Complex Numbers And Quadratic Equations question 43
Question: The number of solutions of the equation $ z^{2}+\bar{z}=0 $ is
Options:
A) 1
B) 2
C) 3
D) 4
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ z=x+iy, $ so that $ \overline{z}=x-iy, $ therefore $ z^{2}+\overline{z}=0\Leftrightarrow (x^{2}-y^{2}+x)+i(2xy-y)=0 $ Equating real and imaginary parts, we get $ x^{2}-y^{2}+x=0 $ …..(i) and $ 2xy-y=0 $
Þ $ y=0 $ or $ x=\frac{1}{2} $ If $ y=0 $ , then (i) gives $ x^{2}+x=0\Rightarrow x=0 $ or $ x=-1 $ If $ x=\frac{1}{2}, $ Then $ x^{2}-y^{2}+x=0 $
Þ $ y^{2}=\frac{1}{4}+\frac{1}{2}=\frac{3}{4} $
Þ $ y=\pm \frac{\sqrt{3}}{2} $ Hence, there are four solutions in all.
BETA
