Question

If $a_r={(\cos 2r\pi +i\sin 2r\pi )}^{1/9}$, then the value of $\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ a_4& a_5& a_6\\ a_7& a_8& a_9\end{array}\right|$ is?

• A. $1$
• B. $-1$
• C. $0$
• D. $2$

Answer 1

The correct option is C $0$

$cos\alpha +isin\alpha =e^{\alpha }$ ...(Euler's representation of complex number)

$a_r=(cos2r\pi +isin2r\pi {)}^{1/9}=(e^{i2r\pi }{)}^{1/9}$

$\therefore a_r=e^{\frac{i2r\pi }{9}}$

$\therefore a_r=e^{r\theta }$ ...$[\theta =\frac{i2\pi }{9}]$

$\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ a_4& a_5& a_6\\ a_7& a_8& a_9\end{array}\right|$

$=\left|\begin{array}{ccc}{e}^{\theta }& e^{2\theta }& e^{3\theta }\\ e^{4\theta }& e^{5\theta }& e^{6\theta }\\ e^{7\theta }& e^{8\theta }& e^{9\theta }\end{array}\right|$

$=\left(e^{3\theta }\right)\left|\begin{array}{ccc}{e}^{\theta }& e^{2\theta }& e^{3\theta }\\ e^{\theta }& e^{2\theta }& e^{3\theta }\\ e^{7\theta }& e^{8\theta }& e^{9\theta }\end{array}\right|$

$=0$

Hence, the answer is option (C).