$. Then $i e^{-i\frac{\pi}{4}}=e^{i\frac{3\pi}{4}}$ and the integral is real. Thus $f(e^{i\theta})-f(1)=C(\theta) e^{i\frac{3\pi}{4}}$ for some $C(\theta)>0$. This shows that the image of the arc of circle is a line segment (with slope $-\frac{3\pi}{4}$) from $\alpha$ to $i\alpha$.

I showed that the boundary of $\mathbb{D}$ is sent to the boundary of the square with vertices $\alpha,i\alpha,-\alpha,-i\alpha$, and 0 is sent to itself. Then there must be a connectivity argument to conclude...

In fact, this problem shows an example of Schwarz-Christoffel mapping ; there is a general formula alike for mapping the unit disc to any polygon.
• Feb 5th 2010, 03:05 PM
Bruno J.
Very good! You are very good, Laurent.

I had this problem to do on an assignment last week, and I knew that it followed from the following more general theorem. This one was also on the assignment and it was by far the hardest problem I've ever had on an assignment so far. I got together with two Ph.D students and 3 hours of discussion got us nowhere. I ended up solving it on my own! (Rock)

So here it is!

Show that in the formula of the Schwarz-Christoffel transformation, the points on the real line which are to be mapped to the vertices of the polygon can be replaced by points on the unit circle, and the resulting function maps the unit disc to a polygon. In other words, if $a_1,\dots,a_n$ are points on the unit circle, and $\alpha_1 \pi, \dots, \alpha_n\pi$ are the corresponding interior angles of the polygon, then

$z \mapsto f(z)= \int_0 ^z \prod_{j=1}^n(t-a_j)^{\alpha_j - 1} \, dt$

maps the unit disc to a polygon having the prescribed angles.

The case above obviously follows right away from this one. Would you know how to show it?
• Feb 5th 2010, 03:48 PM
Laurent
Quote:

Originally Posted by Bruno J.
Would you know how to show it?

Do you mean show the formula for the Schwarz-Christoffel mapping? I wouldn't, but there seems to be a proof here for instance (Itwasntme). (for the half-plane, so one'd need a composition by something like $i\frac{1+z}{1-z}$ to start from the disc...)
• Feb 5th 2010, 04:32 PM
Bruno J.
Well, yes, that was the problem. The hint is to use the transformation you suggest, but I was never able to bring it to look like the original formula again... I ended up finding another proof.
• Feb 6th 2010, 02:05 AM
Laurent
Quote:

Originally Posted by Bruno J.
Well, yes, that was the problem. The hint is to use the transformation you suggest, but I was never able to bring it to look like the original formula again... I ended up finding another proof.

I thought you asked for the usual formula for Schwarz-Christoffel mappings; if all you need is switch to the disc, you can do as follows: the function $z\mapsto g(z)=\int_0^z\prod_{j=1}^n (t-x_j)^{\alpha_j-1}dt$ (where $x_j\in\mathbb{R}$) sends the upper half-plane to a polygon P with the given angles. Then the function $z\mapsto \psi(z)=i\frac{1+z}{1-z}=-i+\frac{2i}{1-z}$ (the second expression is for derivating) maps the unit disc to the upper half-plane. So the function you want is just $f(z)=g(\psi(z))$.

In fact I think the other way around will be more natural... The function $z\mapsto\phi(z)=\frac{z-i}{z+i}=1-\frac{2i}{z+i}$ maps the upper half-plane to $\mathbb{D}$ and we have to check that $g(z)=f(\phi(z))$ is a usual Schwarz-Christoffel mapping with the correct angles. We have $g'(z)=\phi'(z)f'(\phi(z))=-\frac{2i}{(z+i)^2}\prod_{j=1}^n(\frac{z-i}{z+i}-a_j)^{\alpha_j-1}$. Note that $\sum_j (\alpha_j-1)=-2$ (elementary geometry) hence if we reduce to the same denominator in the parentheses, the denominator simplifies with the first factor, and we are left with $g'(z)=-2i\prod_{j=1}^n (z-i-a_j(z+i))^{\alpha_j-1}$. Write $z-i-a_j(z+i)=(1-a_j)z-(1+a_j)i=(1-a_j)(z-i\frac{1+a_j}{1-a_j})$ $=(1-a_j)(z-\psi(a_j))$, hence for some complex constant $K$, $g'(z)=K\prod_{j=1}^n (z-\psi(a_j))^{\alpha_j-1}$. Since $\psi$ maps the unit circle to the real axis, we have a "usual" Schwarz-Christoffel mapping. This concludes. (Side note: $g(0)=f(\phi(0))=f(-1)$ so that $g(z)=f(-1)+\int_0^z g'(t)dt$: there is an additional translation (+ an affinitiy due to K). )
• Feb 6th 2010, 04:31 PM
Bruno J.
Nice! Thank you Laurent. Perhaps if I had followed through with the calculations I would have gotten it in this way. Here is an outline of the solution I found :

Consider $
f(z)= \int_0 ^z \prod_{j=1}^n(t-a_j)^{\alpha_j - 1} \, dt
$
where $a_j=e^{i\beta_j}$ are points on the circle and $\sum_{j=1}^n(\alpha_j - 1) = -2$. First, we want to show that the arcs of the circle corresponding to $\theta \in [\beta_k, \beta_{k+1}]$ are mapped to line segments by $f$. This is equivalent to showing that the tangent vectors to the circle along that arc are mapped to tangent vectors on the boundary $f(S^1)$ of the image domain which do not change direction are $\theta$ stays within the interval $[\beta_k, \beta_{k+1}]$. Let $\tau$ be a vector tangent to the unit circle at the point $e^{i\theta}$. Then $f(\tau)$ is a vector tangent to the curve $f(S^1)$ at the point $f(e^{i\theta})$; moreover $\mbox{Arg }f(\tau) = \mbox{Arg }\tau + \mbox{Arg }f'(e^{i\theta})$. So we want to show $\mbox{Arg }\tau$ is constant when $\theta \in [\beta_k, \beta_{k+1}]$. To do this we write

$f'(e^{i\theta}) = \prod_{j=1}^n(e^{i\theta}-e^{i\beta_k})^{\alpha_j - 1} = \prod_{j=1}^n(2i\sin{\frac{\theta-\beta_k}{2}}e^{i\frac{\theta+\beta_k}{2}})^{\alpha _j - 1}$

and we see that $\mbox{Arg }f'(e^{i\theta}) = K + \sum_{j=1}^n\frac{\theta}{2}(\alpha_j-1) = K - \theta$ for some constant $K$. Moreover $\mbox{Arg }\tau = \theta + \pi/2$ and so $\mbox{Arg }f(\tau) = (K-\theta)+\theta+\pi/2 = K'$ is constant.

To see that $\mbox{Arg }f(\tau)$ suddenly changes when $\theta$ goes over the point $\beta_k$, notice that the factor $\sin{\frac{\theta-\beta_k}{2}}$ changes sign, causing a change of $\pi(\alpha_k-1)$ in $\mbox{Arg }f(\tau)$, and we are left with the same situation as with the original Schwarz-Christoffel mapping.