Complex Numbers And Quadratic Equations question 822
Question: $ \sqrt{-8-6i}= $ [Roorkee 1979; RPET 1992]
Options:
A) $ 1\pm 3i $
B) $ \pm (1-3i) $
C) $ \pm (1+3i) $
D) $ \pm (3-i) $
Show Answer
Answer:
Correct Answer: B
Solution:
Given that $ \sqrt{-8-6i}=x+iy=z $
Þ $ -8-6i={{(x+iy)}^{2}} $
$ \therefore x^{2}-y^{2}=-8 $ …..(i) and $ 2xy=-6 $ …..(ii) Now $ x^{2}+y^{2}=\sqrt{64+36}=\pm 10 $ …..(iii) From (i) and (iii), we get $ x=\pm 1 $ and $ y=\pm 3 $ Hence $ z=\pm (1-3i) $ Trick: Since $ {{{\pm (1-3i)}}^{2}}=-8-6i $
BETA
