Question

Determinati numarul complex z stiind ca | z |=| 1 - z |=| z² |

Answer 1

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Answer 2

[tex]\it\ |z| = |1-z| \Rightarrow z\ne 0 \ \ \ \ (1) \\\;\\ |z| =|z^2| \Rightarrow |z| =|z|^2 \stackrel{(1)}{\Longrightarrow} |z| = 1 \ \ \ \ (2) \\\;\\ Fie\ z = a+bi \Rightarrow \sqrt{a^2+b^2} =|z| =1 \Rightarrow a^2+b^2=1 \ \ \ (3)[/tex]

[tex]\it |z| = |1-z| \Rightarrow 1 = |z-1| \Rightarrow 1= |a+bi-1| \Rightarrow |(a-1)+bi| =1\Rightarrow \\\;\\ \Rightarrow \sqrt{(a-1)^2+b^2} =1 \Rightarrow (a-1)^2+b^2 =1 \Rightarrow \\\;\\ \Rightarrow a^2-2a+1+b^2=1 \Rightarrow a^2+b^2-2a=0 \stackrel{(3)}{\Longrightarrow } 1-2a=0 \Rightarrow \\\;\\ \Rightarrow a = \dfrac{1}{2} \ \ \ \ \ (4)[/tex]

$\it (3),\ (4) \Rightarrow \left(\dfrac{1}{2}\right)^2 +b^2=1 \Rightarrow \dfrac{1}{4}+b^2=1 \Rightarrow b^2= \dfrac{3}{4} \Rightarrow b=\pm\dfrac{\sqrt3}{2}$

Ecuația admite două soluții:

$\it z_1 = \dfrac{1}{2} -\dfrac{\sqrt3}{2}i, \ \ \ z_2 = \dfrac{1}{2}+\dfrac{\sqrt3}{2}i$