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Question Number 139641 by EnterUsername last updated on 30/Apr/21

$$\mathrm{Let}a,b\mathrm{be}\mathrm{non}-\mathrm{zero}\mathrm{complex}\mathrm{numbers}\mathrm{and}{z}_1,z_2\mathrm{be}$$$$\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{equation}{z}^2+az+b=0.\mathrm{If}\mathrm{there}\mathrm{exists}$$$$\lambda ⩾4\mathrm{such}\mathrm{that}{a}^2=\lambda b,\mathrm{then}\mathrm{the}\mathrm{points}{z}_1,z_2\mathrm{and}\mathrm{the}$$$$\mathrm{origin}$$$$\left(\mathrm{A}\right)\mathrm{form}\mathrm{an}\mathrm{equilateral}\mathrm{triangle}$$$$\left(\mathrm{B}\right)\mathrm{form}\mathrm{a}\mathrm{right}-\mathrm{angled}\mathrm{triangle},\mathrm{right}\mathrm{angled}\mathrm{at}\mathrm{the}$$$$\mathrm{origin}$$$$\left(\mathrm{C}\right)\mathrm{are}\mathrm{collinear}$$$$\left(\mathrm{D}\right)\mathrm{form}\mathrm{an}\mathrm{obtuse}-\mathrm{angled}\mathrm{triangle}$$

Answered by MJS_new last updated on 30/Apr/21

$$\lambda ⩾4\Rightarrow \lambda \in \mathbb{R}$$$$\mathrm{let}k\in \mathbb{R}\wedge \lambda =4+k^2$$$$z^2+az+\frac{a^2}{4+k^2}=0$$$$z=-\frac{a}{2}(1\pm \frac{k}{\sqrt{k^2+4}})$$$$1\pm \frac{k}{\sqrt{k^2+4}}\in \mathbb{R}\mathrm{\forall }k\in \mathbb{R}\Rightarrow z_1\mathrm{and}{z}_2\mathrm{are}\mathrm{colinear}$$

Commented by EnterUsername last updated on 30/Apr/21

$$\mathrm{Thank}\mathrm{you}\mathrm{Sir}.\mathrm{Can}\mathrm{it}\mathrm{also}\mathrm{be}\mathrm{done}\mathrm{as}$$$$\frac{z_1-z_0}{z_2-z_0}=\cos \alpha +\mathrm{isin}\alpha .$$$$\mathrm{Then}\mathrm{proving}\mathrm{that}\alpha =\mathrm{k}\pi ,\mathrm{k}\in \mathbb{Z}$$

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