Question

stie cineva sa faca exercitiul 16?

Answer 1

$z = x+yi\\ \\ |z-a| = \sqrt{a^2-b^2} \Rightarrow |(x+yi)-a| = \sqrt{a^2-b^2} \Rightarrow \\ \Rightarrow \Big|(x-a)+yi\Big| = \sqrt{a^2-b^2} \Rightarrow \sqrt{(x-a)^2 + y^2} = \sqrt{a^2-b^2} \Rightarrow \\ \Rightarrow \sqrt{x^2-2ax+a^2+y^2} = \sqrt{a^2-b^2}} \Rightarrow x^2-2ax+a^2+y^2 = a^2-b^2 \Rightarrow \\ \\ \Rightarrow \boxed{x^2+y^2+b^2 = 2ax} \\ \\ \left|\dfrac{b-z}{b+z}\right| = \dfrac{|b-z|}{|b+z|} = \dfrac{\big|b-(x+yi)\big|}{\big|b+(x+yi)\big|} = \dfrac{\big|(b-x)+yi\big|}{\big|(b+x)+yi\big|} = \dfrac{\sqrt{(b-x)^2+y^2}}{\sqrt{(b+x)^2+y^2}} =$

$=\sqrt{\dfrac{(b-x)^2+y^2}{(b+x)^2+y^2}} = \sqrt{\dfrac{b^2-2bx+x^2+y^2}{b^2+2bx+x^2+y^2}} =\\ \\ = \sqrt{\dfrac{(x^2+y^2+b^2)-2bx}{(x^2+y^2+b^2)+2bx}}= \sqrt{\dfrac{2ax-2bx}{2ax+2bx}} =\sqrt{\dfrac{2x(a-b)}{2x(a+b)}} = \sqrt{\dfrac{a-b}{a+b}} \\ \\ \\ \Rightarrow \boxed{\text{e) r\u{a}spuns corect}}$