Question Number 27059 by math solver last updated on 01/Jan/18
Commented by prakash jain last updated on 01/Jan/18
$$\mathrm{1}+{z}+{z}^{\mathrm{2}} ={x}−{xz}+{xz}^{\mathrm{2}} \\ $$$${z}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)+{z}\left(\mathrm{1}+{x}\right)+\left(\mathrm{1}−{x}\right)=\mathrm{0} \\ $$$$\mathrm{Given}\:{z}\:\mathrm{is}\:\mathrm{not}\:\mathrm{real},\:\mathrm{root}\:\mathrm{are} \\ $$$$\mathrm{complex}\:\mathrm{conjugate} \\ $$$$\alpha={z},\beta=\bar {{z}} \\ $$$$\mid{z}\mid^{\mathrm{2}} ={z}\centerdot\bar {{z}}=\alpha\centerdot\beta=\frac{\mathrm{1}−{x}}{\mathrm{1}−{x}}=\mathrm{1} \\ $$
Commented by math solver last updated on 02/Jan/18
$${thank}\:{you}\:{sir}! \\ $$