Question

If points $z_1=a+i,z_2=1+ib$ and $z_0=0+i0$ form an equilateral triangle where $(a,b)\in (0,1),$ then

• A. $a=3-\sqrt{5}$
• B. $a=2-\sqrt{3}$
• C. $b=3-\sqrt{5}$
• D. $b=2-\sqrt{3}$

Answer 1

Answer: (B), (D)

$b=2-\sqrt{3}$

$\because z_0,z_1,z_2$ are the vertices of an equilateral triangle, then
$z_0^2+z_1^2+z_2^2=z_0{z}_1+z_1{z}_2+z_2{z}_0\Rightarrow z_1^2+z_2^2=z_1{z}_2\Rightarrow a^2-1+2ai+1-b^2+2bi=a+abi+i-b\Rightarrow (a^2-b^2)+i(2a+2b)=(a-b)+i(1+ab)$
Comparing the real and imaginary part, we get
$a^2-b^2=a-b\text{and}2a+2b=1+ab$
Now, using
$a^2-b^2=a-b\Rightarrow (a-b)(a+b-1)=0\Rightarrow a=b\text{or}a+b=1$

When $a=b$
$2(a+b)=1+ab\Rightarrow 4a=1+a^2\Rightarrow a^2-4a+1=0\Rightarrow a=\frac{4\pm \sqrt{12}}{2}\Rightarrow a=2-\sqrt{3}(\because a\in (0,1\left)\right)\Rightarrow b=2-\sqrt{3}$

When $a+b=1$
$2(a+b)=1+ab\Rightarrow 2=1+a(1-a)\Rightarrow a^2-a+1=0D<0$
No real solution
Hence, $a=b=2-\sqrt{3}$