geometry: isometries and conjugates

[michelle]

Consider an isometry g such that g(0) =0, g(1)=1 and g(i)=-i. Show that for any z, g(z)= z̅ (namely g is the reflection across the x-axis)
Hint: what that implies in terms of modulus and square all those modulus

 

[pka]

Consider an isometry g such that g(0) =0, g(1)=1 and g(i)=-i. Show that for any z, g(z)= z̅ (namely g is the reflection across the x-axis)
Hint: what that implies in terms of modulus and square all those modulus

It is very hard to know that your post says.

Is it \(\displaystyle g(z)=\overline{z}~?\) If so, there is not much to it. \(\displaystyle |z|=|\overline{z}|\) and \(\displaystyle |z^2|=|z|^2\)