IIT-JEE 2010 · Paper 1 · Mathematics Q53

← IIT-JEE 2010 · Paper 1

MathematicsInteger+3Question 53 of 84

Let $\omega$ be the complex number $\cos\dfrac{2\pi}3+i\sin\dfrac{2\pi}3$ . Then the number of distinct complex numbers z satisfying $\begin{vmatrix}z+1&\omega&\omega^2\\\omega&z+\omega^2&1\\\omega^2&1&z+\omega\end{vmatrix}=0$ is equal to

Your answer

Answer and worked solution

Official answer: 1

Add all columns to the first. Each row then starts with $z+1+\omega+\omega^2=z$ :
$\Delta=z\begin{vmatrix}1&\omega&\omega^2\\1&z+\omega^2&1\\1&1&z+\omega\end{vmatrix}.$
Subtracting the first row from the others and expanding (using $\omega^3=1$ , $1+\omega+\omega^2=0$ ) gives $\Delta=z^3$ .

Roots. $z^3=0$ has only $z=0$ : one distinct complex number.

Answer: 1

Topics:Determinants and properties Roots of unity

← Previous question Next question →

IIT-JEE 2010 · Paper 1, Mathematics question 53