MyQuestionIcon

1

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

Solving Linear Differential Equations of First Order

Value s -i1...

Question

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

A

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

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

B

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

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

C

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

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

D

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

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

Open in App

Solution

The correct options are
A $\frac{\sqrt{3} - i}{2}$
B $\frac{- \sqrt{3} - i}{2}$
$- i$ $= i^{3}$
Hence
$(- i)^{\frac{1}{3}}$ = $(i^{3})^{\frac{1}{3}}$
Now we know that cube root of unity is $1, w, w^{2}$
Thus cube root of $(i^{3})$ will be
$i, i w, i w^{2}$
$= i, \frac{\sqrt{3} - i}{2}, \frac{- \sqrt{3} - i}{2}$

Suggest Corrections

thumbs-up

0

Similar questions

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

arg (z) = \frac{π}{3}

arg (z - 1) = \frac{2 π}{3},

z

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

z_{3} = a i / 3

i^{2} = - 1

are collinear then the value of

a

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

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

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

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

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

Find the real values of x and y, if

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

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Methods of Solving First Order, First Degree Differential Equations

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Solving Linear Differential Equations of First Order

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon