Fie z = a + bi
[tex] \frac{2i-z}{i-z} = \frac{2i-a-bi}{i-a-bi} = \frac{-a+(2-b)i}{-a+(1-b)i} [/tex]
Amplificam cu conjugata:
[tex] \frac{(-a+(2-b)i)(-a-(1-b)i)}{a^{2} +(1-b)^2} = \frac{a^2-(a(1-b)+(2-b))i+(2-b)(1-b)}{a^2+(1-b)^2} [/tex]
Partea reala a fractiei este:
[tex] \frac{a^2+(2-b)(1-b)}{a^2+(1-b)^2} =1\rightarrow a^2+(2-b)(1-b)=a^2+(1-b)^2\\
b^2-3b+2=b^2-2b+1\\
b-1=0 \rightarrow b=1[/tex]
z = a + i, pentru orice a ∈ R
