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Complex Numbers

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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z = - 1

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z = - 1

arg (z^{2 / 3})

z

C o l u m n - I

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c o l u m n - I I

$C o l u m n - I$ (equation in z)	$C o l u m n - I I$ (principal value of arg(z))
$z^{2} - z + 1 = 0$	$- 2 π / 3$
$z^{2} + z + 1 = 0$	$- π / 3$
${2 z}^{2} + 1 + i \sqrt{3} = 0$	$π / 3$
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z = \sqrt{3 + 4 i} + \sqrt{- 3 + 4 i}

\pm \frac{π}{4}, \pm \frac{3 π}{4}

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\sqrt{z} = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \sqrt{\frac{| z | + R e (z)}{2}} + i \sqrt{\frac{| z | - R e (z)}{2}} & , if B > 0 \sqrt{\frac{| z | + R e (z)}{2} - i \sqrt{\frac{| z | - R e (z)}{2}}} & , if B < 0 \end{matrix}

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