If-z-2-z-z-z-2-0-then-locus-of-z-is-

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

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

Answered by ajfour last updated on 03/Sep/17

l e t z = {r e}^{i θ}

\Rightarrow r^{2} e^{2 i θ} + r^{2} e^{i θ} + r^{2} = 0

\Rightarrow e^{2 i θ} + e^{i θ} + 1 = 0

o r e^{i θ} + e^{- i θ} = - 1

o r 2 \cos θ = - 1

\frac{x}{\sqrt{x^{2} + y^{2}}} = - \frac{1}{2}

r e m e m b e r x < 0

f u r t h e r : 4 x^{2} = x^{2} + y^{2}

\Rightarrow y = \pm \sqrt{3} x (x < 0)

o r w e c a n s a y ∣ y ∣= - \sqrt{3} x .

Commented by Tinkutara last updated on 03/Sep/17

Thank you very much Sir!