Question

One of the roots of equation $x^3=j$, where $j$ is the positve square roots of $-1$ is

• A. $j$
• B. $\frac{\sqrt{3}}{2}+\frac{j}{2}$
• C. $\frac{\sqrt{3}}{2}-\frac{j}{2}$
• D. $-\frac{\sqrt{3}}{2}-\frac{j}{2}$

Answer 1

The correct option is B $\frac{\sqrt{3}}{2}+\frac{j}{2}$

$x^3=j$ where $j=\sqrt{-1}$
By verification option (b) wil satisfy the given equation
$x^3=j$
For, $x=\frac{\sqrt{3}}{2}+\frac{j}{2}=cos\left(\frac{\pi }{6}\right)+jsin\left(\frac{\pi }{6}\right)=e^{j\frac{\pi }{6}}$
Then, $x^3={\left(e^{j\frac{\pi }{6}}\right)}^3=e^{j\frac{\pi }{2}}=cos\frac{\pi }{2}+jsin,\frac{\pi }{2}=j$
Hence verified.