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Please solve the following problem.

Express the complex number
z = 8 \text{cis} (2\pi)/(3) in rectangular form
a+bi.

User Windos
by
8.0k points

2 Answers

2 votes

Answer:

- 4 + 4√3 i, where a = -4 & b = 4√3

Explanation:

z = 8 cis(2π/3)

z = 8 [cos(2π/3) + i sin(2π/3) ]

z = 8 [cos(π - π/3) + i sin(π - π/3)]

z = 8 [ - cosπ/3 + i sin(π/3)]

z = 8[ - 1/2 + i √3/2 ]

z = - 4 + 4√3 i [in a + bi form]

Note that: cis x = cosx + i sinx

User Danny Hoeve
by
8.3k points
2 votes

Answer:


a + bi = - 4 + 4 √(3) i

Explanation:

we are given
\displaystyle cis((2\pi)/(3))

recall complex number trigonometric form:


\displaystyle \: r( \cos( \theta) + i \sin( \theta) )

we are already given that
\theta=(2\pi)/(3)

recall complex number rectangular form


\displaystyle \: a + bi

where
\displaystyle a=r\cos(\theta)\: and\: b=r\sin(\theta)

let's work with a:

substitute the value of
\theta and r


\displaystyle \: a = 8 * \cos( (2\pi)/( 3) )

recall unit circle

so
\cos((2\pi)/(3)) should be in Q:II


\displaystyle \: a = 8 * - (1)/(2)

simplify multiplication:


\displaystyle \: a = - 4

let's work with b:

substitute the value of
\theta and r:


\displaystyle \: b = 8\sin( (3\pi)/(2) )

recall unit circle so
\sin((3\pi)/(2)) should be in Q:II


\displaystyle \: b = 8 * ( √(3) )/(2)

simplify:


\displaystyle \: b = 4 √(3)

so


\displaystyle \: b i= 4 √(3) i

hence,


a + bi = - 4 + 4 √(3) i

User RexE
by
8.4k points

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