# Conformal Mappings

The function, $$u(x,0) = \begin{cases} 0, |x|\ge1;\\ 1 & |x|<1 \end{cases}$$ when mapped by the function $$F(z) = \log \frac{1-z}{1+z} = k+il$$

turns out to be $1$ when $l=\pi$ and $0$ when $l=0$ . How? Could anyone please explain me this ?

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I don't understand the question. Are you asking about $F(u)$? –  Ron Gordon Mar 18 '13 at 13:04