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

9 ((z_3 −z_1 )/(z_2 −z_1 ))=+−i ⇒(z_3 −z_1 )^2 =−(z_2 −z_1 )^2 ⇒z_3 ^2 −2z_1 z_3 +z_1 ^2 =−z_2 ^2 −z_1 ^2 +2z_2 z_1 ⇒z_3 ^2 +2z_1 ^2 +z_2 ^2 =2(z_1 z_3 +z_1 z_2 )

$$\mathrm{9} \\ $$$$\frac{{z}_{\mathrm{3}} −{z}_{\mathrm{1}} }{{z}_{\mathrm{2}} −{z}_{\mathrm{1}} }=+−{i} \\ $$$$\Rightarrow\left({z}_{\mathrm{3}} −{z}_{\mathrm{1}} \right)^{\mathrm{2}} =−\left({z}_{\mathrm{2}} −{z}_{\mathrm{1}} \right)^{\mathrm{2}} \\ $$$$\Rightarrow{z}_{\mathrm{3}} ^{\mathrm{2}} −\mathrm{2}{z}_{\mathrm{1}} {z}_{\mathrm{3}} +{z}_{\mathrm{1}} ^{\mathrm{2}} =−{z}_{\mathrm{2}} ^{\mathrm{2}} −{z}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{2}{z}_{\mathrm{2}} {z}_{\mathrm{1}} \\ $$$$\Rightarrow{z}_{\mathrm{3}} ^{\mathrm{2}} +\mathrm{2}{z}_{\mathrm{1}} ^{\mathrm{2}} +{z}_{\mathrm{2}} ^{\mathrm{2}} =\mathrm{2}\left({z}_{\mathrm{1}} {z}_{\mathrm{3}} +{z}_{\mathrm{1}} {z}_{\mathrm{2}} \right) \\ $$

Commented by mind is power last updated on 06/Sep/19

10) us_ e ((z_2 −0)/(z_1 −0))=e^(+_− i(π/3))

$$\left.\mathrm{10}\right)\:{us}_{} {e}\:\frac{{z}_{\mathrm{2}} −\mathrm{0}}{{z}_{\mathrm{1}} −\mathrm{0}}={e}^{\underset{−} {+}{i}\frac{\pi}{\mathrm{3}}} \\ $$

Commented by peter frank last updated on 07/Sep/19

please sir.elaborate and where this ((z_2 −0)/(z_1 −0))=e^(±(π/3)) ?

$${please}\:{sir}.{elaborate}\:{and}\:{where}\:{this}\: \\ $$$$\frac{{z}_{\mathrm{2}} −\mathrm{0}}{{z}_{\mathrm{1}} −\mathrm{0}}={e}^{\pm\frac{\pi}{\mathrm{3}}} \:? \\ $$

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