Answer:
Option B is correct
\(-4+ 4\sqrt{3}i\)
Step-by-step explanation:
Given the complex number:
\((\sqrt{2}(\cos(20^{\circ})+i\sin(20^{\circ})))^6\)
Using D-Moivre's theorem:
if p is the rational; number:
\((\cos \theta +i\sin \theta)^p = \cos p\theta +i\sin\theta\)
then;
Using D-Moivre's theorem:
\((\sqrt{2})^6(\cos(6 \cdot 20^{\circ})+i\sin(6 \cdot 20^{\circ}))\)
\(8((\cos(120^{\circ})+i\sin(120^{\circ}))\)
\(8(-\frac{1}{2}+i \frac{\sqrt{3}}{2}) = -4+ 4\sqrt{3}i\)
Therefore, the given complex in the standard form is \(-4+ 4\sqrt{3}i\)
