Question

How do you find the cube root of $64\left( \cos \left( \dfrac{\pi }{5} \right)+i\sin \left( \dfrac{\pi }{5} \right) \right)$?

Answer 1

In this problem we need to calculate the cube root of the given value which is an imaginary number. For this we are going to use the De Moivre’s theorem to solve the problem. Now we will write the cube root as the ${{\left( \dfrac{1}{3} \right)}^{rd}}$ power of the given value. After that we will apply the De Moivre’s theorem and simplify the obtained equation to get the required result.

Complete step by step solution: Given that, $64\left( \cos \left( \dfrac{\pi }{5} \right)+i\sin \left( \dfrac{\pi }{5} \right) \right)$.
Let $z=64\left( \cos \left( \dfrac{\pi }{5} \right)+i\sin \left( \dfrac{\pi }{5} \right) \right)$.
To calculate the cube root of the above value, applying cube root on both sides of the above equation, then we will get
$\Rightarrow \sqrt[3]{z}=\sqrt[3]{64\left( \cos \left( \dfrac{\pi }{5} \right)+i\sin \left( \dfrac{\pi }{5} \right) \right)}$
Whenever dealing with complex variable equation such as this it is essential to remember that the complex exponential has a period of $2\pi$, so we can equivalently write the above equation as
$\Rightarrow \sqrt[3]{z}=\sqrt[3]{64\left( \cos \left( 2n\pi +\dfrac{\pi }{5} \right)+i\sin \left( 2n\pi +\dfrac{\pi }{5} \right) \right)}$
Now we are writing the cube root as the ${{\left( \dfrac{1}{3} \right)}^{rd}}$ power of the above value, then we will get
$\Rightarrow \sqrt[3]{z}={{\left[ 64\left( \cos \left( 2n\pi +\dfrac{\pi }{5} \right)+i\sin \left( 2n\pi +\dfrac{\pi }{5} \right) \right) \right]}^{\dfrac{1}{3}}}$
From the De Moivre’s theorem we can write the above value as
$\Rightarrow \sqrt[3]{z}={{64}^{\dfrac{1}{3}}}\left( \cos \left( \dfrac{2n\pi +\dfrac{\pi }{5}}{3} \right)+i\sin \left( \dfrac{2n\pi +\dfrac{\pi }{5}}{3} \right) \right)$
We know that the value of $\sqrt[3]{64}=4$. Substituting this value in the above equation, then we will get
$\Rightarrow \sqrt[3]{z}=4\left( \cos \left( \dfrac{2n\pi +\dfrac{\pi }{5}}{3} \right)+i\sin \left( \dfrac{2n\pi +\dfrac{\pi }{5}}{3} \right) \right)$
Hence the value of cube root of the given value $64\left( \cos \left( \dfrac{\pi }{5} \right)+i\sin \left( \dfrac{\pi }{5} \right) \right)$ is $4\left( \cos \left( \dfrac{2n\pi +\dfrac{\pi }{5}}{3} \right)+i\sin \left( \dfrac{2n\pi +\dfrac{\pi }{5}}{3} \right) \right)$ where $n=1,2,3$.

Note: For this we can also calculate the exact value by substituting $n=1,2,3$ in the calculated result and simplify the equation. Then we will get three values which are $4\cos \left( \dfrac{\pi }{15} \right)+i\sin \left( \dfrac{\pi }{15} \right)$, $4\cos \left( \dfrac{7\pi }{15} \right)+i\sin \left( \dfrac{7\pi }{15} \right)$, $4\cos \left( \dfrac{11\pi }{15} \right)+i\sin \left( \dfrac{11\pi }{15} \right)$.