Notice that
[tex]\begin{gathered} z=-4\sqrt[]{3}+4i=8(-\frac{\sqrt[]{3}}{2}+\frac{1}{2}i) \\ \Rightarrow z=8(-\frac{\sqrt[]{3}}{2}+\frac{1}{2}i) \end{gathered}[/tex]
Set,
[tex]\begin{gathered} z=8(\cos x+i\sin x) \\ \Rightarrow8(\cos x+i\sin x)=8(-\frac{\sqrt[]{3}}{2}+\frac{1}{2}i) \\ \Rightarrow\cos x=-\frac{\sqrt[]{3}}{2},\sin x=\frac{1}{2} \end{gathered}[/tex]
Then, angle x is in the second quadrant since cosx is negative and sinx is positive-
On the other hand,
Therefore, using the unitary circle,
Therefore, the angle that produces such values for the cosine and sine function is x=150°; thus,
[tex]\begin{gathered} x=150\degree \\ \Rightarrow z=8(-\frac{\sqrt[]{3}}{2}+\frac{1}{2}i)=8(\cos (150\degree+i\sin (150\degree))) \\ \Rightarrow z=8(\cos (150\degree+i\sin (150\degree))) \end{gathered}[/tex]
The answer is option 4 (top to bottom)
