Question

Hello,

I need help with Problem 4.

Please show all the steps and the solutions of the problem.

Thank you very much.

4. Show that the image of the upper half plane H2 z E C: 3(2) > 0) under the map C(z) =-i is the disk D-{z E C : 려 < 1].

Answer 1

Solution:

$\dpi{100} \small C\left ( z \right )=\frac{z-i}{z&plus;i}$

$\dpi{100} \small w=\frac{z-i}{z&plus;i}$

$\dpi{100} \small \therefore wz&plus;wi=z-i$

$\dpi{100} \small \therefore z\left ( w-1 \right )=-i-iw$

$\dpi{100} \small \therefore z=\frac{i \left ( 1&plus;w \right ) }{1-w}$

$\dpi{100} \small \therefore z=\frac{i \left ( 1&plus;u&plus;iv \right ) }{1-\left ( u-iv \right )}$

$\dpi{100} \small \therefore z=\frac{i \left ( 1&plus;u&plus;iv \right ) }{\left ( 1-u \right )-iv}$

$\dpi{100} \small \therefore z=\frac{i\left [ \left ( 1&plus;u&plus;iv \right ) \right ]\left [ \left ( 1-u\right )&plus;iv \right ] }{\left [ \left ( 1-u \right )-iv \right ]\left [ \left ( 1-u \right )&plus;iv \right ]}$

$\dpi{100} \small \therefore z=\frac{i\left [ \left ( 1-u^{2} \right )&plus;iv\left ( 1&plus;u \right )&plus;iv\left ( 1-u \right )-v^{2}\right ]}{\left ( 1-u \right )^{2}&plus;v^{2}}$

$\dpi{100} \small \therefore z=i\cdot \frac{\left ( 1-u^{2} \right )&plus;2iv-v^{2}}{\left ( 1-u \right )^{2}&plus;v^{2}}$

$\dpi{100} \small \therefore x&plus;iy= \frac{\left ( 1-u^{2} \right )-2v-iv^{2}}{\left ( 1-u \right )^{2}&plus;v^{2}}$

Comparing the imaginary part on both side we get

$\dpi{100} \small \therefore y= \frac{\left ( 1-u^{2} \right )-v^{2}}{\left ( 1-u \right )^{2}&plus;v^{2}}$

But $\dpi{100} \small y> 0$

$\dpi{100} \small \therefore 1-u^{2}-v^{2}>0$

$\dpi{100} \small \therefore u^{2}-v^{2}< 1$

Thus, the image of the upper half plane  $\dpi{100} \small H^{2}$ is the disk  $\dpi{100} \small D^{^{2}}$​​​​​​ .