Question

La 282 cineva vreo idee? :'(

Answer 1

[tex]z=a+bi\\ |z'|= \frac{|z-1|}{|1-\overline{z}|}= \frac{|a-1+bi|}{|1-a+bi|} = \frac{ \sqrt{(a-1)^2+b^2} }{ \sqrt{(1-a)^2+b^2} } =1[/tex]