The function is univalent in the open unit disc, as implies that . As the second factor is non-zero in the open unit disc, must be injective.
One can prove that if and are two open connected sets in the complex plane, and
for all in
Comparison with real functions
given by ƒ(x) = x3. This function is clearly injective, but its derivative is 0 at x = 0, and its inverse is not analytic, or even differentiable, on the whole interval (−1, 1). Consequently, if we enlarge the domain to an open subset G of the complex plane, it must fail to be injective; and this is the case, since (for example) f(εω) = f(ε) (where ω is a primitive cube root of unity and ε is a positive real number smaller than the radius of G as a neighbourhood of 0).