Question

Write the expression in the standard form $a+bi$.

${\left[\sqrt{3}\left(\cos \frac{5\mathrm{\pi }}{18}+i\sin \frac{5\mathrm{\pi }}{18}\right)\right]}^6$

Short answer

The expression in the standard form is $\frac{27}{2}-\frac{27\sqrt{3}}{2}i$.

Step-by-step solution

Step 1. Given information.

Consider the given question,

${\left[\sqrt{3}\left(\cos \frac{5\mathrm{\pi }}{18}+i\sin \frac{5\mathrm{\pi }}{18}\right)\right]}^6$

We know that for a complex number raise to n when it is in the polar form $z^n=r^n\left(\cos \theta +i\sin \left(n\theta \right)\right)......\left(i\right)$

Write the expression in the standard form $z=\left(a+ib\right)$,

$z^6={\left[\sqrt{3}\left(\cos \frac{5\mathrm{\pi }}{18}+i\sin \frac{5\mathrm{\pi }}{18}\right)\right]}^6$

Step 2. Use equation (i) and solve the equation.

Using the equation (i),

$$\begin{gathered} z^6={\left(\sqrt{3}\right)}^6\left[\cos \left(6·\frac{5\mathrm{\pi }}{18}\right)+i\sin \left(6·\frac{5\mathrm{\pi }}{18}\right)\right] \\ {z}^6=3^3\left(\cos \frac{5\mathrm{\pi }}{18}+i\sin \frac{5\mathrm{\pi }}{18}\right) \\ {z}^6=27\left(\frac{1}{2}-\frac{\sqrt{3}}{2}i\right) \\ {z}^6=\frac{27}{2}-\frac{27\sqrt{3}}{2}i \end{gathered}$$