Complex Numbers And Quadratic Equations question 136
Question: Let $ z $ and $ w $ be two complex numbers such that $ |z|\le 1, $ $ |w|\le 1 $ and $ |z+iw|=|z-i\overline{w}|=2 $ . Then $ z $ is equal to [IIT 1995]
Options:
A) 1 or $ i $
B) $ i $ or $ -i $
C) 1 or - 1
D) $ i $ or -1
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ z=a+ib,|z|\le 1 $
Þ $ a^{2}+b^{2}\le 1 $ and $ w=c+id,|w|\le 1 $
Þ $ c^{2}+d^{2}\le 1 $ $ |z+iw|=|a+ib+i(c+id)|=2 $
Þ $ {{(a-d)}^{2}}+{{(b+c)}^{2}}=4 $ ……(i) $ |z-i\overline{w}|=|a+ib-i(c-id)| $
Þ $ {{(a-d)}^{2}}+{{(b-c)}^{2}}=4 $ ……(ii) From (i) and (ii), we get $ bc=0 $
Þ Either $ b=0 $ or $ c=0 $ If $ b=0 $ , then $ a^{2}\le 1 $ . Then, only possibility is $ a=1 $ or $ -1 $ .
BETA
