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# zs and ws on P6 watch

1. How do I do this question?

Show that the transformation

w = (z+1) / (z-1)

maps the circle |z| = 1 in the z-plane to the line |w-1| = |w+i| in the w-plane.

The region |z| in the z-plane is mapped to the region R in the w-plane. Shade the region R on an Argand diagram.
2. (Original post by lesser weevil)
How do I do this question?

Show that the transformation

w = (z+1) / (z-1)

maps the circle |z| = 1 in the z-plane to the line |w-1| = |w+i| in the w-plane.

The region |z| in the z-plane is mapped to the region R in the w-plane. Shade the region R on an Argand diagram.
Rearrange to give z as a function of w. Set |z|=1 and use the fact
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
\left| \frac{a}{b} \right|=\frac{|a|}{|b|
3. I think there's a typo somewhere, because I get |w-1|=|w+1|.
4. no the exam paper definitely says i, not 1. heh hold on but the first is

w = (z - i) / (z + 1)

sorry...

now let me work out the answer and you can tell me if i'm right... :-)
5. (Original post by lesser weevil)
no the exam paper definitely says i, not 1. heh hold on but the first is

w = (z - i) / (z + 1)
w = (z-i)/(z+1)

wz + w = z - i

z = (w+i)/(w-1)

|z|=1 precisely when |(w+i)/(w-1)| = 1 i.e. |w+i| = |w-1|

This is the perpendicular bisector of the line segment 1 to -1 , that is the line x+y=0
6. (Original post by lesser weevil)
The region |z| in the z-plane is ...
This makes no sense.

Do you mean |z|<1? (The interior of the disc)

It will map to one half of the line, which you can work out by seeing which bit the origin (or any other point in the interior) maps to
7. ah yeah, |x| < 1....

sorry my typos...

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