Math 213a: Complex analysis
Problem Set #9 (19 November 2003):
Hypergeometric functions
Recall that the hypergeometric function F(a, b;c;z) (a.k.a. 2F1(a, b;c;z)) is de-
fined by the power series
1 + ab
cz+a(a+ 1)b(b+ 1)
c(c+ 1)
z2
2! +a(a+ 1)(a+ 2)b(b+ 1)(b+ 2)
c(c+ 1)(c+ 2)
z3
3! +· · ·
(|z|<1, c6= 0,−1,−2, . . .), and satisfies the hypergeometric differential equa-
tion
(z−z2)w00 +c−(a+b+ 1)zw0−abw = 0.
1. Prove that if Re(c)>Re(b)>0 then
F(a, b;c;z) = 1
B(b, c −b)Z1
0
tb−1(1 −t)c−b−1(1 −zt)−adt,
and deduce the value of F(a, b;c;1) = limz→1−F(a, b;c;z).
2. Prove the identity F(a, b;c;z) = (1−z)−aF(a, c −b;c;z/(z−1)). What is the
relation between F(a, b;c;z) and F(c−a, c −b;c;z)? [Note that (a, b, c)
and (c−a, c −b, c) yield the same angles π|1−c|, π|c−a−b|, π |a−b|at
z= 0,1,∞.]
3. Suppose 2c=a+b+ 1.
i) Prove that F(a, b;c;z) = F(a/2, b/2; c; 4z(1 −z)), and use this to evaluate
F(a, b;c; 1/2).
ii) Show that (unless a, b, c ∈ {0,−1,−2, . . .})F(a, b;c;z) and F(a, b;c; 1 −z)
constitute a basis of the solutions of the same hypergeometric differential
equation.
4. Assume further that a+b= 0, so c= 1/2. Evaluate F(a, −a; 1/2; z) in closed
form, and verify directly that if 0 < a < 1 then
F(a, −a; 1/2; z)F(a, −a; 1/2; 1 −z)
conformally maps the upper half-plane to a spherical triangle on the Rie-
mann sphere.
