Corresponding Areas in the z and w Planes for Algebraic Equations

[Wishbone]

The question reads:

"what part of the z-plane corresponds to the interior of the unit circle in the w-plane if

a) w = (z-1)/(z+1) b) w = (z-i)/(z+i)"


I really am having problems understanding what the question is asking. I don't understand what the w plane is, and in which plane is the unit circle, the z plane, or the w plane?

 

[quasar987]

w is a function of z. And it's a bijection. So it takes the complex plane and spawns it all over again. This is what we mean by the w-plane: the complex plane, as spawned by the function w. It's no different than the complex plane itself.

What the question is asking is, if you consider the interior of the unit circle in the w-plane (read complex plane), what is its pre-image? I.e. what subset A is such that w(A) = interior of the unit circle. You will have to find the inverse function [itex]w^{-1}[/itex] and apply the unit circle to it. See what it spawns.

Hope that was understandable.

 

[HallsofIvy]

Exactly what quasar987 said. Essentially you need to solve the inequalities
a) |(z-1)(z+1)|< 1 b) |(z-1)/(z+1)|< 1

I would recommend starting by solving the corresponding equations.

 

[Wishbone]

wow hmm...


Are you saying that I need to find the numbers that are within the unit circle in the complex Z plane?

 

[quasar987]

More like "what numbers in the complex z plane correspond (via the w mapping) to numbers that are in the unit circle in the w-plane."

 

[Wishbone]

one question, about solving for z, when I solve those inequaltities for z, what is that tell me? Is that the same thing as r(radius) in the real plane?

 

[Wishbone]

also I cannot seem to solve those equations for z? is it possible?

 

[HallsofIvy]

one question, about solving for z, when I solve those inequaltities for z, what is that tell me? Is that the same thing as r(radius) in the real plane?

Well, no, z is a complex number and so is not a radius!

also I cannot seem to solve those equations for z? is it possible?

Yes, of course it's possible. Setting z= x+ iy, then |(z-1)(z+1)|= |z²- 1|= |(x²-y²)+i(2xy)|.

|(z-1)(z+1)|=|(x²-y²)+i(2xy)|= 1 is the same as
[tex]\sqrt{(x^2-y^2)^2+ 4x^2y^2}= 1[/tex]

What (x,y) satisfy that?