Complex Numbers And Quadratic Equations question 455
Question: If $ k+| k+z^{2} |={{| z |}^{2}}(k\in {R^{-}}) $ , then possible argument of z is
Options:
0
B) $ \pi $
C) $ \pi /2 $
D) none of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ | k+z^{2} |=| z^{2} |+| k |$
$ \Rightarrow $ $ k,z^{2} $ and 0+i0 are collinear
$ \Rightarrow $ arg ( $ z^{2} $ )= arg (k)
$ \Rightarrow $ 2 arg (z) = $ \pi $
$ \Rightarrow $ arg (z) = $ \frac{\pi }{2} $
BETA
