Question

Find all the values of the following expression $\sqrt[6]{-64}$?

Answer 1

Hint: First of all assume $\sqrt[6]{-64}$ as z. Then take to the power of 6 on both the sides we get ${{z}^{6}}$ = -1(64). Then write 64 as some number to the power of 6 which is 26. Now, we can write ${{z}^{6}}$ = -1(64) as ${{z}^{6}}={{e}^{i\pi +i2k\pi }}{{(2)}^{6}}$ then taking to the power of $\dfrac{1}{6}$ on both the sides we get $z={{e}^{\dfrac{i\pi \left( 2k+1 \right)}{6}}}(2)$.

Complete step-by-step answer:

Let us assume $\sqrt[6]{-64}$= z.
Taking to the power of 6 on both the sides in the above equation we get,
${{z}^{6}}=-64$……. Eq. (1)
Now, if we can write 64 as some number to the power of 6 then when we take 6th root on both the sides then 6 in numerator and denominator will be cancelled out. So, we can write 64 as 2×2×2×2×2×2 or we can also write 64 = ${{(2)}^{6}}$. Plugging 64 as ${{(2)}^{6}}$ in eq. (1) we get as shown below:
${{z}^{6}}=-{{(2)}^{6}}$
Taking to the power of $\dfrac{1}{6}$on both the sides we get,
$z={{\left( -1 \right)}^{\dfrac{1}{6}}}(2)$……… Eq. (2)
In the above expression, how can we write ${{\left( -1 \right)}^{\dfrac{1}{6}}}$? So, let us assume ${{\left( -1 \right)}^{\dfrac{1}{6}}}$as ${{z}_{1}}$.
${{z}_{1}}={{\left( -1 \right)}^{\dfrac{1}{6}}}$
Taking to the power of 6 on both the sides we get,
${{z}_{1}}^{6}=\left( -1 \right)$
From complex numbers, we can write -1 as ${{e}^{i\pi }}$. And 1 is also multiplied with -1 so we can write 1 in the form of a complex number as ${{e}^{i2k\pi }}$. Now, plugging these values in the above equation we get,
$z_{1}^{6}={{e}^{i\pi }}({{e}^{i2k\pi }})$
When the base is the same then power of the bases are added.
$\begin{align}
  & z_{1}^{6}={{e}^{i\pi +i2k\pi }} \\
 & z_{1}^{6}={{e}^{i\pi \left( 2k+1 \right)}} \\
\end{align}$
Taking to the power of $\dfrac{1}{6}$on both the sides we get,
${{z}_{1}}={{e}^{\dfrac{i\pi }{6}\left( 2k+1 \right)}}$
In the above equation, k = 0, 1, 2, 3, 4, 5 after k = 5 the value of ${{z}_{1}}$ keeps on repeating the same value from one of k = 0, 1, 2, 3, 4, 5.
Substituting the value of ${{z}_{1}}$ in eq. (2) because as we have assumed above that${{z}_{1}}={{\left( -1 \right)}^{\dfrac{1}{6}}}$, we get:
$z={{e}^{\dfrac{i\pi \left( 2k+1 \right)}{6}}}\left( 2 \right)$
In the above equation, k = 0, 1, 2, 3, 4, 5.
Hence, the value of $\sqrt[6]{-64}$ is $z=(2){{e}^{\dfrac{i\pi \left( 2k+1 \right)}{6}}}$where k = 0, 1, 2, 3, 4, 5.

Note: In this expression,${{z}_{1}}={{e}^{\dfrac{i\pi }{6}\left( 2k+1 \right)}}$we have taken value as k = 0, 1, 2, 3, 4, 5 and said that after k = 5, the value of $z_1$ is keep on repeating and yields the same result as one of k = 0, 1, 2, 3, 4, 5.
Plugging different values of k in the $z_1$ expression we get,
For k = 0,
$\begin{align}
  & {{z}_{1}}={{e}^{\dfrac{i\pi }{6}\left( 1 \right)}} \\
 & {{z}_{1}}={{e}^{\dfrac{i\pi }{6}}} \\
\end{align}$
For k = 1,
$\begin{align}
  & {{z}_{1}}={{e}^{\dfrac{i\pi }{6}\left( 2+1 \right)}} \\
 & \Rightarrow {{z}_{1}}={{e}^{\dfrac{i\pi }{6}\left( 3 \right)}} \\
 & \Rightarrow {{z}_{1}}={{e}^{\dfrac{i\pi }{2}}} \\
\end{align}$
For k = 2,
${{z}_{1}}={{e}^{\dfrac{i\pi }{6}(5)}}$
For k = 3,
${{z}_{1}}={{e}^{\dfrac{i\pi }{6}\left( 7 \right)}}$
For k = 4,
${{z}_{1}}={{e}^{\dfrac{i\pi }{6}\left( 9 \right)}}$
For k = 5,
${{z}_{1}}={{e}^{\dfrac{i\pi }{6}\left( 11 \right)}}$
For k =6,
${{z}_{1}}={{e}^{\dfrac{i\pi }{6}\left( 13 \right)}}$
Now, when k =6, we get the angle as $\dfrac{13\pi }{6}$in which when we subtract 2π from this angle we get $\dfrac{\pi }{6}$because after 2π values of angles of cosine and sine, tan are repeating. As we have seen that in the k = 6 case, after subtracting from 2π we get the angle as$\dfrac{\pi }{6}$which is the same as that we were getting when k = 0.
From the discussion that we just had, we have established that after k = 5, the values of $z_1$ keep on repeating.