Question

How do you find the cube root of $81\left( \cos \left( \dfrac{\pi }{12} \right)+i\sin \left( \dfrac{\pi }{12} \right) \right)$

Answer 1

In this problem we need to calculate the cube root of the given value. We can observe that the given value is an imaginary number which is in the form of $r\left( \cos \theta +i\sin \theta \right)$. We know that De moivre’s theorem states that ${{\left[ r\left( \cos \theta +i\sin \theta \right) \right]}^{n}}={{r}^{n}}\left( \cos n\theta +i\sin n\theta \right)$. So first we will take the cube root of the given value as the ${{\left( \dfrac{1}{3} \right)}^{rd}}$ power of the given value and apply the De moivre’s theorem. Now we will simplify the equation to get the required result.

Complete step by step solution:
Given that, $81\left( \cos \left( \dfrac{\pi }{12} \right)+i\sin \left( \dfrac{\pi }{12} \right) \right)$.
Let $z=81\left( \cos \left( \dfrac{\pi }{12} \right)+i\sin \left( \dfrac{\pi }{12} \right) \right)$.
We can write the cube root of the above value as
$\Rightarrow \sqrt[3]{z}=\sqrt[3]{81\left( \cos \left( \dfrac{\pi }{12} \right)+i\sin \left( \dfrac{\pi }{12} \right) \right)}$
We are going to write the cube root as the ${{\left( \dfrac{1}{3} \right)}^{rd}}$ power of the above value in the above equation, then we will get
$\Rightarrow \sqrt[3]{z}={{\left[ 81\left( \cos \left( \dfrac{\pi }{12} \right)+i\sin \left( \dfrac{\pi }{12} \right) \right) \right]}^{\dfrac{1}{3}}}$
From the De moivre’s theorem have the formula ${{\left[ r\left( \cos \theta +i\sin \theta \right) \right]}^{n}}={{r}^{n}}\left( \cos n\theta +i\sin n\theta \right)$ applying this formula in the above equation, then we will get the cube root value as
$\Rightarrow \sqrt[3]{z}={{81}^{\dfrac{1}{3}}}\left( \cos \left( \dfrac{1}{3}\times \dfrac{\pi }{12} \right)+i\sin \left( \dfrac{1}{3}\times \dfrac{\pi }{12} \right) \right)$
Simplifying the above equation by multiplying the denominators in the above equation, then we will get
$\Rightarrow \sqrt[3]{z}={{81}^{\dfrac{1}{3}}}\left( \cos \dfrac{\pi }{36}+i\sin \dfrac{\pi }{36} \right)$
Hence the cube root of the given value $81\left( \cos \left( \dfrac{\pi }{12} \right)+i\sin \left( \dfrac{\pi }{12} \right) \right)$ is ${{81}^{\dfrac{1}{3}}}\left( \cos \dfrac{\pi }{36}+i\sin \dfrac{\pi }{36} \right)$.

Note: We can also use the value of $\sin \dfrac{\pi }{36}=0.173648$, $\cos \dfrac{\pi }{36}=0.984808$ in the above equation and simplify the value. We can use this procedure for any power of the imaginary number to simplify the value.