Question Number 192160 by universe last updated on 10/May/23
$$\mathrm{if}\:\mathrm{x},\mathrm{y},\mathrm{z}\:\mathrm{are}\:\mathrm{three}\:\mathrm{distinct}\:\mathrm{complex}\:\mathrm{numbers} \\ $$$$\mathrm{such}\:\mathrm{that}\:\frac{\mathrm{x}}{\mathrm{y}−{z}}+\frac{\mathrm{y}}{\mathrm{z}−\mathrm{x}}+\frac{\mathrm{z}}{\mathrm{x}−\mathrm{y}}\:=\:\mathrm{0}\:\mathrm{then}\: \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\Sigma\:\frac{\mathrm{x}^{\mathrm{2}} }{\left(\mathrm{y}−\mathrm{z}\right)^{\mathrm{2}} } \\ $$
Commented by mehdee42 last updated on 09/May/23
$$\Sigma\:? \\ $$
Commented by Frix last updated on 09/May/23
$$\frac{\mathrm{x}}{\mathrm{y}−\mathrm{x}}+\frac{\mathrm{y}}{\mathrm{z}−\mathrm{x}}+\frac{\mathrm{z}}{\mathrm{x}−\mathrm{y}},\:\mathrm{no}\:\mathrm{symmetry},\:\mathrm{really}? \\ $$
Commented by Frix last updated on 09/May/23
$$\frac{{x}}{{y}−{z}}+\frac{{y}}{{z}−{x}}+\frac{{z}}{{x}−{y}}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\frac{{x}^{\mathrm{2}} }{\left({y}−{z}\right)^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{\left({z}−{x}\right)^{\mathrm{2}} }+\frac{{z}^{\mathrm{2}} }{\left({x}−{y}\right)^{\mathrm{2}} }=\mathrm{2} \\ $$
Answered by York12 last updated on 09/May/23
$$\frac{{x}−{z}}{{y}−{x}}=\frac{{y}}{{x}−{z}}\:\rightarrow\:\left({x}−{z}\right)^{\mathrm{2}} ={y}\left({y}−{x}\right) \\ $$$$\frac{{x}−{z}}{{y}−{x}}+\frac{\underset{−} {+}\sqrt{{y}}}{\underset{−} {+}\sqrt{{y}−{x}}}=\frac{\underset{−} {+}\mathrm{2}\sqrt{{y}}}{\underset{−} {+}\sqrt{{y}−{x}}}=\mathrm{0}\:\rightarrow\:{y}=\mathrm{0} \\ $$$$\therefore\:{x}={z} \\ $$$$\therefore\:\underset{{sym}} {\sum}\frac{{x}^{\mathrm{2}} }{\left({y}−{z}\right)^{\mathrm{2}} }=\mathrm{2} \\ $$