Univalent function: Difference between revisions

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== Examples==

show that the function

Consider the application <math>\phi_a</math> mapping the open [[unit disc]] to itself such that

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

:<math>\phi_a(z) =\frac{z-a}{1 - \bar{a}z}.</math>

Is univalent in the ___domain.

We have that <math>\phi_a</math> is univalent when <math>|a|<1</math>.

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==