Complex Numbers And Quadratic Equations question 152
Question: If at least one value of the complex number $ z=x+iy $ satisfy the condition $ |z+\sqrt{2}|=a^{2}-3a+2 $ and the inequality $ |z+i\sqrt{2}|<a^{2} $ , then
Options:
A) $ a>2 $
B) $ a=2 $
C) $ a<2 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
If $ z=x+iy $ is a complex number satisfying the given conditions, then $ a^{2}-3a+2=|z+\sqrt{2}|=|z+i\sqrt{2}+\sqrt{2}-i\sqrt{2}| $ $ \le |z+i\sqrt{2}|+\sqrt{2}|1-i| $ $ <a^{2}+2 $
Þ $ -3a<0\Rightarrow a>0 $ …..(i) Since $ |z+\sqrt{2}|=a^{2}-3a+2 $ represents a circle with centre at $ A(-\sqrt{2},0) $ and radius $ \sqrt{a^{2}-3a+2} $ , and $ |z+\sqrt{2}i| $ $ <a^{2} $ represents the interior of the circle with centre at $ B(0,-\sqrt{2}) $ and radius $ a $ , therefore there will be a complex number satisfying the given condition and the given inequality if the distance $ AB $ is less than the sum or difference of the radii of the two circles, i.e., if $ \sqrt{{{(-\sqrt{2}-0)}^{2}}+{{(0+\sqrt{2})}^{2}}}<\sqrt{a^{2}-3a+2}\pm a $
Þ $ 2\pm a<\sqrt{a^{2}-3a+2} $
Þ $ 4+a^{2}\pm 4a<a^{2}-3a+2 $
Þ $ -a<-2 $ or $ 7a<-2 $
Þ $ a>2 $ or $ a<-\frac{7}{2} $ But $ a>0 $ from (i), therefore $ a>2 $ .
BETA
