Question

Value(s) of $ {{\left( -i \right)}^{\dfrac{1}{3}}} $ is/are:
This question has multiple correct options.
A. $ \dfrac{\sqrt{3}-i}{2} $
B. $ \dfrac{\sqrt{3}+i}{2} $
C. $ \dfrac{-\sqrt{3}-i}{2} $
D. $ \dfrac{-\sqrt{3}+i}{2} $

Answer 1

Hint: Remember that $ i=\sqrt{-1} $ . Therefore, $ {{i}^{2}}=-1 $ and $ {{i}^{3}}=-i $ .
There are three cube roots of any number, one is real and two are complex numbers.
Use the following identity: $ {{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+ab+{{b}^{2}}) $ .
Solve the quadratic equation $ a{{x}^{2}}+bx+c=0 $ by using the quadratic formula: $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $.

Complete step by step answer:
We can write $ {{\left( -i \right)}^{\dfrac{1}{3}}}=z $ .
Cubing both the sides, we will get:
⇒ $ -i={{z}^{3}} $
Using the value $ -i={{i}^{3}} $ and bringing all the terms to the same side of the equation, we get:
⇒ $ {{z}^{3}}-{{i}^{3}}=0 $
On using the factorization $ {{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+ab+{{b}^{2}}) $ , we have:
⇒ $ (z-i)({{z}^{2}}+zi+{{i}^{2}})=0 $
Using $ {{i}^{2}}=-1 $ and the fact that the product of two terms can be 0 only if at least one of them is 0, we conclude the following:
⇒ $ z-i=0 $ OR $ {{z}^{2}}+iz-1=0 $
Using the quadratic formula, we get:
⇒ $ z=i $ OR $ z=\dfrac{-i\pm \sqrt{{{i}^{2}}-4(1)(-1)}}{2(1)} $
⇒ $ z=i $ OR $ z=\dfrac{-i\pm \sqrt{-1+4}}{2} $
⇒ $ z=i $ OR $ z=\dfrac{-i+\sqrt{3}}{2} $ OR $ z=\dfrac{-i-\sqrt{3}}{2} $

∴ The correct answer options are A. $ \dfrac{\sqrt{3}-i}{2} $ and C. $ \dfrac{-\sqrt{3}-i}{2} $ .

Note: There are exactly 'n' nth-roots of any number.
The complex roots of a real number always occur in conjugate pairs.
One can also find the nth power of any complex number using the De' Moivre's formula:
 $ {{\left( \cos \theta +i\sin \theta \right)}^{n}}=\cos n\theta +i\sin n\theta $
A complex number can always be expressed in the polar form:
 $ r{{e}^{i\theta }}=r\left( \cos \theta +i\sin \theta \right) $