The value of - i 1-3 isA- 1-3 i-2B- 1-3 i-2C- -3-i-2D- -3-i-2

The correct options are C D nbsp;Since √(3) i/2= (cosπ/6 +isinπ/6) ⇒ ( √(3) i/2)^3= (cosπ/6 +isinπ/6)^3= i And √(3) i/2=(cosπ/6 +isinπ/6) And ( √(3) i/2) ...

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Byju's Answer

Standard XI

Mathematics

Sign of Trigonometric Ratios in Different Quadrants

The value of ...

Question

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

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Solution

The correct options are
C

D

Since $\frac{- \sqrt{3} - i}{2}$ =- $(c o s \frac{π}{6} + i s i n \frac{π}{6})$

$\Rightarrow$ ${(\frac{- \sqrt{3} - i}{2})}^{3}$ =- ${(c o s \frac{π}{6} + i s i n \frac{π}{6})}^{3}$ =-i

And $\frac{\sqrt{3} - i}{2}$ = $(c o s \frac{π}{6} + i s i n \frac{π}{6})$

And ${(\frac{- \sqrt{3} - i}{2})}^{3}$ = $(c o s \frac{π}{2} + i s i n \frac{π}{2})$ =-i

Hence the result

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Similar questions

∣ ∣ ∣ ∣ \frac{\sqrt{3} + i}{(1 + i) (1 + \sqrt{3} i)} ∣ ∣ ∣ ∣ =

Q. Let

z_{1} = \frac{(\sqrt{3} + i)^{2} . (1 - \sqrt{3} i)}{1 + i}

z_{2} = \frac{(1 + \sqrt{3} i)^{2} . (\sqrt{3} - i)}{1 - i}

Q. If

i^{2} = - 1

, calculate the value of

3 i^{2} + i^{3} - i^{4}

If $x_{n}$ = cos( $\frac{π}{4^{n}}$ ) + i sin( $\frac{π}{4^{n}}$ ) then $x_{1} x_{2} x_{3} . . . . . . . . . \infty =$

Q. If

arg (z) = \frac{π}{3}

and

arg (z - 1) = \frac{2 π}{3},

then

z

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The value of (−i)13 is

The correct options are C D Since −√3−i2=-(cosπ6+isinπ6) ⇒ (−√3−i2)3=-(cosπ6+isinπ6)3=-i And √3−i2=(cosπ6+isinπ6) And (−√3−i2)3 =(cosπ2+isinπ2)=-i Hence the result

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