Answer:
[tex]a + bi = - 4 + 4 \sqrt{3} i[/tex]
Step-by-step explanation:
we are given [tex]\displaystyle cis(\dfrac{2\pi}{3})[/tex]
recall complex number trigonometric form:
[tex] \displaystyle \: r( \cos( \theta) + i \sin( \theta) )[/tex]
we are already given that [tex]\theta=\dfrac{2\pi}{3}[/tex]
recall complex number rectangular form
[tex] \displaystyle \: a + bi[/tex]
where [tex]\displaystyle a=r\cos(\theta)\: and\: b=r\sin(\theta)[/tex]
let's work with a:
substitute the value of [tex]\theta[/tex] and r
[tex] \displaystyle \: a = 8 \times \cos( \frac{2\pi}{ 3} ) [/tex]
recall unit circle
so [tex]\cos(\dfrac{2\pi}{3})[/tex] should be in Q:II
[tex] \displaystyle \: a = 8 \times - \frac{1}{2} [/tex]
simplify multiplication:
[tex] \displaystyle \: a = - 4[/tex]
let's work with b:
substitute the value of [tex]\theta[/tex] and r:
[tex] \displaystyle \: b = 8\sin( \frac{3\pi}{2} ) [/tex]
recall unit circle so [tex]\sin(\dfrac{3\pi}{2})[/tex] should be in Q:II
[tex] \displaystyle \: b = 8 \times \frac{ \sqrt{3} }{2} [/tex]
simplify:
[tex] \displaystyle \: b = 4 \sqrt{3} [/tex]
so
[tex] \displaystyle \: b i= 4 \sqrt{3} i[/tex]
hence,
[tex]a + bi = - 4 + 4 \sqrt{3} i[/tex]
