Question

Fie x€C\{i} astfel încât (z+i)/(1+iz) este număr real. Calculați modulul lui z.

Answer 1

$\displaystyle Fie~z=a+bi,~unde~a,b \in \mathbb{R}. \\ \\ \frac{z+i}{1+iz}= \frac{a+(b+1)i}{1-b+ai}= \frac{ \big(a+(b+1)i \big) \big(1-b-ai \big)}{(1-b+ai)(1-b-ai)}= \\ \\ = \frac{a(1-b)+(1-b)(b+1)i-a^2i+a(b+1)}{(1-b)^2+a^2}= \\ \\ \frac{2a}{a^2+(1-b)^2}+ \frac{ \left(1-b^2-a^2 \right)}{a^2+(1-b)^2} \cdot i. \\ \\ Acest~numar~este~real ~daca~si ~ numai~daca~partea~sa ~imaginara\\ \\~este~egala~cu~0.~Deci:$

$\displaystyle \frac{ 1-b^2-a^2 }{a^2+(1-b)^2}=0 \Rightarrow 1-b^2-a^2=0 \Rightarrow a^2+b^2=1. \\ \\ Dar~a^2+b^2=|z|^2 \Rightarrow |z|^2=1,~si~cum~|z| \in \mathbb{R},~|z| \geq 0 \Rightarrow \boxed{|z|=1 }~.$