Math 411: Honours Complex Variables - University of Alberta

and note that
z−i 1+ z+i
g(f(z)) = i
1− z−i
z+i
z +i+z −i
= i
z +i−z +i
= i 2z
2i
= z
for z ∈ H. Hence, f is injective. Let x 2 +y 2 < 1, and note that
g(x+iy) = i (1+x)+iy
(1−x)−iy
For z ∈ D, we can thus evaluate
= i [(1+x)+iy][(1−x)+iy]
(1−x) 2 +y2 2y
= −
(1−x) 2 +y2 +i 1−(x2 +y2 )
(1−x) 2 +y2 ∈ H.
� �� �
>0
f(g(z)) = i1+z
1−z −i
i 1+z
1−z +i
= 1+z −1+z
= 2z
2
= z.
1+z +1−z
111
Hence, f is also surjective and thus bijective with inverse g. Since f and g
are obviously holomorphic, this means that D and H are biholomorphically
equivalent.
3. Thereisnobiholomorphicmapf: C → Dbecause anyholomorphicmapfromC
to D is bounded and thus constant by Liouville’s theorem. Hence, C and D are
not biholomorphically equivalent.