Exploring the Image of Q Under f(z)=z^2

[latentcorpse]

Let $Q:=\{ z: Re(z)>0, Im(z)>0, |z|<1 \}$ i.e. the quarter disc.

what is the image of Q by the mapping $f(z)=z^2$

by trial and error with various points, my answer is that it takes Q to the semicircle $\{ z: Re(z)>0, |z|<1 \}$

but can't how this explicitly as it's not a mobius transformation with which I am used to dealing with.

 

[n!kofeyn]

Let $Q:=\{ z: Re(z)>0, Im(z)>0, |z|<1 \}$ i.e. the quarter disc.

what is the image of Q by the mapping $f(z)=z^2$

The best way to find the image of mappings like this is to let $z=re^{i\theta}$ and then look at what your mapping does to this polar representation of all z in your region Q.

 

[latentcorpse]

ok so the modulus will square but that's just 1 again and the argument doubles.

our original angle was from 0 to pi/2
so now we go from 0 to pi
so it will be a semicircle in the upper half plane of radius 1?

 

[n!kofeyn]

so it will be a semicircle in the upper half plane of radius 1?

Looks good to me.