Show the pre-image of a vertical line Re(w)=c under f

  • Thread starter Thread starter Raziel2701
  • Start date Start date
  • Tags Tags
    Line Vertical
Raziel2701
Messages
128
Reaction score
0
Show the pre-image of a vertical line Re(w)=c under f...

Homework Statement


a) Show that the pre-image of a vertical line Re(w)=c under f is a circle centered on the line Re(z) = -1 and tangent to the real axis Im(z) = 0.

b) How does such a circle intersect the circle |z| =1?

Here's f by the way: i\frac{1-z}{1+z}

The Attempt at a Solution



So I took f and set it equal to c:

i\frac{1-z}{1+z} =c

Solving for z, I get that the Re(z)

Re(z) = \frac{-c^2 +1}{c^2 +1}

So that was me not knowing what I'm doing and just messing with the given information. I then took the limit as z approaches infinity and I get -1, so I again, not knowing if I'm doing this right simply declared that Re(z) is now equal to -1 as I'm supposed to prove.

So am I doing this right?

For part b, the circles intersect twice at 90 degrees right?
 
Last edited:
Physics news on Phys.org


f(z) is NOT equal to c, its real part is. The real part of any complex number, u, is equal to (u+ \overline{u})/2. Find that for this function
\overline{f}(z)= f(\overline{z})
and set it equal to 0.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Back
Top