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: Try to manipulate the given equation to get into the form of algebraic identity of ${{a}^{3}}-{{b}^{3}}$ and then use algebraic identity:
${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$
By using this find the possible algebraic solutions for the required expression.

Complete step-by-step answer:

Definition of i; can be written as,
The solution of the equation: ${{x}^{2}}+1=0$, is i. i is an imaginary number. Any number which has an imaginary number in its representation is called a complex number.
Definition of a complex equation, can be written as, an equation containing complex numbers in it, is called a complex equation.
It is possible to have a real root for complex equations.
Example: $\left( 1+i \right)x+\left( 1+i \right)=0$, $x=1$ is the root of the equation.
Given expression for which we need value(s) is given by:
${{\left( -i \right)}^{\dfrac{1}{3}}}$
So let the given expression be assumed as x here,
$x={{\left( -i \right)}^{\dfrac{1}{3}}}$
By cubing on both sides of this equation we get:
${{x}^{3}}=-i$
By adding the imaginary number i on both sides, we get:
${{x}^{3}}+i=0$
For making the degree same, we do a change to “i”:
${{x}^{3}}+{{\left( -i \right)}^{3}}=0$
This can be simplified as:
${{x}^{3}}-{{i}^{3}}=0.....\left( 1 \right)$
Now we use general algebraic identity:
${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$
By substituting, $a=x,b=i$, we get:
${{x}^{3}}-{{i}^{3}}=\left( x-i \right)\left( {{x}^{2}}-1+ix \right)$
Substituting this back into equation (1), we get:
$\left( x-i \right)\left( {{x}^{2}}-1+ix \right)=0$
One of the root is, $x=i$.
Other roots of the equation are , ${{x}^{2}}-1+ix=0$.
Roots of a quadratic equation, $a{{x}^{2}}+bx+c=0$ re given by
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
By substituting values of a, b, c as 1, i ,-1 respectively, we get:
$x=\dfrac{-i\pm \sqrt{-1+4}}{2}=\dfrac{-i\pm \sqrt{3}}{2}$
So, the values of ${{\left( -i \right)}^{\dfrac{1}{3}}}$ expression are, $i,\dfrac{-i\pm \sqrt{3}}{2}$.
Options (a), (c) are true.

Note: Be careful while substituting a, b don’t forget to take the ${{3}^{rd}}$ root $\left( x-i \right)$. You get the result even by neglecting that root but if we change options you may miss one root. So always remember to take all possible roots of the equation.