Question

Sa se determine z apartine lui C din egalitatea : Re( (2i-z)/(i-z) ) = 1 . Multumesc mult de tot !!!

Answer 1

Fie z = a + bi

$\frac{2i-z}{i-z} = \frac{2i-a-bi}{i-a-bi} = \frac{-a+(2-b)i}{-a+(1-b)i}$

Amplificam cu conjugata:

$\frac{(-a+(2-b)i)(-a-(1-b)i)}{a^{2} +(1-b)^2} = \frac{a^2-(a(1-b)+(2-b))i+(2-b)(1-b)}{a^2+(1-b)^2}$

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