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

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SUMMARY

The pre-image of a vertical line Re(w)=c under the function f(z) = i(1-z)/(1+z) is a circle centered on the line Re(z) = -1 and tangent to the real axis Im(z) = 0. The calculation of Re(z) yields Re(z) = (-c² + 1)/(c² + 1). The intersection of this circle with the unit circle |z| = 1 occurs at two points, forming a right angle. The analysis confirms the geometric properties of the transformation defined by the function f.

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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:
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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.
 

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