Question

If the vertices of triangle ABC in argand plane are the roots of equation $z^3+i{z}^2+2i=0$, then which of the following is (are) correct
[Note: where $i^2=-1$]

• A. Area of triangle ABC is 2 sq. units
• B. Circumradius of triangle ABC is $\frac{5}{2}$
• C. Triangle ABC is isosceles
• D. Inradius of triangle ABC is $\frac{\sqrt{5}-1}{2}$

Answer 1

The correct option is A Area of triangle ABC is 2 sq. units

Given cubic equation is

$z^3+i{z}^2+2i=0$

Now put $z=i$ in LHS

$i^3+i\times i^2+2i$

$=i\times i^2+i\times i^2+2i$

$=-i-i+2i$ (since$i^2=-1$)

$=-2i+2i$

$=0$

= RHS

So $z=i$ is a factor of the given equation.

Now when we divide $(z^3+i{z}^2+2i)$ by $(z-i)$, we get other two factors $(1-i)$ and $(-1-i)$.

So three roots are $i,1-i,-1-i.$

Now vertices of triangle are $(0,1),(1,-1),(-1,-1)$.

Now are of the triangle is defined as

$\frac{1}{2}\times base\times height$

$=\frac{1}{2}\left(\right(1+1\left)\right(1+1\left)\right)$

$=\frac{1}{2}\left(2\right)\left(2\right)$

$=2$ sq. units

Area of the triangle is $2$ sq. units