Complex analysis - LFT

[sweetvirgogirl]

Verify that the linear fractional transformation
T(z) = (z2 - z1) / (z - z1)
maps z1 to infinity, z2 to 1 and infinity to zero.

^^^ so for problems like these, do I just plug in z1, z2 and infinity in the eqn given for T(z) and see what value they give?
in this case, do i assume 1/ 0 is infinity then?
I just want to make sure I am taking the right approach

(also ... for complex analysis in general, it is safe to assume 1/0 is infinity?)

 

[AKG]

Yes, it's all correct. The thing you're asked to prove is only true when z₁ and z₂ are different, but I suppose the question intended that to be the case, so yes, just do what you said.

 

[matt grime]

in this case, do i assume 1/ 0 is infinity then?

It is defined to be infinity, as is z/0 for any complex number z not equal to 0. 0/0 is not defined.

 

[sweetvirgogirl]

It is defined to be infinity, as is z/0 for any complex number z not equal to 0. 0/0 is not defined.

what do i do with cases infinity/infinity and/or 0/0?

 

[matt grime]

You cannot get LFTs which result in 0/0 or infinity/infinty. It is part of their definition: (az+b)/(cz+d) such that ad-bc=/=0

 

[haritshelat]

in that case,in 0/0 or inf/inf i think u should take limits and use lh rule.

 

[shmoe]

in that case,in 0/0 or inf/inf i think u should take limits and use lh rule.

as already mentioned, 0/0 won't come up with a linear fractional transformation.

If f(z)=(az+b)/(cz+d) with a,c both non zero you will get infinity/infinity if you try to evaluate f(infinity) by a straight substitution. Usualy linear fractional transformations will define f(infinity) either as a limit or, what amounts to the same thing, simple as f(infinity)=a/c.