Find all the values of the given root : ${\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}}$ where $\left( {a \in R} \right)$
Answer
Verified363k+ views
Here, we will solve for the roots by using the polar form of a complex number.
Since, $
{\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}} = {\left( { - 1} \right)^{\dfrac{1}{4}}}{\left( {64} \right)^{\dfrac{1}{4}}}{\left( {{a^4}} \right)^{\dfrac{1}{4}}} = {\left( { - 1} \right)^{\dfrac{1}{4}}}{\left( {{2^6}} \right)^{\dfrac{1}{4}}}\left( a \right) = {\left( { - 1} \right)^{\dfrac{1}{4}}}{\left( 2 \right)^{\dfrac{6}{4}}}\left( a \right) \\
\Rightarrow {\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}} = {\left( { - 1} \right)^{\dfrac{1}{4}}}{\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right){\text{ }} \to {\text{(1)}} \\
$
As we know that $\left( { - 1} \right)$ can be represented in polar form as $ - 1 = \cos \pi + i\left( {\sin \pi } \right)$
Substituting the above value of $\left( { - 1} \right)$ in equation (1), we get
$ \Rightarrow {\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}} = {\left[ {\cos \pi + i\left( {\sin \pi } \right)} \right]^{\dfrac{1}{4}}}{\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right)$
Also, we know that $\cos \theta = cos\left( {2n\pi + \theta } \right)$ and $\sin \theta = \sin \left( {2n\pi + \theta } \right)$
$ \Rightarrow {\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}} = {\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right){\left[ {\cos \left( {2n\pi + \pi } \right) + i\left( {\sin \left( {2n\pi + \pi } \right)} \right)} \right]^{\dfrac{1}{4}}}$
Using identity ${\left( {\cos \theta + i\sin \theta } \right)^n} = {\left( {{e^{i\theta }}} \right)^n} = {e^{i\left( {n\theta } \right)}} = \cos \left( {n\theta } \right) + i\sin \left( {n\theta } \right)$, we can write
$ \Rightarrow {\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}} = {\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right)\left[ {\cos \left[ {\dfrac{{\left( {2n\pi + \pi } \right)}}{4}} \right] + i\left( {\sin \left[ {\dfrac{{\left( {2n\pi + \pi } \right)}}{4}} \right]} \right)} \right]$ where $n = 0,1,2,..$
The required roots can be obtained by putting the different values of $n$
For $n = 0$, ${\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}} = {\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right)\left[ {\cos \left[ {\dfrac{{\left( {2 \times 0 \times \pi + \pi } \right)}}{4}} \right] + i\left( {\sin \left[ {\dfrac{{\left( {2 \times 0 \times \pi + \pi } \right)}}{4}} \right]} \right)} \right] = {\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right)\left[ {\cos \left( {\dfrac{\pi }{4}} \right) + i\left( {\sin \left( {\dfrac{\pi }{4}} \right)} \right)} \right]$
As, $\cos \left( {\dfrac{\pi }{4}} \right) = \sin \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}$
$ \Rightarrow {\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}} = {\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right)\left[ {\left( {\dfrac{1}{{\sqrt 2 }}} \right) + i\left( {\dfrac{1}{{\sqrt 2 }}} \right)} \right] = \left( {2\sqrt 2 } \right)\left( a \right)\left[ {\dfrac{{1 + i}}{{\sqrt 2 }}} \right] = 2a\left( {1 + i} \right)$
For $n = 1$, ${\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}} = {\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right)\left[ {\cos \left[ {\dfrac{{\left( {2 \times 1 \times \pi + \pi } \right)}}{4}} \right] + i\left( {\sin \left[ {\dfrac{{\left( {2 \times 1 \times \pi + \pi } \right)}}{4}} \right]} \right)} \right] = {\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right)\left[ {\cos \left( {\dfrac{{3\pi }}{4}} \right) + i\left( {\sin \left( {\dfrac{{3\pi }}{4}} \right)} \right)} \right]$
As, $\cos \left( {\dfrac{{3\pi }}{4}} \right) = - \dfrac{1}{{\sqrt 2 }}$ and $\sin \left( {\dfrac{{3\pi }}{4}} \right) = \dfrac{1}{{\sqrt 2 }}$
$ \Rightarrow {\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}} = {\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right)\left[ {\left( { - \dfrac{1}{{\sqrt 2 }}} \right) + i\left( {\dfrac{1}{{\sqrt 2 }}} \right)} \right] = \left( {2\sqrt 2 } \right)\left( a \right)\left[ {\dfrac{{ - 1 + i}}{{\sqrt 2 }}} \right] = - 2a\left( {1 - i} \right)$
For $n = 2$, ${\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}} = {\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right)\left[ {\cos \left[ {\dfrac{{\left( {2 \times 2 \times \pi + \pi } \right)}}{4}} \right] + i\left( {\sin \left[ {\dfrac{{\left( {2 \times 2 \times \pi + \pi } \right)}}{4}} \right]} \right)} \right] = {\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right)\left[ {\cos \left( {\dfrac{{5\pi }}{4}} \right) + i\left( {\sin \left( {\dfrac{{5\pi }}{4}} \right)} \right)} \right]$
