Proving Complex Mapping: f(z) Maps Real Axis to Circle of Radius 1

[handiman]

Homework Statement

Let a be a complex number for which Im(a) ≠ 0, and f(z) = (z + conj(a))/(z + a).

Prove f(z) maps the real axis onto the circle lwl = 1.

2. The attempt at a solution

I wrote out f(z) in an a+bi for and then with the Im(a) ≠ 0 I set the equation as

f(a+bi) = (a+a₀-ib)/(a+a₀+ib).

I made a substitution let d = (a+a₀+ib) and conj(d) = (a+a₀-ib)

This gave me d/conj(d). I have exhausted all of the identities I could remember/find and I see no path leading this line of thinking to a circle.

 

[scurty]

If z = a + bi, \quad a,b \in \mathbb{R}, then what is the formula for all the points on the real axis in terms of z? Plug that into f(z) and argue it is the form of a circle.

Edit: I just realized there is a point a in the problem, perhaps it would be better to write out z = x + yi, \quad x,y \in \mathbb{R}

Edit2: This turns out to be a big algebra problem once you plug in for z. It works out nicely so don't get scared by all of the factors that appear.

 

[gestalt]

I thought I had gotten to the point of it being a big algebra problem but the wall I am hitting is what to do to it. Is there some way to connect conj(z)/z to the unit circle?

 

[scurty]

I'm not sure how you got \frac{\bar{z}}{z}. Can you show your work?

Can you answer the first part of my question:

If z=x+yi, \quad x,y \in \mathbb{R}, then what is the formula for all the points on the real axis in terms of z? Plug that into f(z) and argue it is the form of a circle.

 

[handiman]

I will try.

the iy term is zero because the inputs are on the real axis.

f(x+iy) = x+a₀-ib₀/x+a₀+ib₀

Let x+a₀+ib₀ be d
Let x+a₀-ib₀ be conj(d)

then I end up with conj(d)/d

if there is an error in the logic I apologize. I am fairly new to this subject matter.

 

[cng99]

Okay, first of all. I know it's too late to answer but I believe I got it.

Let z=x+iy,a=p+iq and w=u+iv
But on a real line, z=x (y=0 is the line)
Hence z = x
w=u+iv=f(z)=f(x+i0)=f(x)

=\frac{x+p-iq}{x+p+iq}

Now as you can see (and also prove) that |w|=1 no matter what x,p and q are.