Question

How do you find the power ${{\left[ 3\left( \cos \left( \dfrac{\pi }{6} \right)+i\sin \left( \dfrac{\pi }{6} \right) \right) \right]}^{3}}$ and express the result in rectangular form?

Answer 1

Hint: In this problem, we are to find the power of a complex number and express the result in the rectangular form. To start with, we are going to use De Moivre’s theorem to find the solution. De Moivre’s theorem states $r{{\left( \cos \theta +i\sin \theta \right)}^{n}}={{r}^{n}}\left( \cos \left( n.\theta \right)+i\sin \left( n.\theta \right) \right)$for any complex number. Next, we will try to compare the equations and then will put the values to find the simplified value.

Complete step by step solution:
 According to the problem, we are given, ${{\left[ 3\left( \cos \left( \dfrac{\pi }{6} \right)+i\sin \left( \dfrac{\pi }{6} \right) \right) \right]}^{3}}$, and we are to find the power of this.
To solve this we are going to use, De’ Moivre’s theorem.
The theorem says, $r{{\left( \cos \theta +i\sin \theta \right)}^{n}}={{r}^{n}}\left( \cos \left( n.\theta \right)+i\sin \left( n.\theta \right) \right)$
Now, if we are trying to compare the equations we get, r = 3, n =3 and the value of $\theta $ is said to be $\dfrac{\pi }{6}$ .
And, now, while trying to use the theorem we are having,
${{\left[ 3\left( \cos \left( \dfrac{\pi }{6} \right)+i\sin \left( \dfrac{\pi }{6} \right) \right) \right]}^{3}}={{3}^{3}}\left[ \cos \left( 3.\dfrac{\pi }{6} \right)+i\sin \left( 3.\dfrac{\pi }{6} \right) \right]$
Again, we will now try to simplify the equation.
So, we are getting, ${{3}^{3}}\left[ \cos \left( 3.\dfrac{\pi }{6} \right)+i\sin \left( 3.\dfrac{\pi }{6} \right) \right]=27\left[ \cos \left( \dfrac{\pi }{2} \right)+i\sin \left( \dfrac{\pi }{2} \right) \right]$
From the table of trigonometric identities, we also know, $\cos \left( \dfrac{\pi }{2} \right)$ has a value of 0 and $\sin \left( \dfrac{\pi }{2} \right)$has a value of 1.
Hence, putting the values, $27\left[ \cos \left( \dfrac{\pi }{2} \right)+i\sin \left( \dfrac{\pi }{2} \right) \right]=27\left[ 0+i\left( 1 \right) \right]=27i$

Note: In this problem we have used the De Moivre’s theorem to get our solution. Talking about the theorem we can state that by squaring a complex number, the absolute value is squared and the argument is multiplied by 2. For $n\ge 3$ , de Moivre's theorem generalizes this to show that to raise a complex number to the nth power, the absolute value is raised to the nth power and the argument is multiplied by n.