Univalent function: Difference between revisions

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== Examples==

The function <math>f \colon z \mapsto 2z + z^2</math> is univalent in the open unit disc, as <math>f(z) = f(w)</math> implies that <math>f(z) - f(w) = (z-w)(z+w+2) = 0</math>. As the second factor is non-zero in the open unit disc, <math>f</math> must be injective.

show that the function

f(z)=2z+z^2, |z|<1

Is univalent in the ___domain.

Solve..

Let z1 and z2 be any two points in |z|<1. then

f(z1)=f(z2)→(z1-z2)(z1+z2+2)=0

→z1-z2=0 and z1+z2+2≠0 in |z|<1

This is clear from the fact

Re(z1+z2+2)=Rez1+Rez2+2>-1-1+2=0

Thus f is one-one in |z|<1 and hence univalent in the ___domain.

==Basic properties==