Complex Numbers And Quadratic Equations question 87
Question: If $ |z|=4 $ and $ argz=\frac{5\pi }{6}, $ then z = [MP PET 1987]
Options:
A) $ 2\sqrt{3}-2i $
B) $ 2\sqrt{3}+2i $
C) $ -2\sqrt{3}+2i $
D) $ -\sqrt{3}+i $
Show Answer
Answer:
Correct Answer: C
Solution:
$ |z|=4 $ and $ argz=\frac{5\pi }{6}=150^{o} $ Let $ z=x+iy $ , then $ |z|=r=\sqrt{x^{2}+y^{2}}=4 $ and $ \theta =\frac{5\pi }{6}=150^{o} $
$ \therefore $ $ x=r\cos \theta =4\cos 150^{o}=-2\sqrt{3} $ . and $ y=r\sin \theta =4 $ $ \sin 150^{o}=4\frac{1}{2}=2 $
$ \therefore $ $ z=x+iy=-2\sqrt{3}+2i $ Trick: Since $ argz=\frac{5\pi }{6}=150^{o} $ , here the complex number must lie in second quadrant, so (a) and (b) rejected. Also $ |z|=4 $ which satisfies (c) only.
BETA
