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.