Question

The bilinear transformation $w=\frac{z-1}{z+1}$

• A. maps the inside of the unit circle in the z-plane to the left half of the w-plane
• B. maps the outside the unit circle in the z-plane to the left half of the w-plane
• C. maps the inside of the unit circle in the z-plane to right half of the w-plane
• D. maps the outside the unit circule in the z-plane to the right half of the w-plane.

Answer 1

Answer: (A), (D)

maps the outside the unit circule in the z-plane to the right half of the w-plane.

Given $w=\frac{z-1}{z+1}\Rightarrow z=\frac{-w-1}{w-1}$ where $w=f\left(z\right)=u+iv$
Consider $|z|<1$ which represents interior of unit circle $|z|=1$
$\Rightarrow \left|\frac{-w-1}{w-1}\right|<1$
$\Rightarrow |u+iv+1|<|u+iv-1|$
$\Rightarrow |(u+1)+iv{|}^2<|(u-1)+iv{|}^2$
$\Rightarrow (u+1{)}^2+v^2<(u-1{)}^2+v^2$
$\Rightarrow u^2+1^2+2u+v^2<u^2+1-2u+v^2$
$\Rightarrow 4u<0$
$\Rightarrow u<0$
Hence, the function $w=\frac{z-1}{z+1}$ maps the interior of unit circle in the z plane, to the left half of the w plane.



Similarly, option (d) is also correct