Question

Please solve the following problem.

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

Answer 1

- 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

Answer 2

`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`