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

[Raziel2701]

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?

 

[HallsofIvy]

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.