Complex Mapping Proof

  • Thread starter handiman
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  • #1
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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.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
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If [itex]z = a + bi, \quad a,b \in \mathbb{R}[/itex], 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 [itex]z = x + yi, \quad x,y \in \mathbb{R}[/itex]

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.
 
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  • #3
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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?
 
  • #4
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I'm not sure how you got [itex]\frac{\bar{z}}{z}[/itex]. Can you show your work?

Can you answer the first part of my question:

If [itex]z=x+yi, \quad x,y \in \mathbb{R}[/itex], then what is the formula for all the points on the real axis in terms of z? Plug that into [itex]f(z)[/itex] and argue it is the form of a circle.
 
  • #5
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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.
 
  • #6
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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)

=[itex]\frac{x+p-iq}{x+p+iq}[/itex]

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

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