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

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

Answered by ajfour last updated on 03/Sep/17

$$letz={re}^{i\theta }$$$$\Rightarrow r^2{e}^{2i\theta }+r^2{e}^{i\theta }+r^2=0$$$$\Rightarrow e^{2i\theta }+e^{i\theta }+1=0$$$$or{e}^{i\theta }+e^{-i\theta }=-1$$$$or2\cos \theta =-1$$$$\frac{x}{\sqrt{x^2+y^2}}=-\frac{1}{2}$$$$rememberx<0$$$$further:4x^2=x^2+y^2$$$$\Rightarrow y=\pm \sqrt{3}x(x<0)$$$$orwecansay\mid y\mid =-\sqrt{3}x.$$

Commented by Tinkutara last updated on 03/Sep/17

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

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