Question

If a and b are real numbers between 0 & 1 such that the points $z_1=a+i,z_2=1+bi$ &$z_3=0$ form an equilateral triangle, then find $a$ and $b$

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

Answer 1

The correct option is B $a=b=2-\sqrt{3}$

Hence
$|z_3-z_1|=|z_3-z_2|$
$a^2+1=b^2+1$
Or
$a=\pm b$
and
$z_1^2+z_2^2+z_3^2=z_1.z_2+z_3{z}_2+z_3{z}_1$
$a^2-1+2ai+1-b^2+2bi=(a-b)+i(1+ab)$
$(a^2-b^2)+i\left(2\right(a+b\left)\right)=(a-b)+i(1+ab)$
Hence
$2(a+b)=1+ab$
Considering $b=a$, we get
$b^2+1-4b=0$ ....................(1)
$b=\frac{4\pm \sqrt{16-4}}{2}$
$=2\pm \sqrt{3}$
Now
$0<a<1$ and $0<b<1$
Hence
$a=b=2-\sqrt{3}$

Now, considering $a=-b$,

$1-b^2=0$ $\left[\text{Putting a = -b in equation (1)}\right]$

$\Rightarrow (1+b)(1-b)=0$

$\therefore b=1orb=-1$

As $a$ and $b$ are real numbers between $0$ and $1$, above values will be rejected.

Hence, $a=b=2-\sqrt{3}$ is the answer.