As, $\cos \left( {\dfrac{{5\pi }}{4}} \right) = - \dfrac{1}{{\sqrt 2 }}$ and $\sin \left( {\dfrac{{5\pi }}{4}} \right) = - \dfrac{1}{{\sqrt 2 }}$
$ \Rightarrow {\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}} = {\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right)\left[ {\left( { - \dfrac{1}{{\sqrt 2 }}} \right) + i\left( { - \dfrac{1}{{\sqrt 2 }}} \right)} \right] = - \left( {2\sqrt 2 } \right)\left( a \right)\left[ {\dfrac{{1 + i}}{{\sqrt 2 }}} \right] = - 2a\left( {1 + i} \right)$
For $n = 3$, ${\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}} = {\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right)\left[ {\cos \left[ {\dfrac{{\left( {2 \times 3 \times \pi + \pi } \right)}}{4}} \right] + i\left( {\sin \left[ {\dfrac{{\left( {2 \times 3 \times \pi + \pi } \right)}}{4}} \right]} \right)} \right] = {\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right)\left[ {\cos \left( {\dfrac{{7\pi }}{4}} \right) + i\left( {\sin \left( {\dfrac{{7\pi }}{4}} \right)} \right)} \right]$
As, $\cos \left( {\dfrac{{5\pi }}{4}} \right) = \dfrac{1}{{\sqrt 2 }}$ and $\sin \left( {\dfrac{{5\pi }}{4}} \right) = - \dfrac{1}{{\sqrt 2 }}$
$ \Rightarrow {\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}} = {\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right)\left[ {\left( {\dfrac{1}{{\sqrt 2 }}} \right) + i\left( { - \dfrac{1}{{\sqrt 2 }}} \right)} \right] = \left( {2\sqrt 2 } \right)\left( a \right)\left[ {\dfrac{{1 - i}}{{\sqrt 2 }}} \right] = 2a\left( {1 - i} \right)$
For rest of the values of $n$, the roots will repeat so the final four required roots of ${\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}}$ where $\left( {a \in R} \right)$ can be collectively represented as $ \pm 2a\left( {1 \pm i} \right)$.
Note- In these type of problems, the given expression is represented in polar form of a complex number so that the power of that expression can be easily solved by using the formula for ${e^{i\left( {n\theta } \right)}}$ and further various roots can be obtained by putting different values of $n$.
Since, $
{\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}} = {\left( { - 1} \right)^{\dfrac{1}{4}}}{\left( {64} \right)^{\dfrac{1}{4}}}{\left( {{a^4}} \right)^{\dfrac{1}{4}}} = {\left( { - 1} \right)^{\dfrac{1}{4}}}{\left( {{2^6}} \right)^{\dfrac{1}{4}}}\left( a \right) = {\left( { - 1} \right)^{\dfrac{1}{4}}}{\left( 2 \right)^{\dfrac{6}{4}}}\left( a \right) \\
\Rightarrow {\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}} = {\left( { - 1} \right)^{\dfrac{1}{4}}}{\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right){\text{ }} \to {\text{(1)}} \\
$
As we know that $\left( { - 1} \right)$ can be represented in polar form as $ - 1 = \cos \pi + i\left( {\sin \pi } \right)$
Substituting the above value of $\left( { - 1} \right)$ in equation (1), we get
$ \Rightarrow {\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}} = {\left[ {\cos \pi + i\left( {\sin \pi } \right)} \right]^{\dfrac{1}{4}}}{\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right)$
Also, we know that $\cos \theta = cos\left( {2n\pi + \theta } \right)$ and $\sin \theta = \sin \left( {2n\pi + \theta } \right)$
$ \Rightarrow {\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}} = {\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right){\left[ {\cos \left( {2n\pi + \pi } \right) + i\left( {\sin \left( {2n\pi + \pi } \right)} \right)} \right]^{\dfrac{1}{4}}}$
Using identity ${\left( {\cos \theta + i\sin \theta } \right)^n} = {\left( {{e^{i\theta }}} \right)^n} = {e^{i\left( {n\theta } \right)}} = \cos \left( {n\theta } \right) + i\sin \left( {n\theta } \right)$, we can write
$ \Rightarrow {\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}} = {\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right)\left[ {\cos \left[ {\dfrac{{\left( {2n\pi + \pi } \right)}}{4}} \right] + i\left( {\sin \left[ {\dfrac{{\left( {2n\pi + \pi } \right)}}{4}} \right]} \right)} \right]$ where $n = 0,1,2,..$
The required roots can be obtained by putting the different values of $n$
