Complex Numbers And Quadratic Equations question 154
Question: If $ z^{2}+z|z|+|z{{|}^{2}}=0 $ , then the locus of $ z $ is
Options:
A) A circle
B) A straight line
C) A pair of straight lines
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ z^{2}+z|z|+|z{{|}^{2}}=0 $
Þ $ {{( \frac{z}{|z|} )}^{2}}+\frac{z}{|z|}+1=0 $
Þ $ \frac{z}{|z|}=\omega ,{{\omega }^{2}} $
Þ $ z=\omega |z| $ or $ z={{\omega }^{2}}|z| $
Þ $ x+iy=|z|( \frac{-1}{2}+\frac{i\sqrt{3}}{2} ) $ or $ x+iy=|z|( \frac{-1}{2}-\frac{i\sqrt{3}}{2} ) $
Þ $ x=-\frac{1}{2}|z|,y=|z|\frac{\sqrt{3}}{2} $ or $ x=-\frac{|z|}{2},y=-\frac{|z|\sqrt{3}}{2} $
Þ $ y+\sqrt{3}x=0 $ or $ y-\sqrt{3}x=0 $
Þ $ y^{2}-3x^{2}=0 $ .
BETA
