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Question Number 20799 by Tinkutara last updated on 03/Sep/17

If z^2 + z∣z∣ + ∣z∣^2 = 0, then locus of z is

$$\mathrm{If}\:{z}^{\mathrm{2}} \:+\:{z}\mid{z}\mid\:+\:\mid{z}\mid^{\mathrm{2}} \:=\:\mathrm{0},\:\mathrm{then}\:\mathrm{locus}\:\mathrm{of}\:{z}\:\mathrm{is} \\ $$

Answered by ajfour last updated on 03/Sep/17

let z=re^(iθ) ⇒ r^2 e^(2iθ) +r^2 e^(iθ) +r^2 =0 ⇒ e^(2iθ) +e^(iθ) +1=0 or e^(iθ) +e^(−iθ) =−1 or 2cos θ=−1 (x/(√(x^2 +y^2 ))) =−(1/2) remember x < 0 further : 4x^2 =x^2 +y^2 ⇒ y=± (√3)x (x<0) or we can say ∣y∣=−(√3)x .

$${let}\:{z}={re}^{{i}\theta} \\ $$$$\Rightarrow\:\:{r}^{\mathrm{2}} {e}^{\mathrm{2}{i}\theta} +{r}^{\mathrm{2}} {e}^{{i}\theta} +{r}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:\:\:{e}^{\mathrm{2}{i}\theta} +{e}^{{i}\theta} +\mathrm{1}=\mathrm{0} \\ $$$${or}\:\:\:\:\:\:\:\:{e}^{{i}\theta} +{e}^{−{i}\theta} =−\mathrm{1} \\ $$$${or}\:\:\:\:\:\:\:\:\mathrm{2cos}\:\theta=−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{x}}{\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }}\:=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${remember}\:\:\:\:{x}\:<\:\mathrm{0} \\ $$$${further}\::\:\:\:\:\mathrm{4}{x}^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\:\:\:\:{y}=\pm\:\sqrt{\mathrm{3}}{x}\:\:\:\:\:\left({x}<\mathrm{0}\right) \\ $$$${or}\:{we}\:{can}\:{say}\:\:\:\:\mid{y}\mid=−\sqrt{\mathrm{3}}{x}\:. \\ $$

Commented by Tinkutara last updated on 03/Sep/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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