# Question-68189

Question Number 68189 by peter frank last updated on 06/Sep/19
Commented by peter frank last updated on 06/Sep/19
$${Qn}\:\mathrm{9},\:\:\mathrm{10} \\$$
Commented by mind is power last updated on 06/Sep/19
$$\mathrm{9} \\$$$$\frac{{z}_{\mathrm{3}} −{z}_{\mathrm{1}} }{{z}_{\mathrm{2}} −{z}_{\mathrm{1}} }=+−{i} \\$$$$\Rightarrow\left({z}_{\mathrm{3}} −{z}_{\mathrm{1}} \right)^{\mathrm{2}} =−\left({z}_{\mathrm{2}} −{z}_{\mathrm{1}} \right)^{\mathrm{2}} \\$$$$\Rightarrow{z}_{\mathrm{3}} ^{\mathrm{2}} −\mathrm{2}{z}_{\mathrm{1}} {z}_{\mathrm{3}} +{z}_{\mathrm{1}} ^{\mathrm{2}} =−{z}_{\mathrm{2}} ^{\mathrm{2}} −{z}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{2}{z}_{\mathrm{2}} {z}_{\mathrm{1}} \\$$$$\Rightarrow{z}_{\mathrm{3}} ^{\mathrm{2}} +\mathrm{2}{z}_{\mathrm{1}} ^{\mathrm{2}} +{z}_{\mathrm{2}} ^{\mathrm{2}} =\mathrm{2}\left({z}_{\mathrm{1}} {z}_{\mathrm{3}} +{z}_{\mathrm{1}} {z}_{\mathrm{2}} \right) \\$$
Commented by mind is power last updated on 06/Sep/19
$$\left.\mathrm{10}\right)\:{us}_{} {e}\:\frac{{z}_{\mathrm{2}} −\mathrm{0}}{{z}_{\mathrm{1}} −\mathrm{0}}={e}^{\underset{−} {+}{i}\frac{\pi}{\mathrm{3}}} \\$$
Commented by peter frank last updated on 07/Sep/19
$${please}\:{sir}.{elaborate}\:{and}\:{where}\:{this}\: \\$$$$\frac{{z}_{\mathrm{2}} −\mathrm{0}}{{z}_{\mathrm{1}} −\mathrm{0}}={e}^{\pm\frac{\pi}{\mathrm{3}}} \:? \\$$