Conformal mapping

[ferry2]

Can you tell me is my solution true of the next problem.
Find center w_0 and radius R of the circle k, in wich the transformation w=\frac{z+2}{z-2}
converts the line l:\text{Im} z+\text{Re} z=0.

Solution:

2 \to\infty

-2i=(2)^*\to w_0

w_0=w(-2i)=\frac{-2i+2}{-2i-2}=\frac{1-i}{-1-i}*\frac{-1+i}{-1+i}=i - center of k

0\to \frac{0+2}{0-2}=-1\in k

R=|-1-i|=\sqrt{2}

And can you help me with these problems:

1. Find the image of the domain \left{\begin{array}{ll}\text{Re}>0 \\ \text{Im} >0 \end{array}\right, cut along the arc \left{\begin{array}{ll} |z|=1 \\ 0 \le \arg z \le \frac{\pi}{4} \end{array}\right, by transformation w=\frac{1}{z^2}

2. The domain \left{\begin{array}{ll} |z-1|<1 \\ |z-\frac{1}{3}|>\frac{1}{3} \end{array}\right, cut along the segment [1;2], to display conformal in the stripe 0<\text{Im} w<1.

Thanks in advansed :) .

[Hyperbolful]

Is it asking find the cirlce that is mapped to the line under the given transformation?

And the answer to 1. is similar to the map of the inverse transformation, but now the angle has changed.

It might be helpful to think of it as a composition of two operations, inversion, and then squaring. Put z into polar coordinates, and the map should be fairly strightforward

[yus310]

for z=1, then z=e^(i*thetha), so plug into the mapping equation, so you get w in terms of e^(i*thetha), you can do that by getting rid of the bottom by e^(-i*thetha)-2, so you get a real & imaginary part in numerator, and real part in denominator. Thus you get real and imaginary parts and this corresponds to the real and imaginary parts and hence the image of the mapping.. hopefully this helps.

[ferry2]

Thanks for the replies. Already handled with these problems :cool:.