Question

prove that image of
the vertical line x=k with k is not =0 to the circle mode of w-1/2k = mode of 1/2k

Answer 1

$z\rightarrow \frac{1}{z} .$ Hence $(x+iy)\rightarrow \frac{1}{x+iy}=\frac{x-iy}{x^2+y^2}$ . Let $u=\frac{x}{x^2+y^2}, v= \frac{-y}{x^2+y^2}.$ Then $(u-\frac{1}{2k})^2+ v^2= u^2+v^2-\frac{u}{k} +\frac{1}{4k^2} = \frac{x^2+y^2}{(x^2+y^2)^2}-\frac{u}{k}+\frac{1}{4k^2}$ $=\frac{1}{x^2+y^2}-\frac{u}{k}+\frac{1}{4k^2}$ But here $x=k$ is the given line. Hence $u=\frac{k}{x^2+y^2}\Rightarrow \frac{u}{k}=\frac{1}{x^2+y^2}$ . Hence RHS= $\frac{1}{4k^2}$. Let $w=u+iv.$ Then $|w-\frac{1}{2k}|=\sqrt{(u-\frac{1}{2k})^2+v^2}= \frac{1}{2k .}$

Hence the image of the line is the circle.