Nevanlinna's criterion

A univalent function h on the unit disk satisfying h(0) = 0 and h'(0) = 1 is starlike, i.e. has image invariant under multiplication by real numbers in [0,1], if and only if ${\displaystyle zh^{\prime }(z)/h(z)}$ has positive real part for |z| < 1 and takes the value 1 at 0.

Note that, by applying the result to a•h(rz), the criterion applies on any disc |z| < r with only the requirement that f(0) = 0 and f'(0) ≠ 0.

Let h(z) be a starlike univalent function on |z| < 1 with h(0) = 0 and h'(0) = 1.

For t < 0, define[1]

${\displaystyle f_{t}(z)=h^{-1}(e^{-t}h(z)),\,}$

a semigroup of holomorphic mappings of D into itself fixing 0.

Moreover h is the Koenigs function for the semigroup f_t.

By the Schwarz lemma, |f_t(z)| decreases as t increases.

Hence

${\displaystyle \partial _{t}|f_{t}(z)|^{2}\leq 0.}$

But, setting w = f_t(z),

${\displaystyle \partial _{t}|f_{t}(z)|^{2}=2\Re \,{\overline {f_{t}(z)}}\partial _{t}f_{t}(z)=2\Re \,{\overline {w}}v(w),}$

where

${\displaystyle v(w)=-{h(w) \over h^{\prime }(w)}.}$

Hence

${\displaystyle \Re \,{\overline {w}}{h(w) \over h^{\prime }(w)}\geq 0.}$

and so, dividing by |w|²,

${\displaystyle \Re \,{h(w) \over wh^{\prime }(w)}\geq 0.}$

Taking reciprocals and letting t go to 0 gives

${\displaystyle \Re \,z{h^{\prime }(z) \over h(z)}\geq 0}$

for all |z| < 1. Since the left hand side is a harmonic function, the maximum principle implies the inequality is strict.

Conversely if

${\displaystyle g(z)=z{h^{\prime }(z) \over h(z)}}$

has positive real part and g(0) = 1, then h can vanish only at 0, where it must have a simple zero.

Now

${\displaystyle \partial _{\theta }\arg h(re^{i\theta })=\partial _{\theta }\Im \,\log h(z)=\Im \,\partial _{\theta }\log h(z)=\Im \,{\partial z \over \partial \theta }\cdot \partial _{z}\log h(z)=\Re \,z{h^{\prime }(z) \over h(z)}.}$

Thus as z traces the circle ${\displaystyle z=re^{i\theta }}$, the argument of the image ${\displaystyle h(re^{i\theta })}$ increases strictly. By the argument principle, since ${\displaystyle h}$ has a simple zero at 0, it circles the origin just once. The interior of the region bounded by the curve it traces is therefore starlike. If a is a point in the interior then the number of solutions N(a) of h(z) = a with |z| < r is given by

${\displaystyle N(a)={1 \over 2\pi i}\int _{|z|=r}{h^{\prime }(z) \over h(z)-a}\,dz.}$

Since this is an integer, depends continuously on a and N(0) = 1, it is identically 1. So h is univalent and starlike in each disk |z| < r and hence everywhere.

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