a) Since
as
, it makes sense to consider
)
(i.e. the integral converges). In addition,
\in(0,+\infty))
.
b) Symmetry of the image: Let
. Considering the radial path
from 0 to
, we get
=\int_0^1\frac{iz\ dt}{\sqrt{1-(itz)^4}})
^4}}=i f(z))
. Thus, it suffices to prove that one quarter of disc is sent to one quarter of square.
c) Of course,
=0)
. For
)
, we even have
=\int_0^{u}\frac{dt}{\sqrt{1-t^4}}\in(0,\alpha))
, hence
=if(u)\in i(0,\alpha))
.
d) Let's consider the image of the arc of circle from 1 to
. For
)
, we have (considering the path from 0 to 1 and then to
along the unit circle)
=f(1)+\int_0^\theta \frac{ie^{i\gamma}\ d\gamma}{\sqrt{1-e^{4i\gamma}}})
. And
e^{2i\gamma}=2\sin(2\gamma)e^{2i\ga mma+\frac{-i\pi}{2}})
hence
}e^{i\gamma}e^{-i\frac{\pi}{4}})
. Thus
simplify and
=f(1)+ie^{-i\frac{\pi}{4}}\int_0^\theta\frac{d\gamma}{\sqrt{2 \sin(2\gamma)}})
. Then
and the integral is real. Thus
-f(1)=C(\theta) e^{i\frac{3\pi}{4}})
for some
>0)
. This shows that the image of the arc of circle is a line segment (with slope
) from
to
.
I showed that the boundary of
is sent to the boundary of the square with vertices
, 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.
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Originally Posted by Bruno J. 
. (for the half-plane, so one'd need a composition by something like 
