If z=-1, th...

Question

If $z = - 1$ , then the principal value of the arg $(z^{2 / 3})$ is equal to-

\frac{π}{3}

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\frac{2 π}{3}

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\frac{π}{2}

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π

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Solution

The correct option is B $\frac{2 π}{3}$
Given, $z = - 1$
Thus $z = c o s θ + i s i n θ$

and $R e (z) = - 1$ therefore $c o s θ = - 1$
$∴ θ = π$

$∴ z = c o s π + i s i n π$

$∴ (z^{\frac{2}{3}}) = (c o s π + i s i n π)^{\frac{2}{3}}$

Using De Moivre's Theorem:

$∴ (z^{\frac{2}{3}}) = (c o s \frac{2 π}{3} + i s i n \frac{2 π}{3})$

= $a r g (z^{\frac{2}{3}}) = \frac{2 π}{3}$

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