Value(s) of $(- i)^{1 / 3}$ is/are

\frac{\sqrt{3} - i}{2}

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\frac{\sqrt{3} + i}{2}

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\frac{- \sqrt{3} - i}{2}

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\frac{- \sqrt{3} + i}{2}

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Solution

The correct options are
A $\frac{- \sqrt{3} - i}{2}$
B $\frac{\sqrt{3} - i}{2}$
Let $x^{3} = - i$
or
$\Rightarrow (x^{3} + i) = 0$
$\Rightarrow (x^{3} + (- i)^{3}) = 0$
$\Rightarrow (x^{3} - (i)^{3}) = 0$
$\Rightarrow (x - i) (x^{2} - 1 + i x) = 0$
$∴ x = i$
and
$\Rightarrow x^{2} + i x - 1 = 0$
$\Rightarrow x = \frac{- i \pm \sqrt{- 1 + 4}}{2}$
$= \frac{- i \pm \sqrt{3}}{2}$

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Similar questions

The value of $(- i)^{\frac{1}{3}}$ is

Q. Find the real values of x and y, if
(i)

(x + i y) (2 - 3 i) = 4 + i

(ii)

(3 x - 2 i y) (2 + i)^{2} = 10 (1 + i)

(iii)

\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i}

(iv)

(1 + i) (x + i y) = 2 - 5 i

Q. If

arg (z) = \frac{π}{3}

and

arg (z - 1) = \frac{2 π}{3},

then

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Find the real values of x and y, if

$(i) (x + i y) (2 - 3 i) = 4 + i (i i) (3 x - 2 i y) (2 + i)^{2} = 10 (1 + i) (i i i) \frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i (i v) (1 + i) (x + i y) = 2 - 5 i$

Q. If

z_{1} = 1 + i, z_{2} = - 2 + 3 i

and

z_{3} = a i / 3

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Value(s) of (−i)1/3 is/are

The correct options are A −√3−i2 B √3−i2Let x3=−ior ⇒(x3+i)=0⇒(x3+(−i)3)=0⇒(x3−(i)3)=0⇒(x−i)(x2−1+ix)=0∴x=iand ⇒x2+ix−1=0⇒x=−i±√−1+42=−i±√32

Value(s) of $(- i)^{1 / 3}$ is/are