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JEE MainMathematicsModerateComplex Numbers

If z2+z∣z∣+∣z2∣=0,z^{2} + z|z| + \left| z^{2} \right| = 0, then the locus
of z is

Show answer and solution
(z∣z∣)2+(z∣z∣)+1=0\left( \frac{z}{|z|} \right)^{2} + \left( \frac{z}{|z|} \right) + 1 = 0
⇒z∣z∣=ω,ω2\Rightarrow \frac{z}{|z|} = \omega,\omega^{2}
⇒z=∣z∣ω,∣z∣ω2\Rightarrow z = |z|\omega,|z|\omega^{2}
Let z = x + iy
⇒y+3x=0 or y−3x=0\Rightarrow y + \sqrt{3}x = 0\ or\ y - \sqrt{3}x = 0
⇒y2−3x2=0\Rightarrow y^{2} - 3x^{2} = 0

Source: Grass-G / M-18 Mathematics Paper 27 Nov Third.docx

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