Complex Numbers And Quadratic Equations question 315
Question: If $ a $ and $ b $ are real numbers between 0 and 1 such that the points $ z_1=a+i,z_2=1+bi $ and $ z_3=0 $ form an equilateral triangle, then [IIT 1989]
Options:
A) $ a=b=2+\sqrt{3} $
B) $ a=b=2-\sqrt{3} $
C) $ a=2-\sqrt{3},b=2+\sqrt{3} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Since the triangle with vertices $ z_1=a+i,z_2=1+bi $ and $ z_3=0 $ is equilateral, we have $ z_1^{2}+z_2^{2}+z_3^{2}=z_1z_2+z_2z_3+z_3z_1 $
Þ $ {{(a+i)}^{2}}+{{(1+ib)}^{2}}+0=(a+i)(1+ib)+0+0 $
Þ $ a^{2}-b^{2}+2i(a+b)=a-b+i(1+ab) $ Equating real and imaginary parts, $ a^{2}-b^{2}=a-b $ ??(i) and $ 2(a+b)=1+ab $ ….. (ii) From (i), $ (a-b)[(a+b)-1]=0 $
Þ either $ a=b $ or $ a+b=1 $ Taking $ a=b $ , we get from (ii) $ 4a=1+a^{2} $ or $ a^{2}-4a+1=0 $
$ \therefore a=\frac{4\pm \sqrt{16-4}}{2}=2\pm \sqrt{3} $ Since $ 0<a<1 $ and $ 0<b<1, $ we have $ a=b=2-\sqrt{3} $ Taking $ a+b=1 $ or $ b=1-a, $ we get from (ii) $ 2=1+a(1-a) $ or $ a^{2}-a+1=0 $ , which gives imaginary values of $ a $ . Hence $ a=b=2-\sqrt{3} $ is the required value of $ a $ and $ b $ .
BETA
