Answer
$(\frac{-27}{2}+i.\frac{27\sqrt 3}{2})$
Work Step by Step
De Moivre’s Theorem states that when $ r (\cos\theta+i \sin\theta)$ is a complex number, and if $n$ is any real number, then the following relationship holds.
$[ r (\cos\theta+i \sin\theta)]^{n}=[ r^{n} (\cos n\theta+i \sin n\theta)]$
In compact form, this is written
$[ r cis\theta]^{n}=[ r^{n} (cis \theta)]$
$[ 3 cis40^{\circ}]^{3}=[ 3^{3} (cis 40^{\circ})]$
$[27 (\cos 40^{\circ}+i \sin 40^{\circ})]$
$=[ 27(\frac{-1}{2}+i.\frac{\sqrt 3}{2})]$
$=(\frac{-27}{2}+i.\frac{27\sqrt 3}{2})$
