Solveeit Logo
Login

Question

Question: Find all the values of the given root : \({\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}}\) where \(\le...

Find all the values of the given root : (−64a4)14{\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}} where (a∈R)\left( {a \in R} \right)

Explanation

Solution

Here, we will solve for the roots by using the polar form of a complex number.

Since, (−64a4)14=(−1)14(64)14(a4)14=(−1)14(26)14(a)=(−1)14(2)64(a) ⇒(−64a4)14=(−1)14(2)32(a) →(1)  {\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 (−1)\left( { - 1} \right) can be represented in polar form as −1=cos⁡π+i(sin⁡π) - 1 = \cos \pi + i\left( {\sin \pi } \right)
Substituting the above value of (−1)\left( { - 1} \right) in equation (1), we get
⇒(−64a4)14=[cos⁡π+i(sin⁡π)]14(2)32(a)\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⁡θ=cos(2nπ+θ)\cos \theta = cos\left( {2n\pi + \theta } \right) and sin⁡θ=sin⁡(2nπ+θ)\sin \theta = \sin \left( {2n\pi + \theta } \right)
⇒(−64a4)14=(2)32(a)[cos⁡(2nπ+π)+i(sin⁡(2nπ+π))]14\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 (cos⁡θ+isin⁡θ)n=(eiθ)n=ei(nθ)=cos⁡(nθ)+isin⁡(nθ){\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
⇒(−64a4)14=(2)32(a)[cos⁡[(2nπ+π)4]+i(sin⁡[(2nπ+π)4])]\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,..n = 0,1,2,..
The required roots can be obtained by putting the different values of nn
For n=0n = 0, (−64a4)14=(2)32(a)[cos⁡[(2×0×π+π)4]+i(sin⁡[(2×0×π+π)4])]=(2)32(a)[cos⁡(π4)+i(sin⁡(π4))]{\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⁡(π4)=sin⁡(π4)=12\cos \left( {\dfrac{\pi }{4}} \right) = \sin \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}
⇒(−64a4)14=(2)32(a)[(12)+i(12)]=(22)(a)[1+i2]=2a(1+i)\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=1n = 1, (−64a4)14=(2)32(a)[cos⁡[(2×1×π+π)4]+i(sin⁡[(2×1×π+π)4])]=(2)32(a)[cos⁡(3π4)+i(sin⁡(3π4))]{\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⁡(3π4)=−12\cos \left( {\dfrac{{3\pi }}{4}} \right) = - \dfrac{1}{{\sqrt 2 }} and sin⁡(3π4)=12\sin \left( {\dfrac{{3\pi }}{4}} \right) = \dfrac{1}{{\sqrt 2 }}
⇒(−64a4)14=(2)32(a)[(−12)+i(12)]=(22)(a)[−1+i2]=−2a(1−i)\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=2n = 2, (−64a4)14=(2)32(a)[cos⁡[(2×2×π+π)4]+i(sin⁡[(2×2×π+π)4])]=(2)32(a)[cos⁡(5π4)+i(sin⁡(5π4))]{\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⁡(5π4)=−12\cos \left( {\dfrac{{5\pi }}{4}} \right) = - \dfrac{1}{{\sqrt 2 }} and sin⁡(5π4)=−12\sin \left( {\dfrac{{5\pi }}{4}} \right) = - \dfrac{1}{{\sqrt 2 }}
⇒(−64a4)14=(2)32(a)[(−12)+i(−12)]=−(22)(a)[1+i2]=−2a(1+i)\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=3n = 3, (−64a4)14=(2)32(a)[cos⁡[(2×3×π+π)4]+i(sin⁡[(2×3×π+π)4])]=(2)32(a)[cos⁡(7π4)+i(sin⁡(7π4))]{\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⁡(5π4)=12\cos \left( {\dfrac{{5\pi }}{4}} \right) = \dfrac{1}{{\sqrt 2 }} and sin⁡(5π4)=−12\sin \left( {\dfrac{{5\pi }}{4}} \right) = - \dfrac{1}{{\sqrt 2 }}
⇒(−64a4)14=(2)32(a)[(12)+i(−12)]=(22)(a)[1−i2]=2a(1−i)\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 nn, the roots will repeat so the final four required roots of (−64a4)14{\left( { - 64{a^4}} \right)^{\dfrac{1}{4}}} where (a∈R)\left( {a \in R} \right) can be collectively represented as ±2a(1±i) \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 ei(nθ){e^{i\left( {n\theta } \right)}} and further various roots can be obtained by putting different values of nn.