Question

Demonstrati ca daca |z|=1,atunci$\frac{z-1}{z+1}$ este pur imaginar

Answer 1

[tex]z=a+bi\\ |z|=1=\ \textgreater \ \sqrt{a^2+b^2} =1=\ \textgreater \ a^2+b^2=1\\ \displaystyle \frac{z-1}{z+1} = ^{(a+1)-bi)}\frac{(a-1)+bi}{(a+1)+bi} =\\ = \frac{a^2-1+abi+bi-abi+bi+b^2}{a^2+2a+1+b^2}=\\ = \frac{2bi}{2(a+1)} = \frac{b}{a+1}i \in C-R [/tex]