You visited us 1 times! Enjoying our articles? Unlock Full Access!

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}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2 π}{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{π}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

π

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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:

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

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

Suggest Corrections

Similar questions

^{'} a^{'}

^{'} b^{'}

f (x)

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} x + a \sqrt{2} sin x, & 0 \leq x < \frac{π}{4} 2 x cot x + b, & \frac{π}{4} \leq x \leq \frac{π}{2} a cos 2 x - b sin x, & \frac{π}{2} < x \leq π \end{matrix}

x = \frac{π}{4}, \frac{π}{2}

z = - 1

arg (z^{2 / 3})

z = - 1

arg (z^{2 / 3})

z

C o l u m n - I

a r g (z)

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))
$z^{2} - z + 1 = 0$	$- 2 π / 3$
$z^{2} + z + 1 = 0$	$- π / 3$
${2 z}^{2} + 1 + i \sqrt{3} = 0$	$π / 3$
${2 z}^{2} + 1 - i \sqrt{3} = 0$	$2 π / 3$

z = \sqrt{3 + 4 i} + \sqrt{- 3 + 4 i}

\pm \frac{π}{4}, \pm \frac{3 π}{4}

\sqrt{- 1} = i

\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}

Join BYJU'S Learning Program