For $n = 0$, ${\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}} = {\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right)\left[ {\cos \left[ {\dfrac{{\left( {2 \times 0 \times \pi + \pi } \right)}}{4}} \right] + i\left( {\sin \left[ {\dfrac{{\left( {2 \times 0 \times \pi + \pi } \right)}}{4}} \right]} \right)} \right] = {\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right)\left[ {\cos \left( {\dfrac{\pi }{4}} \right) + i\left( {\sin \left( {\dfrac{\pi }{4}} \right)} \right)} \right]$
As, $\cos \left( {\dfrac{\pi }{4}} \right) = \sin \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}$
$ \Rightarrow {\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}} = {\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right)\left[ {\left( {\dfrac{1}{{\sqrt 2 }}} \right) + i\left( {\dfrac{1}{{\sqrt 2 }}} \right)} \right] = \left( {2\sqrt 2 } \right)\left( a \right)\left[ {\dfrac{{1 + i}}{{\sqrt 2 }}} \right] = 2a\left( {1 + i} \right)$
For $n = 1$, ${\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}} = {\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right)\left[ {\cos \left[ {\dfrac{{\left( {2 \times 1 \times \pi + \pi } \right)}}{4}} \right] + i\left( {\sin \left[ {\dfrac{{\left( {2 \times 1 \times \pi + \pi } \right)}}{4}} \right]} \right)} \right] = {\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right)\left[ {\cos \left( {\dfrac{{3\pi }}{4}} \right) + i\left( {\sin \left( {\dfrac{{3\pi }}{4}} \right)} \right)} \right]$
As, $\cos \left( {\dfrac{{3\pi }}{4}} \right) = - \dfrac{1}{{\sqrt 2 }}$ and $\sin \left( {\dfrac{{3\pi }}{4}} \right) = \dfrac{1}{{\sqrt 2 }}$
$ \Rightarrow {\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}} = {\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right)\left[ {\left( { - \dfrac{1}{{\sqrt 2 }}} \right) + i\left( {\dfrac{1}{{\sqrt 2 }}} \right)} \right] = \left( {2\sqrt 2 } \right)\left( a \right)\left[ {\dfrac{{ - 1 + i}}{{\sqrt 2 }}} \right] = - 2a\left( {1 - i} \right)$
For $n = 2$, ${\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}} = {\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right)\left[ {\cos \left[ {\dfrac{{\left( {2 \times 2 \times \pi + \pi } \right)}}{4}} \right] + i\left( {\sin \left[ {\dfrac{{\left( {2 \times 2 \times \pi + \pi } \right)}}{4}} \right]} \right)} \right] = {\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right)\left[ {\cos \left( {\dfrac{{5\pi }}{4}} \right) + i\left( {\sin \left( {\dfrac{{5\pi }}{4}} \right)} \right)} \right]$
As, $\cos \left( {\dfrac{{5\pi }}{4}} \right) = - \dfrac{1}{{\sqrt 2 }}$ and $\sin \left( {\dfrac{{5\pi }}{4}} \right) = - \dfrac{1}{{\sqrt 2 }}$
$ \Rightarrow {\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}} = {\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right)\left[ {\left( { - \dfrac{1}{{\sqrt 2 }}} \right) + i\left( { - \dfrac{1}{{\sqrt 2 }}} \right)} \right] = - \left( {2\sqrt 2 } \right)\left( a \right)\left[ {\dfrac{{1 + i}}{{\sqrt 2 }}} \right] = - 2a\left( {1 + i} \right)$
For $n = 3$, ${\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}} = {\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right)\left[ {\cos \left[ {\dfrac{{\left( {2 \times 3 \times \pi + \pi } \right)}}{4}} \right] + i\left( {\sin \left[ {\dfrac{{\left( {2 \times 3 \times \pi + \pi } \right)}}{4}} \right]} \right)} \right] = {\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right)\left[ {\cos \left( {\dfrac{{7\pi }}{4}} \right) + i\left( {\sin \left( {\dfrac{{7\pi }}{4}} \right)} \right)} \right]$
As, $\cos \left( {\dfrac{{5\pi }}{4}} \right) = \dfrac{1}{{\sqrt 2 }}$ and $\sin \left( {\dfrac{{5\pi }}{4}} \right) = - \dfrac{1}{{\sqrt 2 }}$
$ \Rightarrow {\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}} = {\left( 2 \right)^{\dfrac{3}{2}}}\left( a \right)\left[ {\left( {\dfrac{1}{{\sqrt 2 }}} \right) + i\left( { - \dfrac{1}{{\sqrt 2 }}} \right)} \right] = \left( {2\sqrt 2 } \right)\left( a \right)\left[ {\dfrac{{1 - i}}{{\sqrt 2 }}} \right] = 2a\left( {1 - i} \right)$
For rest of the values of $n$, the roots will repeat so the final four required roots of ${\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}}$ where $\left( {a \in R} \right)$ can be collectively represented as $ \pm 2a\left( {1 \pm i} \right)$.
Note- In these type of problems, the given expression is represented in polar form of a complex number so that the power of that expression can be easily solved by using the formula for ${e^{i\left( {n\theta } \right)}}$ and further various roots can be obtained by putting different values of $n$.
Last updated date: 16th Sep 2023
•
Total views: 363k
•
Views today: 4.63k
